答:
Carlos told you, how you can solve (c). Here is another method: by the rank-nullity- theorem, we have

$T$ is an isomorphism $ \iff ker(T)= \{0\}.$

If $p$ is a plynomial of degree $2$, hence $p(x)=ax^2+bx+c$, such that $p(0)=p'(0)=p''(0)=0$, what follows for $a,b$ and $c$ ?


问:
Calculate the Indefinite integral
I've been stuck with calculating the intgral of the following problem. Can you help me?

$$\int e^{x^2} \mathrm{d}x$$

I know that, by Using the definition of gamma and beta functions, we have $\int_{-\infty}^{+\infty} e^{-x^2} \mathrm{d}x = \sqrt{\pi}$.

Thanks in advance.

答:
Among the best things we can do about this one is to expand the function into the Taylor series and formally integrate term by term:
$$\int e^{x^2}dx= \sum_{k=0}^\infty \int\frac{x^{2k}}{k!}dx= \sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)k!}+C$$

ContentLicense: CC BY-SA 4.0

Title: Calculate the Indefinite integral


问:
What is the name of the quadrilateral shape described by two radii and two arcs?
What is the name of the four sided shape described by two radii and two concentric arcs?
Like each black and white section taken individually on this image

Or the double score / triple score areas on a dart board.

I'm attempting to create each section individually, programmatically and need help on the best thing to google to get help actually making this.

答:
Since I'm not a native English speaker, I was a bit reluctant to post something I made up myself as an answer. But Googling a little and the comment by amWhy encouraged me (showing me that I didn't really make it up, after all):

I think the word you're looking for is

Annular Sector

in analogy with a circular sector. I think that everyone knowing what a circular sector is, should be able to figure out what an annular sector is. An alternative word would be sector of an annulus.

You can specify an annular sector by giving its center $C$, the opening angle $\theta$ and the radii $0 \lt r \lt R$ of the two circular arcs bounding it:

Both these terms, annular sector and sector of an annulus give quite a few hits in Google. For example Googling for area annular sector brings up this link where you can find the formula

$$\textbf{Area of an angular sector:} \qquad \qquad A_{r,R,\theta} = \frac{(R^{2} - r^2)}{2}\cdot \theta$$

for the area $A_{r,R,\theta}$ of an annular sector with radii $0 \lt r \lt R$ and angle $\theta$ (measured in radians). This is because an annular sector is the circular sector of radius $R$ minus the circular sector of radius $r$ whose areas are $\dfrac{R^2}{2} \cdot \theta$ and $\dfrac{r^2}{2}\cdot\theta$, respectively. If the opening angle $\theta$ is given in degrees, you can convert it to radians using the formula $$\operatorname{radians} = \operatorname{degrees} \times \frac{\pi}{180}.$$

The length of the circumference $c_{r,R,\theta}$ of an annular sector with radii $0 \lt r \lt R$ and angle $\theta$ is

$$ \textbf{Circumference of an annular sector:} \qquad c_{r,R,\theta} = R \cdot (\theta + 2) + r \cdot (\theta-2)$$

where again $\theta$ is measured in radians. This is obtained by adding the lengths $R \theta$ and $r \theta$ of the two circular arcs and twice the length $(R-r)$ of the straight side.

ContentLicense: CC BY-SA 3.0


问:
For which values is $x^3$ less than or equal to $3x$?
The title says it all. The answers say:

$x\le -\sqrt{3}$ and $0\le x\le \sqrt{3}$

(can someone edit this so all the $<$ have an 'or equal to' sign. Edit the roots as well please.

I'm not sure how to attempt this question. When I simplify, I get $x^2\le 3$, so $x\le  \pm\sqrt{3}$.

Thanks!

答:
The equality can be solved very easily.

$$x^3< 3x\iff  x(x^2-3)< 0 $$

If $x<0, x^2-3>0$

As $x^2-3>0\implies$ either $x>\sqrt3$ or $x<-\sqrt3$

and as $x<0$ the required region will be $x<-\sqrt3$

Similarly, if $x>0, x^2-3<0\implies -\sqrt3<x<\sqrt3$

and as $x>0$ the required region will be $0<x<\sqrt3$

Alternatively,

HINT:

$$x^3< 3x\iff  x(x-\sqrt3)(x+\sqrt3)< 0 $$

This will hold true if odd number (one or three) of factors $< 0$

Now check for the ranges $(-\infty,-\sqrt3);[-\sqrt3,0);[0,\sqrt3);[\sqrt3, \infty)$


问:
Find the curvature if $x= \cos t$ and $y=\ln 2t$ at $t=\pi$

Find the curvature of the following curve:
$$ \begin{cases} x= \cos t \\ y=\ln(2t) \end{cases}$$
at $t=\pi$.

Can anyone help me with how to solve the question? Because I am confused if I need to put the $\pi$ earlier or after I solve the $x$ and $y$ equation by using the curvature formula.

答:
Hint.

First calculate $x'(t),y'(t),x''(t),y''(t)$. Then plug in $t=\pi$ to get $x'(\pi),y'(\pi),x''(\pi),y''(\pi)$. Then apply the curvature formula for plane curves.


问:
How one can find $y$ in terms of $x$ and $z$
Let us consider a congruence of the form:
$$x≡(xy) \bmod z$$

Then the question is: How one can find $y$ in terms of $x$ and $z$

答:
Well, if $x$ is not a zero divisor in $\Bbb Z_z$, i.e., $gcd(x,z)=1$ and so $x$ is a unit in $\Bbb Z_z$, the shortening rule can be applied in $\Bbb Z_z$:

$x\cdot 1 = x\cdot y \Rightarrow y=1$

or you can multiply $x=xy$ in $\Bbb Z_z$ with $x^{-1}$.


问:
Explicit formula for number of n-digit ternary sequences with no consecutive 2's
I'm looking for an explicit formula for $a_n$ which denotes the number of ternary sequences with no consecutive 2's. I've managed to find the recursive formula
$$a_n = 2a_{n-1} + 2a_{n-2}$$
I've solved the characteristic equation and found the explicit formula
$$a_n = A * (1+\sqrt{3})^n + B * (1-\sqrt{3})^n$$
And the basic conditions being $a_1 = 3$ and $a_2 = 8$. However, I cannot solve the system of equations to find the coefficients. I'd appreciate any help.

答:
From the recursion: $a_2 = 2\cdot a_1 + 2\cdot a_0$, so $8 = 2\cdot 3 + 2a_0$, which gives $a_0 = 1$. Therefore we can write

$$a_1 = A\cdot(1+\sqrt{3}) + B\cdot(1-\sqrt{3}) = 3$$
$$a_0 = A\cdot(1+\sqrt{3})^0 + B\cdot(1-\sqrt{3})^0 = A + B = 1$$

We have a 2x2 linear system over $A,B$ with non null determinant (you can check), so it has a unique solution: $A=1/2 + \sqrt{3}/3, B = 1/2 -\sqrt{3}/3$


问:
Is it true that a monotic, differentiable function with non-zero derivative has a continuous inverse?
Is it true that a strictly monotic, differentiable function on $\mathbb{R}$ with non-zero derivative has a continuous inverse?
This is a small caveat in a problem I'm working on, if this is true then I'm all good.</p>

答: Something stronger is true. A strictly monotone and continuous function on $\mathbb R $ has a continuous inverse. The function doesn't even have to have a derivative. </p>


问:
Order of calculation about nested absolute values
In case of nested absolute values, is there a difference between working first on the outer abs value and the opposite?
$$ \frac{|2x^2-3|x|+1|}{|x|+1} = 1 $$
Should I consider first the inner one?

My real solutions for this equation are $\left \{0, \pm 2\right \}$, and I've found those working first with the inner one.

答:
It is generally easier to start from the inside and then go outside. The usual procedure would be to make 2 cases: $x\geq0$ and $x<0$ which will give you:
$$\frac{|2x^2-3x+1|}{x+1} = 1 ~ \text{for} ~x \geq 0 $$
$$\frac{|2x^2+3x+1|}{-x+1} = 1 ~ \text{for} ~x < 0 $$
Now you can solve these two equations separately by further dividing each case into sub-cases of $2x^2+3x+1 \geq 0$  and $2x^2+3x+1<0$

Also note that for this particular question, $2x^2+3x+1 \geq 0 ~\forall ~x \geq0$ so, it reduces a case.


答:
By definition, the curvature is given by

$$\kappa = \frac{|\dot{x}\ddot{y} - \dot{y}\ddot{x}|}{(\dot{x}^2 + \dot{y}^2)^{3/2}}$$

Where $\dot{x}$ means derivative wrt $t$ (and similarly for $y$).

hence we have

$$\dot{x} = -\sin(t) ~~~~~~~~~~~ \ddot{x} = -\cos(t)$$
$$\dot{y} = \frac{1}{t} ~~~~~~~~~~~ \ddot{y} = -\frac{1}{t^2}$$

$$\kappa = \frac{\Big|\dfrac{\sin(t)}{t} + \dfrac{\cos(t)}{t^2}\Big|}{\left( \sin^2(t) + \dfrac{1}{t^2}\right)^{3/2}}$$

You don't neet to arrange, just plug $t = \pi$ and you get

$$\kappa = \pi$$

Title: Find the curvature if $x= \cos t$ and $y=\ln 2t$ at $t=\pi$

Body:
Find the curvature of the following curve:
$$ \begin{cases} x= \cos t \\ y=\ln(2t) \end{cases}$$
at $t=\pi$.

Can anyone help me with how to solve the question? Because I am confused if I need to put the $\pi$ earlier or after I solve the $x$ and $y$ equation by using the curvature formula.


问:
How can I derive the integral like this one?
I have a problem with integral like this one:

$$\int_{0}^{2\pi} \frac{\text{d}x}{\cos x+a}$$

when $a$ is from $1$ to $+\infty$.

Could anyone help me please?

答:
The constant $a$ must be strictly greater than $1.$

In that case, let $z=e^{i x}$ and $\cos x = \frac{1}{2}( e^{i x}+e^{-i x}) = \frac{1}{2} (z+1/z)$, etc.

After evaluating the integral by computing the residue at the pole inside the contour, you should obtain:
$$\int_0^{2\pi} \frac{dx}{a+\cos x} = \frac{2\pi}{\sqrt{a^2-1}}$$


答:
As long as you do it correctly, you can do whatever way you want.

I would simply do it as follows.

Let $y=|x|$. Then
$$
2y^2-3y+1=y+1,\quad y\ge 0\tag{1}
$$
or
$$
2y^2-3y+1=-(y+1),\quad y\ge 0\tag{2}
$$

Now solve (1) and (2); you can then easily find solutions to (1) and find that (2) has no solution.

Title: Order of calculation about nested absolute values

Body:
In case of nested absolute values, is there a difference between working first on the outer abs value and the opposite?
$$ \frac{|2x^2-3|x|+1|}{|x|+1} = 1 $$
Should I consider first the inner one?

My real solutions for this equation are $\left \{0, \pm 2\right \}$, and I've found those working first with the inner one.


答:
AD's point is right, however it can be written down in terms of non-elementary functions so, we make new function as its evaluation.

$\operatorname{erf} (x) $ is the error function, defined as $\operatorname{erf}(z)= \frac {2}{\sqrt{\pi}} \int_0^z e^{-t^2} dt$

If $I(x)=\int e^{-x^2}dx$,
$$\operatorname{erf}(x)= \frac {2}{\sqrt{\pi}} (I(x)-I(0)) \\
I(x)=\int e^{-x^2}dx=\frac{\sqrt{\pi}}{2} \operatorname{erf}(x)+I(0) = \frac{\sqrt{\pi}}{2} \operatorname{erf}(x)+ C$$

Title: Calculate the Indefinite integral

Body:
I've been stuck with calculating the intgral of the following problem. Can you help me?

$$\int e^{x^2} \mathrm{d}x$$

I know that, by Using the definition of gamma and beta functions, we have $\int_{-\infty}^{+\infty} e^{-x^2} \mathrm{d}x = \sqrt{\pi}$.

Thanks in advance.


问:
A few questions about the function sending a Borel set U to integration over U
Let $\mu$ be a measure on $\mathbb{R}^n$ and $\mathcal{F}:= \{f: \mathbb{R^n \to \mathbb{R}} \mid f\text{ is integrable w.r.t. } \mu\}$. Define a map $\iota:\mathcal{B}(\mathbb{R}^n)\setminus\mathcal{N}\mapsto \mathcal{F}^*, U \mapsto \int_U (\cdot)d\mu$, where $\mathcal{N}$ is the set of $\mu$-null-sets. My questions are:

I assume $\iota$ is not surjective. Is that so ?

Under which circumstances does an arbitraty functional $L\in\mathcal{F}^*$ have a preimage under $\iota$ ?

Can $\mathcal{B}(\mathbb{R}^n)\setminus\mathcal{N}$ be equipped with a topology/smooth structure such that $\iota$ becomes an embedding ?

Are there interesting resources concerning this topic ?

答:
Your space $\mathcal F$ is usually denoted  $L^1(\mathbb R^n,\mu )$ and, if you are denoting by $\mathcal{F}^*$ its dual
as a normed space, equipped with the the norm
$$
  \|f\|_1 = \int_{{\mathbb R}^n}|f(x)|\,d\mu(x),
  $$
then $L^1(\mathbb R^n,\mu )^*\simeq L^\infty (\mathbb R^n,\mu )$.  According to this isomorphism your functional $\iota _U$ corresponds
to the characteristic function $\chi_U$.  So, another way to view $\iota $ is
$$
  \iota:U\in  \mathcal{B}(\mathbb{R}^n) \mapsto \ \chi _U\in  L^\infty (\mathbb R^n,\mu ).
  $$

Your question (1) is therefore asking whether or not every bounded measurable function coincides with a
characteristic function a.e.,
and I suppose it has now become  obvious that the answer is no.

The answer to (2) can also be deduced  from the above formulation, as it is essentially asking which measurable
functions are characteristic functions.

Regarding (3), I have not seen any discussion on topology/smooth structures making $\iota $ an embedding, but one could
easily define an ad-hoc one by borrowing from the corresponding structures of $L^\infty (\mathbb R^n,\mu )$.


问:
Will conditional independence imply independence under this assumption?
Suppose the events $A$ and $B$ are conditionally independent given $C$. Also, suppose the events $A$ and $B$ are conditionally independent given $C^c$. Under this assumption, will the events $A$ and $B$ be independent? Can you help me either prove or disprove it?

答:
Imagine that I have two coins in my pocket: a fair coin, and a coin that is biased to land heads $\frac23$ of the time. I take a coin from my pocket and flip it twice. Let:

$A$ be the event that the coin lands heads the first time;

$B$ be the event that the coin lands heads the second time;

$C$ be the event that I took the fair coin from my pocket.

Then $A,B$ are conditionally independent both given $C$, and given $C^c$: these are two separate flips of the same coin.

But if you use the law of total probability to compute $\Pr[A]$, $\Pr[B]$, and $\Pr[A \cap B]$, you will find that $A$ and $B$ are not independent. The intuition is that the result of one coinflip gives you some hint as to whether the coin is fair, so you gain some information about the other coinflip.


问:
Deadly disease and probability to die
NHS records show that of patients suffering from a certain disease, 75% die of it.
In a specific hospital, there are 100 patients. What is the probability that 25 to 65 will die?

Clearly this is an example of use of binomial distribution.

I don't know, however, how to deal with the range 25 to 65. If the question was for exactly 25 patients to die, it would be

$P = \binom {100}{25}0.75^{25}.0.25^{75}$

but I don't know how to take into account the range.

Thank you.

答:
I feel like I've got this completely wrong, because if I'm right then it's quite elementary maths and someone would have answered it by now. I'll give my answer anyway and look forward to the reason why I'm wrong.

I will interpret the question as:

NHS records show that of patients suffering from a certain disease, $75$% die of it. In a specific hospital, there are $100$ patients with this disease. What is the probability that $25$ to $65$ (inclusive) will die from this disease?

Otherwise the question doesn't contain enough information to answer it.

Let $X$ be the random variable, "number of people who will die from the disease in the specific hospital".

Then assuming $(*)$ the four following assumptions are true:

Each outcome is either a "success" or a "failure": here I define success to mean the person dies. Not because I'm sadistic, but because that's how Binomial distribution works. We need to define terms. We could instead define "success" as the person not dying from the disease if we wanted: that would lead to different calculations, but the same answer of course.

The probability of "success", $p = 0.75$, is constant: it does not vary from person to person;

The "trials", namely seeing whether or not someone will die from the disease, are independent from one another;

The number of trials, $100$, is fixed/constant (i.e. it is $100$ and cannot be anything else/doesn't change);

then we may model $X$, the number of people who will die from the disease in the specific hospital, by a Binomial distribution  $X \sim B(100, 0.75). $

We then need to calculate $p(25 \leq X \leq 65)$. Any decent calculator should allow you to calculate this via the cumulative Binomial distribution function, either directly, or by using:

$$p(X \leq 65) - p(X \leq 24) = p(25 \leq X \leq 65).$$

I got: $p(25 \leq X \leq 65) = 0.0164\ (3sf).$

$(*)$ Some assumptions are more reasonable than others: only the 4th assumption is clearly a reasonable assumption. The 1st assumption is not a bad one, but doesn't take into consideration that the cause of death can be more than one disease, which is in fact often the case. The second assumption may or may not be reasonable: some people may fight the disease better than others - or not: perhaps this disease is immune system-independent - we don't know. I'm not sure how reasonable the 3rd assumption is. Anyway, the question ignores all of this for the sake of you learning how the Binomial distribution works.

Title: Deadly disease and probability to die


问:
How many natural numbers can't be written as 5p+7q, where p and q are natural numbers?
If $A = \{5p+7q|p,q \in \mathbb{N}\}$, determine $|\mathbb{N} \setminus A|$.

答:
For two prime numbers p,q the set which can not be represented as pa+qb with $a,b \in N$ is finite. There is a maximal non-representable number which is pq-p-q. This is known as the Frobenius number. There is a formula for the number of non-representable numbers $\frac{(p-1)(q-1)}{2}$.

Refs:

Ramírez-Alfonsín, J. L. The Diophantine Frobenius Problem. Oxford, England: Oxford University Press, 2005.


问:
is $\sin(a)=\sqrt{1-\cos^2(a)}$ derived from $\sin^2(a)+\cos^2(a)=1$
I have a problem of $\cos(a)=t for $270\leq a\leq360$

$a)\cos(90+a)$
for obvious , my answer is the answer is
$\cos(90)\cos(a)-\sin(90)\sin(a)$ and left me with
$=-\sin(a)$

 But what is the proof $\sin(a)=\sqrt{(1-\cos^2(a)}$
My solution:

$\sin^2(a)+\cos^2(a)=1$
$\sin^2(a)=1-\cos^2(a)$
$\sin(a)=\sqrt{1-\cos^2(a)}$
I have a feeling that this is false since I can't find this proof anywhere
Were this a real solution or proof to the $\sin(a)=\sqrt{(1-\cos^2(a)}$?

答: It is true that $\sin^2(a)+\cos^2(a)=1$, and therefore $\sin(a)$ is a square root of $1-\cos^2(a)$. But this number has two square roots (unless $\cos(a)=\pm1$.) So, it depends upon the sign of $\sin(a)$. If $0^\circ<a<180^\circ$, then indeed $\sin(a)=\sqrt{1-\cos^2(a)}$. And, if $180^\circ<a<360^\circ$, then $\sin(a)=-\sqrt{1-\cos^2(a)}$.


问:
How are $x \equiv -\frac {JZ}{LX}\pmod {p^D},y \equiv -\frac {J^2YZ}{LX^2}\pmod {p^D}$ calculated?
In this question

For what values of $l$ and $j$ (simultaneously) is the following congruence solvable?

I do not understand how are $$x \equiv -\frac {JZ}{LX}\pmod {p^D},y \equiv -\frac {J^2YZ}{LX^2}\pmod {p^D}$$ calculated?

Also, I do not understand how is  $$  xp^{s}j p^{a-b} + yp^t-l(xp^s)^2=xJp^A+yp^B-Lx^2p^C$$
this
$$ -\frac {J^2Z}{LX^2}(Xp^A+Yp^B+Zp^C )\equiv 0\pmod {p^D}.$$Could someone show me the details of the calculations please?

Here is the part of the answer containing those parts:

The transformation

Let $l=Lp^U$ and $j=Jp^V$ where $LJ \not\equiv 0\pmod p$. Then let $$A=a-b+s+V,B=t,C=2s+U,D=b.$$

Theorem 2

For non-negative integers $a\ge b,s$ and $t$, the equation
$$l (xp^s)^2 \equiv xp^{s}j p^{a-b} + yp^t\pmod{p^b}$$
can be solved with $xy \not\equiv 0\pmod p$ if and only if the equation
$$Xp^A+Yp^B+Zp^C \equiv 0\pmod {p^D}$$
can be solved with $XYZ \not\equiv 0\pmod p$.

Proof

If the $x,y$ equation can be solved then simply take $$X=xJ,Y=y,Z=-Lx^2.$$

Conversely, suppose we have a solution of the $X,Y,Z$ equation. Since $X$ and $L$ are coprime to $p$ we can solve
$$x \equiv -\frac {JZ}{LX}\pmod {p^D},y \equiv -\frac {J^2YZ}{LX^2}\pmod {p^D}.$$
Then $$  xp^{s}j p^{a-b} + yp^t-l(xp^s)^2=xJp^A+yp^B-Lx^2p^C$$
and this is
$$ -\frac {J^2Z}{LX^2}(Xp^A+Yp^B+Zp^C )\equiv 0\pmod {p^D}.$$

答:
Q1.

$LX$ and $p^D$ are coprime. Therefore there are integers $\alpha$ and $\beta$ such that $\alpha LX+\beta p^D=1$.

Then  $\alpha LX\equiv 1 \pmod {p^D}$  and so  $-\alpha JZ LX\equiv -JZ \pmod {p^D}$. Then take $x=-\alpha JZ$.

Similarly for the other congruence.

Q2.

Substitute $j=Jp^V$ and $l=Lp^U$ into $xp^{s}j p^{a-b} + yp^t-l(xp^s)^2$ and you will obtain $xJp^{s+a-b+V}+yp^t-x^2Lp^{s+2U}=xJp^A+yp^B-Lx^2p^C.$

Now substitute the expressions $x \equiv -\frac {JZ}{LX}\pmod {p^D},y \equiv -\frac {J^2YZ}{LX^2}\pmod {p^D}$ to obtain
$$-\frac {J^2Z}{LX^2}(Xp^A+Yp^B+Zp^C ),$$
where we know that $Xp^A+Yp^B+Zp^C\equiv 0 \pmod p^D$.


答:
As stated, it is false. The only thing you can deduce is that
$$\sin a=\pm\sqrt{1-\cos^2a}$$
and the sign depends on which interval $a$ lives ($a$ is in radians):

if $a\in[2k\pi,(2k+1)\pi]$ for some $k\in\mathbf Z$, $\sin a \ge 0$, so you have a $+$ sign;

if $a\in[(2k-1)\pi,2k\pi]$ for some $k\in\mathbf Z$, $\sin a \le 0$, so you have a $-$ sign.

Title: is $\sin(a)=\sqrt{1-\cos^2(a)}$ derived from $\sin^2(a)+\cos^2(a)=1$

Body:
I have a problem of $\cos(a)=t for $270\leq a\leq360$

$a)\cos(90+a)$
for obvious , my answer is the answer is
$\cos(90)\cos(a)-\sin(90)\sin(a)$ and left me with
$=-\sin(a)$

 But what is the proof $\sin(a)=\sqrt{(1-\cos^2(a)}$
My solution:

$\sin^2(a)+\cos^2(a)=1$
$\sin^2(a)=1-\cos^2(a)$
$\sin(a)=\sqrt{1-\cos^2(a)}$
I have a feeling that this is false since I can't find this proof anywhere
Were this a real solution or proof to the $\sin(a)=\sqrt{(1-\cos^2(a)}$?


问:
Integrate $\int_0^{2\pi}\frac{1}{4\cos^2(t)+9\sin^2(t)}\mathrm{d}t$ with complex functions or Fourier transformations.

What is the value of the following integral:
$$\int_0^{2\pi}\frac{1}{4\cos^2(t)+9\sin^2(t)}\mathrm{d}t$$
$\frac\pi9$ ; $\frac\pi6$ ; $\frac\pi3$ ; $\frac\pi2$ or $\frac\pi4$?

a full solution for this problem would be much appreciated

答:
Integrate with Fourier series as follows

$$\int_0^{2\pi}\frac{1}{4\cos^2 t+9\sin^2 t}{d}t
=\int_0^{2\pi}\frac{2}{13-5\cos 2t} {d}t\\
=\int_0^{2\pi}\left( \frac16 + \frac13\sum_{n=1}^\infty \frac{1}{5^n} \cos (2n t) \right)dt
=\frac\pi3
$$


问:
How to define unit matrix?
We have unit vector,w.r.t $2$-norm. I mean $$e_v = \frac{v}{\|v\|_2},$$ so $\|e_v\|_2=1$.

Do we have unit matrix w.r.t different matrix norm?

For example, unit matrix w.r.t Frobenius norm, $\|X\|_F = (\text{tr }(X^TX))^{1/2}$, so $$e_X = \frac{X}{\|X\|_F}.$$

Can we define the unit matrix in this way?  Does it have the same properties as unit vector?

Any suggestions or reference?   Thanks so much!

答:
In general, whenever you have a normed vector space $(V,\|\|)$ and a nonzero vector $v\in V$, you can always define a vector "with unit length" by
$$
w=\frac{v}{\|v\|}
$$

Title: How to define unit matrix?


问:
Is this correct regarding subtraction and addition of numbers or functions
Let us say we have

$$2=2$$

Then I would do is $$2-2=0$$i.e. I shifted $2$ from RHS to LHS. Now, my textbook says that this is not true and is a wrong concept. What my textbook tells me is this:

$$ 2-2=2-2$$

This means that when two equalities are equal , we cannot shift one from RHS to LHS but can subtract the same amount on both sides.

$$f(x) = g(x) \implies f(x)-g(x) = g(x)-g(x)$$

So, I wanted to confirm if this is how we really do it or both the methods are correct?

答:
As a note, let $R$ be a commutative ring like the ring of integers.

Subtraction is defined in terms of addition (so its a derived operation):

$r - s  := r+(-s)$

i.e., the subtraction $r-s$ means that you add $r$ and $-s$, the additive inverse of $s$.

E.g. $3 - 2 = 3 +(-2) = 1$.


答:
$$\int_0^{2\pi} \frac{\text{d}x}{a + \cos(x)}$$

Stepping into the complex plane:

$$z = e^{ix} ~~~~~~~ \cos(x) \to \frac{1}{2}\left(z + \frac{1}{z}\right) ~~~~~~~ \text{d}x = \frac{\text{d}z}{iz}$$

The integral becomes

$$\int_{|z|=1} \frac{\text{d}z}{iz\left(a + \frac{1}{2}\left(z + \frac{1}{z}\right)\right)}$$

Now we arrange it into

$$ \frac{2}{i} \int_{|z|=1} \frac{dz}{z^2 + 2az + 1}
= \frac{2}{i} \int_{|z|=1} \frac{dz}{(z-z_1)(z-z_2)}$$

where the circle $|z|=1$ is parametrized counterclockwise and
\begin{align*} z_1 = -a - \sqrt{a^2-1}  && z_2 = -a + \sqrt{a^2-1}.
\end{align*}

are the poles of the integrand function.

Clearly $z_1 < -a < -1$, so $z_1$ is outside of the unit circle. As for $z_2$, it is convenient to notice that
\begin{align*}
z_1 z_2 = 1.
\end{align*}
Thus, since $z_1$ is outside the unit circle, its inverse $z_2$ is inside the unit circle. So, applying the residue theorem, we have

\begin{align*}
\frac{2}{i} \int_{|z|=1} \frac{dz}{(z-z_1)(z-z_2)}
&= \frac{2}{i} \ 2 \pi i \ \mathrm{Res}\left( \frac{1}{(z-z_1)(z-z_2)}, z_2\right) \\
&= 4 \pi \frac{1}{z_2 - z_1} \\
&= 4 \pi \frac{1}{2 \sqrt{a^2 -1}} \\
&= \boxed{\frac{2 \pi}{\sqrt{a^2-1}}}
\end{align*}

Title: How can I derive the integral like this one?

Body:
I have a problem with integral like this one:

$$\int_{0}^{2\pi} \frac{\text{d}x}{\cos x+a}$$

when $a$ is from $1$ to $+\infty$.

Could anyone help me please?


答:
Here is a solution using differentiation under integral sign. Consider a more general integral, namely:

\begin{align*}
I_1(\alpha,\beta) & = \int_{0}^{2\pi}\frac{dx}{\alpha \cos^2x+\beta \sin^2x} \\
                  & = \int_{0}^{2\pi} \frac{\sec^2x}{\alpha + \beta \tan^2x} \,\mathrm{d}x\\
                  & = \frac{1}{\beta} \int_0^{2\pi} \frac{1}{\left(\sqrt{\frac{\alpha}{\beta}}\right)^2 + \tan^2 (x)}\; \mathrm{d}(\tan x)\,\\
& = \frac{1}{\sqrt{\alpha \beta}} \left[ \left(\tan^{-1}\left(\sqrt{\frac{\beta}{\alpha}}\tan (x)\right)\right) \right]_0^{2\pi} = \frac{2\pi}{\sqrt{\alpha\beta}}
\end{align*}

Title: Integrate $\int_0^{2\pi}\frac{1}{4\cos^2(t)+9\sin^2(t)}\mathrm{d}t$ with complex functions or Fourier transformations.

Body:
What is the value of the following integral:
$$\int_0^{2\pi}\frac{1}{4\cos^2(t)+9\sin^2(t)}\mathrm{d}t$$
$\frac\pi9$ ; $\frac\pi6$ ; $\frac\pi3$ ; $\frac\pi2$ or $\frac\pi4$?

a full solution for this problem would be much appreciated


问:
Transformation of Random Variables: How to relate these two approaches?
In the literature, I have found two approaches for the transformation of random variables

$$p_{X}(x) =\int_{-\infty}^{\infty}p_{Y}(y)\delta\left(x-f(y)\right)~\mathrm{d}y$$

and

$$p_{X}(x)=p_{Y}(f^{-1}(x))\left|\frac{\mathrm{d}f^{-1}(x)}{\mathrm{d}y}\right|.$$

How do you manage to reconcile these two?

My feeble attempt

$$p_{X}(x) =\int_{-\infty}^{\infty}p_{Y}(y)\delta\left(x-f(y)\right)~\mathrm{d}y=\int_{-\infty}^{\infty}p_{Y}(y)\delta\left(y-f^{-1}(x)\right)~\mathrm{d}y=p_{Y}\left(f^{-1}(x)\right)$$

is missing a term. Sorry my foolish question and feeble attempt but it really bothers me.

答:
$\newcommand{\dd}{\mathop{}\!\mathrm{d}}$
By properties of the Dirac $\delta$ and the inverse function theorem, we have
$$\delta(x-f(y))=\frac{\delta\left(y-f^{-1}(x)\right)}{\lvert f'(f^{-1}(x))\rvert}=\delta\left(y-f^{-1}(x)\right)\cdot\left\lvert(f^{-1})'(x)\right\rvert.$$
The above is essentially applying a "change of variables" type formula to the Dirac $\delta$. It obviously generalises the scaling and shifting properties. See e.g. this math.SE thread.

So plugging this form in,
$$p_X(x)=\int_{\infty}^\infty p_Y(y)\delta\left(y-f^{-1}(x)\right)\cdot\left\lvert(f^{-1})'(x)\right\rvert\dd y=p_Y\left(f^{-1}(x)\right)\cdot\left\lvert(f^{-1})'(x)\right\rvert,$$
which is as desired.


问:
Is the Arabic "4" a value, numeral, or numerical expression?

What I understand:

A number is an abstract concept. We use numerals such as names (i.e. cuatro), symbols (i.e. Arabic "4"), and numerical expressions (i.e. "9 - 5") to represent numbers.

Where I get tripped up:

"Value" is defined as the number a numeral represents, where "4" is a value of "3 + 1". Doesn't this contradict the author's earlier definition of "4" being a symbol?

Is "4" a number or a numeral?

Bonus Question

"4" is then defined as a numerical expression.
Is a numerical expression a type of numeral, or are they synonymous to one another?

I value your insight!

答:
I think this is mostly a semantic issue. If someone asked me who I was, I would say 'I'm Joe'; I wouldn't say 'I am the person represented by the name Joe'. It's good that the author initially draws a distinction between how we represent a number and what that number actually is. This is because some people overlook the fact that there is no difference whatsoever between $1/2$, $\frac{1}{2}$, $0.5$, $0.4\overline{9}$, and $0.4999...$—they all represent the same number in a kind of platonic sense. In practice, however, we would just say that they're the same number, and avoid quibbling over details. Mathematics would be far too cumbersome if every statement we make had to be $100\%$ precise.

So, as David has already mentioned in the comments, really we should say 'the number represented by $3+1$ is the same as the number represented by $4$', but no one bothers. And it is perfectly correct to write $3+1=4$, since everyone understands the meaning of those symbols, and it would be impossible to write mathematics without any shared notation.


问:
Why does the sign of the result determine the location of the point?
I had learned this in school:- if we have an equation of a line (say x+y-6=0) if the result from substituting a point comes out to be:

negative = then the point is above the line

positive = then the point is below the line

So If I put in the point (3,2), then it would be 3+2-6 = -1 = negative = point above the line.

I attempted to prove it but wasn't able to think of any theorems/logic that can aid this.

Can anyone explain the proof or any logic that explains why this works??
Or is this determined experimentally?

Example:

答:
This is not true.

The above graph shows the lines $x+y-6=0$ and $-x-y-6=0$. If we put in $(3,2)$ into each of the lines, we get $$3+2-6=-1<0$$ and $$-3-2-6=-11<0$$ respectively. Both give a negative result, but the point is below the first line and above the second.


答: Another approach writes $2\int_0^\pi\frac{dx}{\cos x+a}$ with $t=\tan\tfrac{x}{2}=\sqrt{\tfrac{a+1}{a-1}}\tan y$ as$$\int_0^\infty\frac{4dt}{a+1+(a-1)t^2}=\int_0^{\pi/2}\frac{4dy}{\sqrt{a^2-1}}=\frac{2\pi}{\sqrt{a^2-1}}.$$


问:
Prove $P(A|B)+P(\bar{A}|B)=1$
How can I prove the following statement:

$$P(A|B)+P(\bar{A}|B)=1$$

I did the following:

$P(A|B)+P(\bar{A}|B)= \frac{P(A\cap B)}{P(B)} + \frac{P(\bar{A} \cap B)}{P(B)} = \frac{P(A\cap B)+ P(\bar{A} \cap B)}{P(B)}$

But I don't know how to connect the fact that since $A$ and $\bar{A}$ are mutually exclusive, it becomes $\frac{P(B)}{P(B)} = 1$

答:
According to The Probability of Not Mutually Exclusive Events:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
When two events are mutually exclusive, $A \cap B = \varnothing$, $P(A \cap B) = 0$, hence:
$$P(A \cup B) = P(A) + P(B)$$
Now that $(A \cap B) \cap (\bar A \cap B) = [(A \cap B) \cap \bar A] \cap B = [(A \cap \bar A) \cap B] \cap B = (\varnothing \cap B) \cap B = \varnothing$, $A \cap B$ and $\bar A \cap B$ are two mutually exclusive events, naturally:
$$P(A \cap B) + P(\bar A \cap B) = P[(A \cap B) \cup (\bar A \cap B)]$$
Because $(A \cap B) \cup (\bar A \cap B) = (A \cup \bar A) \cap B = B$,
we have $P(A \cap B) + P(\bar A \cap B) = P(B)$.

Eventually, we have $P(A \mid B) + P(\bar A \mid B) = 1$.

Title: Prove $P(A|B)+P(\bar{A}|B)=1$


问:
Can anyone show me, how to solve these system of Equations:
$$\begin{align*}
x+y+z &= 2 \\
(x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y) &= 1 \\
x^2(y+z)+y^2(z+x)+z^2(x+y) &= -6
\end{align*}$$

Can anyone explain me the solution. I asked it in mathoverflow but god knows why it was closed. Please help and detail the process in step by step. It is a RMO question (The Pre- Indian team selection exam for IMO)

Body:
$$x+y+z=2$$

$$x^2+y^2+z^2+2(xy+zx+yz)=4$$

The second equation is $x^2+y^2+z^2+3(xy+zx+zy)=1 \implies zx+yz+yx=-3$

You get $x^2+y^2+z^2=10$

Using the $x+y+z=2$ in the third equation.

$x^2(2-x)+y^2(2-y)+z^2(2-z)=-6 \implies 2(x^2+y^2+z^2)-x^3+z^3+z^3=-6$

A little re-arranging would give

$26 =x^3+y^3+z^3 $

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$26-3xyz=(2)(10-(-3)) \implies 26-26=3xyz \implies $ one of the $x,y,z$ is zero.

Consider $z=0$ (Generality is not lost) Now you have $x^2+y^2=10$ and $yx=-3$

You get solutions as $(x,y,z)=(3,-1,0),(0,-1,3)(0,3,-1),(-1,0,3),(3,0,-1)$ and $(-1,0,3)$.


问:
$x^∗ ∈ C$ is an optimal solution of the problem iff $〈∇f (x),x∗ − x〉 ≤ 0$ for all $x ∈ C$.

Let $f$ be a continuously differentiable convex function over a closed and convex
set $C ⊆ \mathbb{R}^n$. Show that $x∗ ∈ C$ is an optimal solution of the problem
(P) min$\{ f (x) : x ∈ C\}$

if and only if

$〈∇f (x),x∗ − x〉 ≤ 0$ for all $x ∈ C$

.

The first part where I assume that $x^*$ is optimal solution and then showing the inequality I've managed to show , at the second part I've tried assuming that $x^*$ is not optimal solution and then show the inequality doesn't hold but got stuck. any hint?

答:
Since $f$ is convex, then by the Gradient Inequality $$f(\mathbf{x})\geq f(\mathbf{x}^*)+\nabla f(\mathbf{x}^*)^T(\mathbf{x}-\mathbf{x}^*), \quad \forall\mathbf{x}\in C.$$
Next, we assume that $\nabla f(\mathbf{x}^*)^T(\mathbf{x}^*-\mathbf{x})\leq 0$, which is the same as $\nabla f(\mathbf{x}^*)^T(\mathbf{x}-\mathbf{x}^*)\geq 0$. Using this assumption in the Gradient Inequality means that $$f(\mathbf{x})\geq f(\mathbf{x}^*), \quad \forall\mathbf{x}\in C$$
which proves the optimality of $\mathbf{x}^*$.


问:
Confused with a proof about the simplicity of $A_n$.
I am currently self-studying group theory and I am trying to understand why $A_n$ is simple for $n \geq 5$ and I have not made any progress for hours. I have searched for proofs everywhere (including this website) and every proof I've found has confused me. At the moment I am reading a proof found on this website here: https://groupprops.subwiki.org/wiki/Alternating_group_is_simple.
The proof uses induction and the website has a separate article when $n=5$. Luckily, I understand the proof of the base case. The rest of the proof goes as follows:

Let $N$ be a normal subgroup of $A_{n+1}$. Let $H_i$ denote the subgroup of $A_{n+1}$ that stabilizes the letter $i$. Then, each $H_i$ consists of the even permutations on $n$ letters (excluding $i$) and is hence isomorphic to $A_n$. Thus, each $H_i$ is simple. Now, since normality satisfies transfer condition, $N \cap H_i$ is normal in $H_i$ for every $i$. By simplicity of $H_i$, either $N$ contains $H_i$, or $N$ intersects $H_i$ trivially.

Suppose there exists $i$ for which $N$ contains $H_i$. Then, by fact (3) (which says that $A_n$ is a contranormal subgroup inside $A_{n+1}$), $H_i$ is contranormal inside $A_{n+1}$, i.e., its normal closure is $A_{n+1}$. Since $N$ is normal, this forces $N = A_{n+1}$ and we are done.

Otherwise, $N \cap H_i$ is trivial for every $i$. Thus, no nontrivial element of $N$ fixes any letter. Let's use this to show that $N$ can have no nontrivial elements. Suppose $\sigma \in N$ is nontrivial. Then, first observe that in the cycle decomposition of $\sigma$, every element must be in a cycle of the same length $k$ (otherwise, som power of $\sigma$ would fix a letter)(+). Thus, $\sigma$ has $r$ cycles each of size $k$, where $kr = n+1$ (+). Now, if $n \geq 5$, then $n + 1 \geq 6$. Choose an $i$ and a double trransposition $\tau \in A_{n+1}$ such that $\tau$ fixes both $i$ and $\sigma(i)$, but such that $\tau$ does not commute with $\sigma$ (this is possible because $n+1 \geq 6$)(+). Then $\sigma^{-1}\tau\sigma\tau^{-1} \in N$ is a nontrivial element of $N$ fixing both $i$ and $\sigma(i)$, giving the required contradiction. $\square$

The parts that I do not understand in this proof are marked by (+). Why does every element in the cycle decomposition of $\sigma$ have the same length? If that's not the case then how does it come that some power of $\sigma$ would fix a letter? If $\sigma$ has $r$ cycles then why is $kr = n+1$? Lastly, why is it possible to find such a $\tau$ when $n + 1 \geq 6$? It feels like the proof goes way too fast for me and I simply can not understand everything. I am begging for someone to explain this to me. I do not have any teacher or friend that can help me with this so I would seriously appreciate any answer or a link to an easier to understand proof of this theorem!!

答:
$\newcommand{\supp}{\text{supp}}$

Every element in the cycle decomposition of $\sigma$ must have the same length.

Suppose that this is not the case. For simplicity, suppose $\sigma = \sigma_1 \sigma_2$, where $\sigma_1, \sigma_2$ are disjoint cycles of respective length $n_1 < n_2$ (and we have $n_1 +n_2 = n+1$ because $\sigma$ has no fixed point).

Then, $\sigma^{n_1} = \sigma_1^{n_1}\sigma_2^{n_1} = \sigma_2^{n_1} \neq \text{id}_{N}$ because $n_1 < n_2$ and $\sigma_2$ is of order $n_2$. Because $\sigma^{n_1} \in N$, this is a contradiction.

And it is easy to see that in the case where $\sigma = \sigma_1\sigma_2...\sigma_r$ disjoint cycles of the same length $p$, then $\sigma^m$ is either the identity (if $m \equiv 0 \mod r$), or a full-support permutation.

If \sigma has $r$ cycles, then $kr = n+1$

This simply comes from the previous point, and the fact that $\sigma$ has no fixed point, i.e. $|\supp(\sigma)| = n+1$. If we write $\sigma = \sigma_1\sigma_2...\sigma_r$ the decomposition in disjoint cycles, we just proved that they all the same length, and:

$\begin{align*}
\underbrace{|\supp(\sigma)|}_{= n+1} &= \sum_{i=1}^{r} |\supp(\sigma_i)| &\text{because they are disjoint cycles}\\
&= r |\supp(\sigma_1)| &\text{because all the cycles have the same length}\\
&= kr
\end{align*}$

Existence of $\tau$

This is a bit harder to see immediately. You need to find $4$ elements (not $i$ or $\sigma(i)$) $a, b, c, d \in \{1, \dots, n+1\}$ so that $\tau = (ab)(cd)$ does not commute with $\sigma = \sigma_1...\sigma_r$, i.e. we don't want $\t^{-1}\sigma\tau = \sigma$

I don't have the time to write a proper proof now, I will try to come back later - a "nice" way to see this is to understand what the decomposition in cycles of $\tau^{-1}\sigma\tau$ looks like.


问:
The point $c$ from mean value theorem
The mean value theorem says that

Let $a<b$ be real numbers. If $f:[a,b]\rightarrow\mathbb{R}$ is continuous on the closed interval $[a,b]$ and differentiable on open interval $(a,b)$, then
there exists $c\in (a,b)$ such that
\begin{align*}
         f(b)-f(a)=f'(c) (b-a)
       \end{align*}

My question/answer: Suppose now that $f$ is differentiable on $[a,b]$. Can the point $c$ be an endpoint of the interval $[a,b]$? Since the mean value theorem is a corollary of Rolle's theorem, it follows that $c$ cannot be an endpoint of the interval $[a,b]$. Otherwise, the function $f$ would be constant. Am I right? Does exist a nonconstant function such that $c$ be an endpoint of the interval $[a,b]$?

答: Well, you can consider the function $y=x$ restricted on $[0,1]$.

Title: The point $c$ from mean value theorem


问:
Representatives of a totally indiscernible sequence of hyperimaginaries
Let $X$ be a $0$-type definable set and $E$ a $0$-type definable equivalence relation.

One can prove (see for example proposition 16.3 in Enrique Casanovas' stability notes http://www.ub.edu/modeltheory/documentos/stability.pdf) that if we have an indiscernible sequence $([a_i]_E)_{i\in I}$ of hyperimaginaries. There exist a sequence $(b_i)_{i\in I}$ of representatives that is also indiscernible.

Are there any well-known results if we substitute indiscernibility by total indiscernibility? (At least for real parameters?).

答:
This is not true in general, even for imaginaries.

Consider $M = (\mathbb{Q};<,E)$, where $E$ is an equivalence relation with infinitely many classes such that every $E$-class is dense with respect to the order $<$. This is the Fraïssé limit of the class of all finite linear orders equipped with an arbitrary equivalence relation.

Any sequence of distinct $E$-imaginaries is totally indiscernible, but no sequence of representatives is totally indiscernible, due to the ordering.

To prove the claim about total indisceribles in $\mathbb{Q}/E$, it suffices to show  that for any distinct $a_1,\dots,a_n\in\mathbb{Q}/E$ and distinct $b_1,\dots,b_n\in\mathbb{Q}/E$, there is an automorphism of $M$ mapping each $a_i$ to $b_i$ (i.e., the induced structure on $\mathbb{Q}/E$ is that of an infinite set with equality). Pick a representative $a_i'$ for each $a_i$ such that $a_1'<a_2'<\dots< a_n'$.  This is possible since each $E$-class is dense. Similarly, pick a representative $b_i'$ for each $b_i$ such that $b_1'<b_2'<\dots< b_n'$. Then the tuples $(a_1',\dots,a_n')$ and $(b_1',\dots,b_n')$ have the same quantifier-free type in $M$ (the type of an increasing sequence of pairwise non-$E$-equivalent elements), so by ultrahomogeneity of $M$ there is an automorphism mapping each $a_i'$ to $b_i'$ and hence each $a_i$ to $b_i$.

This example shows that an infinite quotient of an unstable set can be stable. It seems unlikely to me that we could prove any general theorem on pulling back total indiscernibles in $X/E$ to total indiscernibles in $X$, except in the trivial case that $X$ itself is stable (in which case all indiscernibles in $X$ are total indiscernibles).


问:
Show operator is not adjoint
I have read that

$$T: \ell^1 \to \ell^1, \quad (x_n)_n \mapsto \left(\sum_{n=1}^\infty x_n,0,0,...\right)$$

is not an adjoint operator. I know that the dual of $c_0$ is actually $\ell^1 $, but I don't seem to get it since the dual of $\ell^1$ is $\ell^\infty$. Any help is appreciated thanks.

答: Assume that $S : c_0 \to c_0$ is the pre-adjoint of $T$, that is, $S^* = T$. Then $S$ would be the restriction of $S^{**} = T^* : \ell^{\infty} \to \ell^{\infty}$ to $c_0$. It is not difficult to see that the adjoint of $T$ is given by the operator that maps $(x_n)_{n \in \mathbb{N}}$ to the sequence $(x_1,x_1,x_1,\ldots)$. Now clearly this operator does not map $c_0$ to $c_0$, hence $S$ cannot exist.

Title: Show operator is not adjoint


问:
Evaluate integral $ \int_{\mathbb{R} } \frac{\sin^4(\frac{t}{2}) }{t^2}$
Let $f(x)= (1-2|x|)\chi_{[-\frac{1}{2}, \frac{1}{2}]} (x)  $. Its Fourier transform is given by
$ \hat{f} (x) = \frac{8\sin^2(\frac{t}{4})}{t^2} $. Based on this, I need to evaluate the integral $ \int_{\mathbb{R} } \frac{\sin^4(\frac{t}{2}) }{t^2}$, but I don't know where to start.  I tried Parseval's identity - I wrote  $\frac{\sin^4(\frac{t}{2}) }{t^2} = \frac{\sin^2(\frac{t}{2}) }{t^2} \cdot \sin^2(\frac{t}{2})$, and then I would need to find the function whose Fourier transform is $\sin^2(t)$. I don't know if this is right way (seems too complicated), so is there any simple way to evaluate integral? I would be very grateful for help.

答:
Let's consider $$I(k,q)=\int_0^\infty\frac{\sin^k(x)}{x^q}dx$$ where $k$ is positive integer and $q$ is rational.

Using the formula $\frac{1}{x^q}=\frac{1}{\Gamma(x)}\int_0^\infty{t}^qe^{-xq}dx$

we get $I(k,q)=\frac{1}{\Gamma(x)}\int_0^\infty{t}^qdt\int_0^\infty\sin^k(x){e}^{-xq}dx$, where $\Gamma(x)$ is Gamma-function.

Integrating by part and grouping the terms with the same power of $\sin(x)$ we get:

$$I(k,q)=\frac{k!}{\Gamma(q)}\int_0^{\infty}\frac{t^{q-2}}{(t^2+2^2)(t^2+4^2)...(t^2+k^2)}dt, k=2n, q>1, n=1, 2,3...$$

$$I(k,q)=\frac{k!}{\Gamma(q)}\int_0^{\infty}\frac{t^{q-1}}{(t^2+1^2)(t^2+3^2)...(t^2+k^2)}dt, k=2n-1, q>0, n=1, 2,3...$$

To evaluate the integrals let's consider the following contour: along the cut from $0$ to ${\infty}$ (along the axis), big circle of radius $R$ (counter clockwise), along the other bank of the cut back to $0$ and along a small circle $r$ around $0$ (around a brunch point of $t^q$ - $q$ is rational) to the starting point.

Making a full round, on the lower bank of the cut the integrand gets additional factor $e^{2\pi{iq}}$. Integral along this closed contour is equal to the sum of residuals in simple poles, and integrals along big and small circle set to zero as $R\to\infty$ and $r\to0$:
$$I(k,q)(1-e^{2\pi{iq}})=2\pi{i}\frac{k!}{\Gamma(q)}\sum{Res}\frac{t^{q-2}}{(t^2+2^2)(t^2+4^2)...(t^2+k^2)}, k=2n$$
$$I(k,q)(1-e^{2\pi{iq}})=2\pi{i}\frac{k!}{\Gamma(q)}\sum{Res}\frac{t^{q-1}}{(t^2+1^2)(t^2+3^2)...(t^2+k^2)}, k=2n-1$$
Finally, for even ($k=2n$), $q>1$ ($q$ - rational, not necessarily integer); $a=2, b=4, ...=k$
$$I(k,q)=-\frac{\pi{k!}}{2\Gamma(q)\cos\frac{\pi{q}}{2}}\left(\frac{a^{q-3}}{(b^2-a^2)(c^2-a^2)..(k^2-a^2)}+\frac{b^{q-3}}{(a^2-b^2)(c^2-b^2)..(k^2-b^2)}+..\right)$$

... and for odd ($k=2n-1$), $q>0$, $a=1, b=3, ...=k$
$$I(k,q)=\frac{\pi{k!}}{2\Gamma(q)\sin\frac{\pi{q}}{2}}\left(\frac{a^{q-2}}{(b^2-a^2)(c^2-a^2)..(k^2-a^2)}+\frac{b^{q-2}}{(a^2-b^2)(c^2-b^2)..(k^2-b^2)}+..\right)$$

Now we can easily evaluate $\int_{\mathbb{R} } \frac{sin^4(\frac{t}{2})}{t^2}=I(k=4,q=2)=\int_0^\infty\frac{sin^4(t)}{t^2}dt$

$k$ is even and $k=4$, therefore we have only $a=2, b=4$.
$$I(4,2)=-\frac{\pi{4!}}{2\Gamma(2)\cos\frac{\pi{2}}{2}}\left(\frac{2^{-1}}{(4^2-2^2)}+\frac{4^{-1}}{(2^2-4^2)}\right)=\frac{12\pi}{4^2-2^2}(\frac{1}{2}-\frac{1}{4})=\frac{\pi}{4}$$

Some additional comments:

While calculating residuals right phases to poles should be appointed: $i=e^{\frac{\pi{i}}{2}}$ and $-i=e^{\frac{3\pi{i}}{2}}$ (one quarter and three quarters of total turn counter clockwise correspondingly).  $i^q=e^{\frac{\pi{iq}}{2}}$ and $(-i)^q=e^{\frac{3\pi{iq}}{2}}$

At some $q$ there are uncertainties (when $\cos\frac{\pi{q}}{2}$ or $\sin\frac{\pi{q}}{2}=0$ in the denominators) that need to be evaluated. For instance, $I(k=3,q=2)=\int_0^\infty\frac{\sin^3(x)}{x^2}dx$. We have $a=1, b=3$, we set
$q=2+\epsilon$ and $$I(3,2)=\lim_{\epsilon\to0}\frac{6\pi}{2}\frac{1}{\sin\frac{\pi(2+\epsilon)}{2}}\left(\frac{1^\epsilon}{3^2-1^2}+\frac{3^\epsilon}{1^2-3^2}\right)=$$$$=-\lim_{\epsilon\to0}\frac{6}{\epsilon}\left(\frac{1^0}{3^2-1^2}+\frac{\epsilon\log(1)}{3^2-1^2}+\frac{3^0}{1^2-3^2}+\frac{\epsilon\log(3)}{1^2-3^2}\right)=\frac{3\log3}{4}$$

Any integral $I(k,q)=\int_0^\infty\frac{\sin^k(x)}{x^q}dx$ for not too big integer $k$ can easily be evaluated. Just few examples:

$I(k=4,q=4)=\int_0^\infty\frac{\sin^4(x)}{x^4}dx=-\frac{4!\pi2^2}{2^4\Gamma(4)}\left(\frac{1}{2^2-1^2}+\frac{2}{1^2-2^2}\right)=-\frac{\pi}{3}(1-2)=\frac{\pi}{3}$
$I(k=3,q=\frac{1}{2})=\int_0^\infty\frac{\sin^3(x)}{\sqrt{x}}dx=\frac{6\pi\sqrt{2}}{2\sqrt{\pi}}\left(\frac{1^-\frac{3}{2}}{3^2-1^2}+\frac{3^-\frac{3}{2}}{1^2-3^2}\right)=-\frac{\sqrt\pi}{4\sqrt6}(3\sqrt3-1)$
$I(k=5,q=\frac{3}{2})=\int_0^\infty\frac{\sin^5(x)}{x^\frac{3}{2}}dx=\frac{5!\pi}{2\Gamma(\frac{3}{2})\sin\frac{3\pi}{4}}\left(\frac{1}{\sqrt1(3^2-1^2)(5^2-1^2)}+\frac{1}{\sqrt3(1^2-3^2)(5^2-3^2)}+\frac{1}{\sqrt5(1^2-5^2)(3^2-5^2)}\right)=\sqrt{\frac{5\pi}{2}}\frac{2\sqrt5-\sqrt15+1}{24}$

etc...


问:
Rewriting M in terms of omega?
Given a vector $v \in \mathbb{R}^n$ then show that we can express $\sum_{k} \omega_kv^2_k$ as a matrix product of the form $v^TMv$. Give an expression for $M$ in terms of $\omega = [\omega_1 ... \omega_n]^T$.

I understand that here, the product of a row vector, matrix, and column vector (in that order) is a scalar. However, how do I write $M$ in terms of $\omega = [\omega_1 ... \omega_n]^T$?

答:
Hint: Note that in general, we have
$$
v^TMv = \sum_{i=1}^n \sum_{j=1}^n m_{ij} v_i v_j.
$$
I would recommend that you verify this for the cases $n = 2,3$. Now, notice that the expression $\sum_k \omega_k v_k^2$ contains no terms of the form $\alpha\,v_iv_j$ (for a scalar $\alpha$).

Title: Rewriting M in terms of omega?


问:
Finding $v_y$ in system of equations.
$$\begin{cases}u^2+v^2+x^2-y^2=1 \\ u^2-v^2-x^2-y^2=4 \end{cases}$$

I want to know $v_y$.

where $u(x,y),v(x,y)$ are functions of $x,y$ and smooth.

To do this I first did:

$$\begin{cases}2uu_x+2vv_x+2x=0 ,& 2uu_y+2vv_y-2y=0 \\ 2uu_x-2vv_x-2x=0 ,& 2uu_y-2vv_y-2y=0 \end{cases}$$

Maybe I can solve it now but I dont think this is the proper way?

答:
By subtraction,

$$2v^2+2x^2=-3$$ and $v$ does not depend on $y$.

Title: Finding $v_y$ in system of equations.


问:
"Pathfinding" in a Category
Suppose you are given the "action" of a morphism, some specification of what it does to elements; in $\textbf{Set}$ or related categories it would be the graph of a function. Is it possible to determine, in theory or practice, (whether there exists) a set of arrows whose composition matches the given action? If there are many such sets, is it possible to determine the one(s) requiring minimal compositions?

My intuition, stemming from the digraph axiomatization of a category, suggests that this is a simple example of pathfinding. However, most categories not only have infinite or large object sets, but infinite or large hom-sets. It's not clear how the standard algorithms would work on an infinite graph, let alone a graph with infinitely many edges between each vertex. I couldn't find any information on generalizations of pathfinding to such cases. However, it seems like this would be a common enough question in the subject to have been asked before. Is there something I'm missing?

答:
There isn't a universal answer to this. Many problems (NP problems) could be easily written in such a formulation. For example : Hamiltonian cycle.

Your category is the set of nodes you've visited +
the node you're actually in. Your action is to move to a neighboring
vertex and mark it as visited. You have an Hamiltonian cycle iff the
shortest path from $\emptyset \times {x}$ to $V \times {x}$ is of
length exactly $|V|$)

So each answer will have to be designed specifically to the problem at hand, with the help of the geometry of the problem.

But this isn't a lost cause ! The edit distances (Levenshtein distance, tree edit distance, time-warp edit distance, etc) are generally defined as pathfinding distances on huge sets, but computed with dynamic algorithm that run in polynomial time of the input size.

If you can't find an "analytic" solution, maybe a pathfinding algorithm like A* can help you  find a solution faster. The idea is to have a lower-bound heuristic ("the path from X to Y is at least n"), and use this kind of bound to orient your search. This could help you explore the neighbors in an interesting order and not explore a huge part of the category.

Title: "Pathfinding" in a Category


问:
Eigenvalues of a linear transformation
So I've been given a question that basically looks like this:
Let $T:V \to V$ be a linear transformation with $T^3 = T \circ T \circ T = T$, Show that $T$ can only have eigenvalues from the set $\{-1,0,1\}$.

So I understand the question and how to calculate eigenvalues 'normally' but in this case the problem I have is what to actually write. I get the idea of the answer (I think) which is basically that only the values of $-1$, $0$ and $1$ can be cubed and get the same value back ($-1$ cubed is $-1$, $0$ cubed is $0$, etc.). I'm pretty much just not sure how I would 'prove' this to show what I need to show.
Thanks in advance.

答: Your idea is correct. The underlying idea is that if $\lambda$ is an eigenvalue of $T$ then $\lambda^3$ is an eigenvalue of $T^3$. Can you prove this?

Title: Eigenvalues of a linear transformation


答:
Let $\lambda,x$ be an eigenvalue/vector pair of $T$. Then $T x = \lambda x$. Then note that $\lambda x = Tx = T^3 x = T^2 (Tx) = T^2 (\lambda x) = \lambda T(Tx) = \lambda^2 T x = \lambda^3 x$.

Therefore $\lambda^3 = \lambda$ or equivalently $\lambda(\lambda^2-1)=0$ with solutions $\lambda = -1,0,1$.

Title: Eigenvalues of a linear transformation

Body:
So I've been given a question that basically looks like this:
Let $T:V \to V$ be a linear transformation with $T^3 = T \circ T \circ T = T$, Show that $T$ can only have eigenvalues from the set $\{-1,0,1\}$.

So I understand the question and how to calculate eigenvalues 'normally' but in this case the problem I have is what to actually write. I get the idea of the answer (I think) which is basically that only the values of $-1$, $0$ and $1$ can be cubed and get the same value back ($-1$ cubed is $-1$, $0$ cubed is $0$, etc.). I'm pretty much just not sure how I would 'prove' this to show what I need to show.
Thanks in advance.


问:
Twisted ribbons and centre of gravity
NOTE
After posting this question, I realised I had not defined the overall shape of the form. The final note at the bottom was intended to correct that omission. I am no longer certain how this affects things. On the face of it, it defeats my argument (see answer by Ethan Bolker). Nevertheless, I'll let the question stand (see comment by  Joshua Wang under EB's answer).

I have looked for a previous thread on this (including under Mobius strip) and have not found the precise same question.

I am not a mathematician so please forgive any incorrect terminology.

I'll define a bracelet as a flat ribbon or band of metal, joined to itself end-to-end with any number of half-twists. Assume that during manufacture the twisting is made uniform along the length of the bracelet.

I'm interested in the wheel-like rotational symmetry about the axis and the placement of the centre of gravity. Intuitively it seems obvious to me that if twisting is uniform, all bracelets have a balance point at the centre** except for the one with a single half-twist.

How can I prove (a) that the C of G is non-central for the single-twist example and (b) that the C of G is central for all other numbers of twists whether zero, odd or even.***

*** Negative twists are not considered distinct for these purposes.

** Let us say that the bracelet is formed in such a way that a central line drawn within and along the length of its volume forms a perfect circle. The centre is then the centre of this circle.

答:
Are you sure the center of gravity is not the central point in all cases? For a single half twist the bracelet is vertical (that is, parallel to the axis) at one end of one of the diagonals and horizontal at the other end with the center line on the circle. The moments will be the same.

I think this kind of argument shows that the center of gravity will always be at the center of the circle you describe.

Title: Twisted ribbons and centre of gravity


问:
Let $n=am+r$ with $m \ge a>5$ and $r \ge 0$. Prove that $\phi(n) \ge m$
I understand that for all $n>6$, we have $\phi(n) \ \ge  \ \sqrt n $. This fact however is not getting me anywhere towards the proof of the claim in question.
($a, m, r$ are all integers)

答:
This is false (for $a>4$ as in the original post, but also for $a>5$ with a different example). Take for example $a=5$ and $m=6006$, $r=0$. Then
$$
\phi(30030)=\phi(5\cdot 6006)=\phi(5)\cdot \phi(6006)=4\cdot 1440=5760,
$$
which is not bigger or equal to $m=6006$. In fact, we have
$$
\lim \inf \frac{\phi(n)}{n}=0.
$$


答:
This is easily understood from a continuity argument.

The line is the locus of the points of coordinates such that $f(x,y)=ax+by+c=0$. By continuity (Intermediate Value theorem), you cannot move from a point where $f$ is positive to a point where it is negative without crossing the line. Hence the line splits the plane by sign.

(For a more correct description, you should replace "above" and "below" by "left" and "right", where the orientation depends on the signs of the coefficients.)

Title: Why does the sign of the result determine the location of the point?

Body:
I had learned this in school:- if we have an equation of a line (say x+y-6=0) if the result from substituting a point comes out to be:

negative = then the point is above the line

positive = then the point is below the line

So If I put in the point (3,2), then it would be 3+2-6 = -1 = negative = point above the line.

I attempted to prove it but wasn't able to think of any theorems/logic that can aid this.

Can anyone explain the proof or any logic that explains why this works??
Or is this determined experimentally?

Example:


问:
When does l'Hospital's rule work for series?
This question stems from a response a colleague of mine received as a (flawed) solution to a problem on his calculus exam. The student was to determine convergence of a series of the form $\displaystyle\sum_{n=1}^\infty \frac{f(n)}{g(n)}$. The student instead considered the series  $\displaystyle\sum_{n=1}^\infty \frac{f'(n)}{g'(n)}$ and thought that this series and the original must behave the same way. This, of course, is false in general.

What conditions on $f$ and $g$ are required so that the series  $\displaystyle\sum_{n=1}^\infty \frac{f(n)}{g(n)}$ converges iff  $\displaystyle\sum_{n=1}^\infty \frac{f'(n)}{g'(n)}$ converges?

答:
Suppose $f$ and $g$ are meromorphic in a neighbourhood $U$ of $\infty$ with poles at $\infty$ (so $\lim_{z \to \infty} f(z) = \lim_{z \to \infty} g(z) = \infty$), having orders $p$ and $q$ respectively.  Then $f'$ and $g'$ have order $p-1$ and $q-1$ respectively at $\infty$.
If $\{N, N+1, \ldots\} \subset U$,
then $\sum_{n=N}^\infty \frac{f(n)}{g(n)}$ converges iff $q \ge p + 2$
iff $\sum_{n=N}^\infty \frac{f'(n)}{g'(n)}$ converges. </p>


问:
Wealth indicator function for bidder agent logic
I want to create a wealth indicator function used by the logic of a bidder agent, that tells the agent if he's rich (in comparison to others).

Given:

Total number of competitors $n$

Amount of all the money $m$ in the system

mean (displaying money per agent) $a$. (E.g. arithmetic mean etc.)

Searching for:
$w(x) \in \mathbb{R}, x \in \mathbb{N}$

so that:

$w(m) = 1$

$w(a) = \frac{1}{2}$

$w(0) = 0$

(doesn't need to be an odd function, although it would be nice)

I had a linear function, but that didn't fit my needs, because most of the values were around ~ 0.5, also had a quadratic and a kubic function that didn't fit (created an equoation and solved it with wolfram alpha).

Exponential growth seems to be the solution (assuming money is normally distributed, the desired function would be a logistic curve), but I'm having problems defining
a monotonically nondecreasing function. Would be glad if you could help me out.

答:
I created a function based on root.

May $f_{money}$ be a function that depicts amount of money an agent possesses,
so that $f_{money}\left(u\right) \in \mathbb{N}, u \in \mathbb{A}$ where $\mathbb{A}$ is the amount of all agents.

Furthermore money in the system $m = \sum_{i=0}^{n-1}f_{money}\left(u_i\right) = 2500$.
and number of agents $n = 5$
Mean of money (harmonic mean)  $z =  \left(\frac{1}{n} \cdot \sum_{i=0}^{n-1} f_{money}\left(u_i\right)^{-1} \right)^{-1} = 277$.

Then wealth indicator $w(x)=a\sqrt{x} + bx$, where

$a = 3.505259*10^{-2}, b = -3.010518*10^{-4}$

Generic formula for $a$ and $b$

$
\begin{pmatrix}
\sqrt{m} & m \\
\sqrt{z} & z
\end{pmatrix} \cdot \begin{pmatrix}
a  \\
b
\end{pmatrix} = \begin{pmatrix}
1  \\
\frac{1}{2}
\end{pmatrix}$

$a = \frac{\frac{1}{2}m-z}{m\sqrt{z}-z\sqrt{m}}, b=\frac{\sqrt{z}-\frac{1}{2}\sqrt{m}}{m\sqrt{z}-z\sqrt{m}}$

$m\not=0, m\sqrt{z}-z\sqrt{m} \not=0$


问:
Horst Herrlich, Topology I: $T_{3 \frac{1}{2}}$-Reflection and continuous maps.
Given the following defintion of a $T_{3 \frac{1}{2}}$ Space:

$\underline{Z}$ is called a $T_{3 \frac{1}{2}}$ Space if for every point $z \in Z$ and every set $A \subset Z$, with $z \not \in A$
there exists a continuous map $m: \underline X \rightarrow \underline{[0,1]}$ with $m(z) = 0$ and $m[A] \subset \{1\}$.

In the chapter about $T_{3 \frac{1}{2}}$ Spaces the author defines $\underline Y$ as a s.c. $T_{3 \frac{1}{2}}$-Reflection of a space $\underline X = (X, T_x) $ like this:

$S = \{f^{-1} [B] \ |\ f : \underline X → \underline {[0,1]}  \text{ continuous map }, B \in T_{[0,1]} \}$

$\underline Y = (X, T_y), \text{where} \  T_y$ is a Topology generated by the Subbasis $S$.

If my understanding is right, then he just defines an Initial Topology on a set of all continuous maps from $\underline X$ to $\underline{[0,1]}$ and calls it $\underline Y$.

What I don't understand is the following. He proceeds by stating that it is "obvious" that for some $f$ it holds that:

$f: \underline Y \rightarrow \underline{[0,1]}$ is continuous $ \iff $ $f: \underline X \rightarrow \underline{[0,1]}$ is continuous.

While I easily get why the $\Leftarrow$ holds, I just can't understand why $\Rightarrow$ is supposed to hold.

If $f: \underline Y \rightarrow \underline{[0,1]}$ is continuous, then why can't it be that $f: \underline X \rightarrow \underline{[0,1]}$ is a discontinuous map?

答:
The identity $1_{X,Y}: \underline{X} \to \underline{Y}$ is continuous because the subbase $\mathcal{S}$ is by definition is subfamily of $\mathcal{T}_X$, so that the topology on $\underline{Y}$ is coarser than $\mathcal{T}_X$.

So if $f: \underline{Y} \to [0,1]$ is continuous, the map $f: \underline{X} \to [0,1]$ (same function) can be seen as $f \circ 1_{X,Y}$ and is continuous as a composition of continuous maps.


问:
Posterior distribution of binomial likelihood and mixture of beta and uniform prior
Let $X=X_1,X_2,...,X_n$ be a random sample from the binomial distribution such that $X_i|\theta\sim Bin(m,\theta)$ with m known. If $\theta\sim p*Unif(0,1)+(1-p)*Beta(a,b)$ with $a,b,p$ known, I'd like find the posterior $\theta|X$. I know that I have to use the Bayes theroem but I don't know how to deal with the fact that the distribution of $\theta$ is a mixture.

答:
Let $\bar{X}$ be the mean of the sample.

The posterior is a normalizing constant times

$$\left( p+(1-p) \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\theta^{a-1}(1-\theta)^{b-1} \right)\prod_{i=1}^{n}\left[ \theta^{X_i} (1-\theta)^{m-X_i} \right]$$
$$=p \theta^{n\bar{X}}(1-\theta)^{n(m-\bar{X})}+(1-p) \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\theta^{a+n\bar{X}-1}(1-\theta)^{b+n(m-\bar{X})-1}$$
$$=p
\frac{\Gamma(n\bar{X}+1)\Gamma(n(m-\bar{X})+1)}{\Gamma(n\bar{X}+1+n(m-\bar{X})+1)}
\frac{\Gamma(n\bar{X}+1+n(m-\bar{X})+1)}{\Gamma(n\bar{X}+1)\Gamma(n(m-\bar{X})+1)}
\theta^{n\bar{X}+1-1}(1-\theta)^{n(m-\bar{X})+1-1}+(1-p) \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}
\frac{\Gamma(n\bar{X}+a)\Gamma(n(m-\bar{X})+b)}{\Gamma(n\bar{X}+a+n(m-\bar{X})+b)}
\frac{\Gamma(n\bar{X}+a+n(m-\bar{X})+b)}{\Gamma(n\bar{X}+a)\Gamma(n(m-\bar{X})+b)}
\theta^{a+n\bar{X}-1}(1-\theta)^{b+n(m-\bar{X})-1}$$

This is a mixture of two $Beta$ random variables.

The first is $Beta(n\bar{X}+1,n(m-\bar{X})+1)$ and the second is $Beta(n\bar{X}+a,n(m-\bar{X})+b)$.

The mixture probabilities (weights) are proportional to
$$p
\frac{\Gamma(n\bar{X}+1)\Gamma(n(m-\bar{X})+1)}{\Gamma(n\bar{X}+1+n(m-\bar{X})+1)}
=p
\frac{\Gamma(n\bar{X}+1)\Gamma(n(m-\bar{X})+1)}{\Gamma(nm+2)}$$
and
$$(1-p) \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}
\frac{\Gamma(n\bar{X}+a)\Gamma(n(m-\bar{X})+b)}{\Gamma(n\bar{X}+a+n(m-\bar{X})+b)}
=(1-p) \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}
\frac{\Gamma(n\bar{X}+a)\Gamma(n(m-\bar{X})+b)}{\Gamma(nm+a+b)}$$

The weights have to be multiplied by a constant so they add up to 1; i.e. divide each of them by their sum to find the normalized weights.


问:
Lie algebra of abelian Lie group is compact?
Is it true that the Lie algebra of and abelian group is compact? For a compact Lie algebra I mean a Lie algebra of some compact Lie group.</p>

答:
Connected abelian Lie groups can be classified in that all of them are $\mathbb{R}^a \times (S^1)^b$ for some numbers $a$ and $b$. (If you want to prove this on your own, it helps to investigate the exponential map from the Lie algebra to the Lie group more closely.) Now, do you know the Lie algebra of this Lie group? Do you know of a compact Lie group with that same Lie algebra?

Edit: As Eric Wofsey pointed out my answer is more roundabout than is really needed. If $G$ is an abelian group, then the Lie algebra $\mathfrak{g}$ is also abelian in the sense that the commutator bracket is trivial. So your $\mathfrak{g}$ is just some $\mathbb{R}^N$ with trivial bracket and now the question reduces to finding a compact (abelian) Lie group with that particular Lie algebra. Do you see what it must be?


问:
Find mean, var and sd of the random variable T where T =4S-3H
The random variable $S$ is the score when a fair ordinary dice is rolled once, and the random variable $H$ is the number of heads obtained when two unbiased coins are tossed once. Find the mean, variance and standard deviation of the random variable $T$ where $T=4S-3H$.

This question is a question in my Math A levels probability and statistics in the topic of sum and difference of independent random variables. I understand the concepts $E(X+Y)=E(X)+E(Y)$. However I do not know how do do this questions as I am given no numbers. Any tips or advice would be much appreciated! Thank you

答:
Hint:

$$E(T)=a_1E(X_1)+a_2E(X_2)+\ldots+a_nE(x_n)$$

$$\text{Var}(T)=(a_1)^2\text{Var}(X_1)+(a_2)^2\text{Var}(X_2)+\ldots+(a_n)^2\text{Var}(X_n)$$

Treat the $a_i$'s as constants. For example, for $T=4S-3H$, $$E(T)=E(4S)-E(3H)=4E(S)-3E(H)$$

$S$ follows the Bernoulli Distribution because our experiment of rolling a dice once has only two outcomes: either you roll a certain number (success) or you don't (failure). For a Bernoulli Distribution,

$$E(S)=p=\frac{1}{6}$$

Can you determine the distribution of $H$ and find $E(H)$ from here?


问:
How can I derive the integral like this one?
I have a problem with integral like this one:

$$\int_{0}^{2\pi} \frac{\text{d}x}{\cos x+a}$$

when $a$ is from $1$ to $+\infty$.

Could anyone help me please?

答:
A common method for integrals of this form is to use the Weierstrass substitution which transforms the integral as so:
$$t=\tan(x/2)$$
$$\cos x=\frac{1-t^2}{1+t^2}$$
$$dx=\frac{2}{1+t^2}dt$$
you may also want to consider changing your domain


问:
Calculate a point's position from 4 semi-accurate distances
I've got 4 sensors with known positions in a square. Each sensor reports the distance, but not vector, between itself and an object (point P). Unfortuanately the sensors aren't perfectly accurate.

Point P can be anywhere inside or outside the square formed by the sensors.

I'm trying to calculate the best-guess position of the point P.

Diagram of sensors and point

Just to make it a bit more real - the sides of the sensor square are 2 meters and the sensor inaccuracy is between 0 and 15cm.

Thinking about it visually - it feels like it could be the geometric center of the area intersection between donuts drawn around each sensor point.

Doing it visually with donuts

The grey area is where point P could be.

答: Your intuition is almost right. Draw a circle around each of the points A, B, C, D with a radius equal to the distance from the points, if we account for the possible inaccuracy you would want two more circles for each point with a radius $\displaystyle \pm$15cm. The intersection of all the circles would tell you an area of where the position of the point might be however you can not guarantee the exact position of the point. This is a lot like how gps works and maybe understanding that from a youtube video might help you understand this.


答:
First a simple substitution
$$\int_{\Bbb{R}}\frac{\sin^4(t/2)}{t^2}\mathrm{d}t=\int_{0}^\infty \frac{\sin^4(s)}{s^2}\mathrm{d}s$$
There is a known formula
$$\int_{0}^\infty\frac{\sin(x)^m}{x^n}\mathrm{d}x$$ $$=\frac{\pi^{1-p}(-1)^{\lfloor(m-n)/2\rfloor}}{2^{m-p}(n-1)!}\sum_{k=0}^{\lfloor m/2\rfloor-p}(-1)^k \mathrm{C}(m,k)(m-2k)^{n-1}\log(m-2k)^p$$
$p=\operatorname{mod}(m-n,2)$, which holds for $m,n\in\Bbb{N}$ and $m\geq n>p$. When using this expression we make the definition $0^0:=1$. In our case, $m=4,n=2,p=0$ so
$$\int_{0}^\infty \frac{\sin^4(s)}{s^2}\mathrm{d}s=\frac{-\pi}{2^4}\sum_{k=0}^{2}(-1)^k \mathrm{C}(4,k)(4-2k)$$
$$=\frac{-\pi}{2^4}\cdot(-4)=\frac{\pi}{4}$$

Title: Evaluate integral $ \int_{\mathbb{R} } \frac{\sin^4(\frac{t}{2}) }{t^2}$

Body:
Let $f(x)= (1-2|x|)\chi_{[-\frac{1}{2}, \frac{1}{2}]} (x)  $. Its Fourier transform is given by
$ \hat{f} (x) = \frac{8\sin^2(\frac{t}{4})}{t^2} $. Based on this, I need to evaluate the integral $ \int_{\mathbb{R} } \frac{\sin^4(\frac{t}{2}) }{t^2}$, but I don't know where to start.  I tried Parseval's identity - I wrote  $\frac{\sin^4(\frac{t}{2}) }{t^2} = \frac{\sin^2(\frac{t}{2}) }{t^2} \cdot \sin^2(\frac{t}{2})$, and then I would need to find the function whose Fourier transform is $\sin^2(t)$. I don't know if this is right way (seems too complicated), so is there any simple way to evaluate integral? I would be very grateful for help.


答:
I would guess it's a typo and there was supposed to be a $t^4$ in the denominator and you were supposed to use Plancherel's theorem. But you can do the original integral as follows:

Note that the integrand is even and that $\sin^2 (t/2) = {1 - \cos t \over 2}$. So your integral is
$$2\int_0^{\infty} {(1 - \cos t)^2 \over 4t^2}\,dt$$
You can integrate this by parts, integrating ${1 \over 4t^2}$ and differentiating $(1 - \cos t)^2$ (being careful that the improper integrals converge). Your integral becomes
$$2 \int_0^{\infty} {2(1 - \cos t)\sin t \over 4t}\,dt$$
Using $\sin 2t = 2\sin t \cos t$, this is
$$\int_0^{\infty} {\sin t \over t}\,dt -  {1 \over 2} \int_0^{\infty}{\sin 2t \over t}\,dt$$
Changing variables to $2t$ in the latter integral, this becomes
$$\int_0^{\infty} {\sin t \over t}\,dt -  {1 \over 2} \int_0^{\infty}{\sin t \over t}\,dt$$
$$= {\pi \over 2} - {\pi \over 4}$$
$$= {\pi \over 4}$$


问:
The independent random variable X and Y have standard deviation 4 and 7 respectively. calculate the standard deviation of: 2X+3Y
For this question,in probability and statistics, the answer is 22.5 however I don't know how to get to it. I tried variance=2(4)^2+3(7)^2= 179 and to get standard deviation rooted 179 and got 13.37 which is wrong. Any tips or help would be great!</p>

答:
Given $V(X)$ the variance of $X$, then
$$V(2X + 3Y)=2^{2}V(X)+3^{2}V(Y)$$
assuming $X$ and $Y$ to be independent random variables. Now, if the standard deviation of $X$ is $4$ then $V(X)=4^{2}$, and we are over.


问:
Find the maximum error of the following polynomial interpolation, but the values are too small

$$f(x) = \log_{10}(x)$$

Error formula:

$$e < max[1.35;1.45] |(x-1.35)(x-1.37)(x-1.40)(x-1.45)| max[1.35;1.45] |\frac{-6/\ln(10) * 1/x^4}{4!}| = \\
max[1.35;1.45] |(x-1.35)(x-1.37)(x-1.40)(x-1.45)|*0.0327$$

The problem is that in that range, the value of $(x-1.35)(x-1.37)(x-1.40)(x-1.45)$ is so small my calculator only displays zeros.

I solved this another way by calculating the derivative and then factoring it, and using the factors on the zeroes. For factoring I used WolframAlpha.https://www.wolframalpha.com/input/?i=4%28x%5E3-4.1775x%5E2%2B5.81575x+-+2.69817%29+factor

The result I got then way $$e < 2.933*10^{-6} *0.0327 = 9.59091*10^-8$$

Is this correct?

Is there any way of solving this without WolframAlpha?

答:
For the four data points, the interpolating polynomial is

$$P_3(x) = 0.416667 x^3 - 1.85 x^2 + 3.03996 x - 1.62717$$

The formula for the error bound is given by:

$$E_n(x) = {f^{(n+1)}(\xi(x)) \over (n+1)!} \times (x-x_0)(x-x_1)...(x-x_n)$$

So, we have

$$E_3(x) = {f^{(4)}(\xi(x)) \over 4!} \times (x-1.35)(x-1.37)(x-1.4)(x-1.45)$$

The fourth derivative of $f(x)$ is

$$f^{(4)} = -\dfrac{6}{x^4 \ln(10)}$$

The error is given by

$$E_3(x) = \max_{1.35 \le x \le 1.45}\left|-\dfrac{6}{24 \ln(10) x^4}\right| \max_{1.35 \le x \le 1.45}\left| (x-1.35)(x-1.37)(x-1.4)(x-1.45)\right|$$

For the first max, it is clear that it is at the left endpoint

$$\max_{1.35 \le x \le 1.45}\left|-\dfrac{6}{24 \ln(10) x^4}\right| \approx 0.0326881 $$

We find the second max at $x\approx1.43287$ as

$$\max_{1.35 \le x \le 1.45}\left| (x-1.35)(x-1.37)(x-1.4)(x-1.45)\right| \approx 2.93358×10^{-6}$$

Note, we use calculus by finding the derivative, setting it to zero and then testing the absolute value at each critical point and the endpoints.

Thus

$$ E_3(x) \le 9.5893156398 \times 10^{-8}$$

If we plot the interpolating polynomial and points, we have

If we overlay the $P_3(x)$ and $\log_{10}(x)$ plots, we have

The error seems to match the results and this makes sense because a log plot looks linear.


答:
$\def\rk{\mathop{\mathrm{rank}}}$We have, as $\rk A = \rk A^t$ holds for every matrix, that
\begin{align*}
  \dim U &= \rk \begin{pmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{pmatrix}\\
         &= \rk \begin{pmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{pmatrix}^t\\
         &= \rk \begin{pmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3 \end{pmatrix}\\
    &= \dim W.
\end{align*}</p>

Title: How do I prove $\dim{U} = \dim{W}$ when...

Body:
$\mathbf{U} = sp\{(a_1 a_2 a_3),(b_1 b_2 b_3),(c_1 c_2 c_3)\}$

$\mathbf{W} = sp\{(a_1 b_1 c_1),(a_2 b_2 c_2),(a_3 b_3 c_3)\}$

$\mathbf{U}$ and $\mathbf{W}$ are subspaces of $\mathbb{R}^3$


问:
When change making problem has an optimal greedy solution?
A well-known Change-making problem, which asks

how can a given amount of money be made with the least number of coins of given denominations

for some sets of coins (50c, 25c, 10c, 5c, 1c) will yield an optimal solution by using a greedy algorithm (grab the highest value coin). For some other sets one have to use a dynamic programming.

Is there any way to prove whether for a given set of coins a greedy solution will always yield an optimal solution? Coin denomination can be any natural number (not only smaller then 100) and there can be any number of different coin denominations.

答: The paper by Pearson, "A Polynomial-Time Algorithm for the Change-Making Problem", Operation Research Letters 33:3 (may 2005), pp. 231-234 (an earlier technical report is here) gives the most efficient algorithm to check this to date.</p>


问:
Taylor series of $e^{-{1}/{x}}$
Does the integral converge?
\begin{align*}
\int_0^\infty (1-e^{-{1}/{x}})dx
\end{align*}

Idea: First I tried to expand the $e^{-{1}/{x}}$ using the Maclaurin series and evaluated the integral but it was not a good result and after I realized the result is not correct, because we cannot expand $e^{-{1}/{x}}$ by the Maclaurin series (because all derivatives of $e^{-{1}/{x}}$ at $x=0$ are zero but $e^{-{1}/{x}}$ is not a zero function so the approximate centered at $0$ is not work)

We can find the Taylor series expansion of $e^{-{1}/{x}}$ centered at some point (like $x=1$). According to that, the integral is divergent. I just want to know

Is my understanding correct? and

Can we expand $e^{-{1}/{x}}$ by the Taylor series centered at some point (except zero) even the function is non-analytic?

and is this the right way to evaluate this kind of integral? It will be a great pleasure to me if someone can give a little explanation.

答:
You have that $$e^{x}=1+x+\frac{x^2}{2}+o(x^3)$$

so $$1-e^{-1/x} = \frac{1}{x}-\frac{1}{2x^2}-O\left(\frac{1}{x^3}\right) < \frac{1}{x}-\frac{1}{2x^2}$$

And because $\int_1^{\infty}\left(\frac{1}{x} - \frac{1}{2x^2}\right) dx$ diverges, $\int_1^{\infty}\left(1-e^{-1/x}\right)dx$ will diverge as well.

Edit: I dropped the integral from $0$ to $1$ because the integrand is bounded by $1$, so it is convergent for that interval.

Title: Taylor series of $e^{-{1}/{x}}$


问:
Extension of the series expansion of $\ln(1+x)$ to complex numbers
You can prove, without too much trouble, that :
$$\ln(1+x) = x- \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n}$$
This is true if $|x|<1$ and $x \in \Re$ but also if $x$ is a complex number. I know how to prove this with real numbers, but how can I extend this definition to the complex plane ? Can I simply say it works the same way, by replacing the real values with complex ones ? Or are there specific things one has to pay attention to ?

I have a start of university math level, by the way.

答:
This conditions for $z$ is from this book Classical Algebra, G. Paria, page 221:

The series holds true for complex quantity $z=x+iy$ with following conditions:

I. The modulus of $z$ is less than unity, or

II. The modulus of $z$ is equal to unity but the argument of $z$ is not equal to an odd multiple of $\pi$.

By definition we have:  $\operatorname{Ln}(1+z)=2n\pi i+\ln(1+z)$, it follows that:

$$\operatorname{Ln}(1+z)=2n\pi i+z-\frac{z^2}2 +\frac{z^3}3- \cdots,$$

where $n$ is zero or an integer (positive or negative).

Note that $\operatorname{Ln}(1+z)$ is not the same as $\ln(1+z)$ which is:

$$\ln(1+z)=z-\frac{z^2}2+\frac{z^3}3 - \cdots$$

and is called principal value of $\operatorname{Ln}(1+z)$ and results in when $n=0$.


问:
Continuity of an $\mathbb {R}^2$ function
Let $f$ be an $\mathbb{R}^2$ endomorphism and $N:\mathbb{R}^2\to\mathbb{R}^+$
defined by $$\forall u \in \mathbb {R }^2, N(u) = ||f(u)|| $$

I need to show $N$ is continuous.

The problem is that $N$ is only a seminorm, otherwise it would have verified Lipschitz criterion and I'd be done.

Thanks for your help.

答:
Let's prove that a linear function is continuous. Recalling the definition:

A function $f:\Bbb{R}^n\to\Bbb{R}^k$ is continuous iff for each $x\in\Bbb{R}^n$ and for each $\epsilon>0$ there exists a $\delta>0$ such that for all $y\in\Bbb{R}^n$ with $||y-x||<\delta$ we have $||f(y)-f(x)||<\epsilon$.

Now if $f$ is linear, then it takes the form $f(x)=Ax$ for some $k\times n$ matrix $A$. Then

$$||f(y)-f(x)||=||Ay-Ax||=||A(y-x)||,$$

and we want to show that this is less than $C||y-x||$ for some constant $C$ dependent on $A$ only. This would finish the proof, because for any $\epsilon$, if we let $\delta=\epsilon/C$ we have that

$$||f(y)-f(x)||=||A(y-x)||\le C||y-x||<C\delta=\epsilon,$$

and so $f$ is continuous.

But let's back up the claim that for all $x\in\Bbb{R}$, $||Ax||\le C||x||$ for some $C$. (The smallest upper bound for $C$ here is conventionally called the induced matrix norm $||A||$.) An easy-to-prove upper bound here is the Frobenius norm, defined by $||A||_F^2=\sum_{i=1}^k\sum_{j=1}^nA_{ij}^2$.

Writing out all these norms as sums (and squaring both sides for convenience):
$$||Ax||^2=\sum_{i=1}^k\left(\sum_{j=1}^nA_{ij}x_j\right)^2$$
$$||A||_F^2||x||^2=\left(\sum_{i=1}^k\sum_{j=1}^nA_{ij}^2\right)\sum_{k=1}^nx_k^2=\sum_{i=1}^k\left(\sum_{j=1}^nA_{ij}^2\cdot\sum_{k=1}^nx_k^2\right)$$

We wish to prove that $||Ax||^2\le ||A||_F^2||x||^2$, but given the above expressions, this reduces to

$$\left(\sum_{j=1}^nA_{ij}x_j\right)^2\le\sum_{j=1}^nA_{ij}^2\cdot\sum_{k=1}^nx_k^2,$$

which is exactly the famous Cauchy–Schwarz inequality. This completes the proof.

Now that we know that linear functions are continuous, let's return to the problem. We want to know that an endomorphism of $\Bbb{R}^2$, i.e. a linear function $f:\Bbb{R}^2\to\Bbb{R}^2$, composed with the norm operation $||\cdot||:\Bbb{R}^2\to\Bbb{R}$, is continuous. If it is accepted that the norm operation is continuous and the composition of continuous functions is continuous, then the above proof is sufficient to prove that $N(x)=||f(x)||$ is continuous.

Although the above was proven in full generality of $\Bbb{R}^n\to\Bbb{R}^k$, for application to the problem we need only $\Bbb{R}^2\to\Bbb{R}^2$, and the only linear functions $\Bbb{R}^2\to\Bbb{R}^2$ are functions of the form $f(x)=Ax$ for $A$ a $2\times2$ matrix; this yields the representation $f(x)=(ax_1+bx_2,cx_1+dx_2)$. Note that these functions (choosing arbitrary constant $a,b,c,d\in\Bbb{R}$) are the only endomorphisms of $\Bbb{R}^2$.


问:
Positive-semidefiniteness of combination of two positive-semidefinite matrices
I am struggling for a while in proving the positive-semidefiniteness

( in the usual sense, i.e., $A$ is Hermitian and $x^*Ax\geq 0\,$ for all
$\,x \in \mathbb{C}^2\,\big)$

of the following matrix
$$A :=\,BC^\top + CB^\top+BB^\top,$$
under the hypotheses that both $B$ and $C$ have real entries and are positive-semidefinite in $\mathbb R^2 $: in the sense that
$$x^\top B x\geq 0\quad\forall\, x\in\mathbb{R}^2$$
and likewise for $C$. Furthermore, $B,C$ are not necessarily symmetric.

Numerical tests suggest that this holds true whenever the corresponding non-diagonal elements of $B$ and $C$ share the same sign, that is, if we set $B=\begin{bmatrix} b_1 & b_2\\b_3 & b_4\end{bmatrix}$ and $C=\begin{bmatrix} c_1 & c_2\\c_3 & c_4\end{bmatrix}$ then $sign(b_2)=sign(c_2)$ and $sign(b_3)=sign(c_3)$.

My most promising attempt has been effectively computing matrix $A$ and checking the principal minors. We have that
$$A=\begin{bmatrix}
 2(l_1u_1+l_2u_2)+l_1^2+l_2^2 & l_1u_3+l_3u_1+l_4u_2+l_2u_4+l_1l_3+l_2l_4\\
 l_1u_3+l_3u_1+l_4u_2+l_2u_4+l_1l_3+l_2l_4 & 2(l_3u_3+l_4u_4)+l_3^2+l_4^2
 \end{bmatrix},$$
and the non-negativity of the diagonal elements of $A$ comes from the shared sign of the off-diagonal elements of $B$ and $C$. It remains to ensure that $\det(A)\geq 0$. My bet is on the following computations:

\begin{equation*}
\begin{aligned}
\det(A)&=4(b_1c_1+b_2c_2(b_3c_3+b_4c_4)+2b_1^2(b_3c_3+b_4c_4)+2b_2^2(b_3c_3+b_4c_4)+2b_3^2(b_1c_1+b_2c_2)+2b_4^2(b_1c_1+b_2c_2)\\
&\quad +b_1^2b_3^2+b_1^2b_4^2+b_2^2b_3^2+b_2^2b_4^2-(b_1c_3+b_3c_1+b_4c_2+b_2c_4+b_1b_3+b_2b_4)^2\\
&=4(b_1c_1+b_2c_2)(b_3c_3+b_4c_4)+2b_1^2({b_3c_3}+b_4c_4)+2b_2^2(b_3c_3+{b_4c_4})+2b_3^2({b_1c_1}+b_2c_2)+2b_4^2(b_1c_1+{b_2c_2})\\
&\quad +{b_1^2b_3^2}+b_1^2b_4^2+b_2^2b_3^2+{b_2^2b_4^2}-b_1^2c_3^2-b_3^2c_1^2-b_4^2c_2^2-b_2^2c_4^2-{b_1^2b_3^2}-{b_2^2b_4^2}\\
&\quad-2b_1c_3b_3c_1-2b_1c_3b_4c_2-2b_1c_3b_2c_4-{2b_1^2c_3b_3}-2b_1c_3b_2b_4-2b_3c_1b_4c_2-2b_3c_1b_2c_4-{2b_3^2b_1c_1}-2b_3c_1b_2b_4\\
&\quad-2b_4c_2b_2c_4-2b_4c_2b_1b_3-{2b_4^2c_2b_2}-2b_2c_4b_1b_3-{2b_2^2c_4b_2}-2b_1b_3b_2b_4\\
&=4b_1c_1b_3c_3+4b_1c_1b_4c_4+4b_2c_2b_3c_3+4b_2c_2b_4c_4+2b_1^2b_4c_4+2b_2^2b_3c_3+2b_3^2b_2c_2+2b_4^2b_1c_1+b_1^2b_4^2+b_2^2b_3^2\\
&\quad -b_1^2c_3^2-b_3^2c_1^2-b_4^2c_2^2-b_2^2c_4^2-2b_1c_3b_3c_1-2b_1c_3b_4c_2-2b_1c_3b_2c_4-2b_1c_3b_2b_4-2b_3c_1b_4c_2-2b_3c_1b_2c_4\\
&\quad -2b_3c_1b_2b_4-2b_4c_2b_2c_4-2b_4c_2b_1b_3-2b_2c_4b_1b_3-2b_1b_3b_2b_4\\
&=2b_1c_1b_3c_3+4b_1c_1b_4c_4+4b_2c_2b_3c_3+2b_2c_2b_4c_4+\underline{2b_1^2b_4c_4+2b_2^2b_3c_3+2b_3^2b_2c_2+2b_4^2b_1c_1}+\underline{(b_1b_4-b_2b_3)^2}\\
&\quad -b_1^2c_3^2-b_3^2c_1^2-b_4^2c_2^2-b_2^2c_4^2-2b_1c_3b_4c_2-2b_1c_3b_2c_4-2b_3c_1b_4c_2-2b_3c_1b_2c_4\\
&\quad \underline{-2b_1c_3b_2b_4-2b_3c_1b_2b_4-2b_4c_2b_1b_3-2b_2c_4b_1b_3}\\
&=2b_1c_1b_3c_3+4b_1c_1b_4c_4+4b_2c_2b_3c_3+2b_2c_2b_4c_4-b_1^2c_3^2-b_3^2c_1^2-b_4^2c_2^2-b_2^2c_4^2\\
&\quad -2b_1c_3b_4c_2-2b_1c_3b_2c_4-2b_3c_1b_4c_2-2b_3c_1b_2c_4\\
&\quad +(b_1b_4-b_2b_3)(b_1b_4-b_2b_3+2b_1c_4+2c_1b_4-2b_2c_3-2c_2b_3),
\end{aligned}
\end{equation*}
where I have underlined the terms involved in the final equality in order to facilitate the reading.

And now I am stuck. I don't know if there is any obscure equality to gather the remaining terms, and can not even ensure that
$$b_1b_4-b_2b_3+2b_1c_4+2c_1b_4-2b_2c_3-2c_2b_3\geq 0,$$
although intuition tells me it should be true.

If you have any ideas for the follow up from here, I would be happy to hear them.

答: In case $B$ and $C$ are positive semidefinite in mathematicians' definition, they are symmetric and $A=X^2-Y^2$, where $X:=B+C\succeq Y:=C\succeq0$. However, it is known that $X\succeq Y\succeq0$ in general does not imply that $X^2\succeq Y^2$ (see here for instance). Therefore $A$ is not always positive semidefinite.


问:
Limit to $e^2$.
I have to show
$$
\lim_{x \to \infty} {\left(\frac{x^2 + 1}{1 - x^2}\right)}^{x^2} = e^2,
$$
but I don't get the trick to see it, I suppose I can use something like
$$
{\left(\lim_{x \to \infty} {\left(1 + \frac1x\right)}^x\right)}^2 = e^2,
$$
but I do not see how I can apply to the problem. On the other hand,
$$
\frac{x^2 + 1}{1 - x^2} = 1 + \frac{2 x^2}{1 - x^2}.
$$
Any hint?

答: Hint: divide both denominator and numerator by $x^2$ and rewrite the quotient as a product where each factor tends to $e$.

Title: Limit to $e^2$.


问:
Proof of the second isomorphism theorem for rings
I want to understand the proof my textbook provides of the second isomorphism theorem for rings.

Ler $R$ a ring, $I$ and $J$ two ideals of R such that $I \subseteq J$. Then

$(R/I)/(J/I) \simeq R/J$

This theorem seems basically a corollary of the fundamental homomorphism theorem for rings. So it reduces to proving the existence of an epimorphism whose kernel is exaclty $J$; my book choose this $\pi = \pi_2 \circ \pi_1$ from $R$ on $(R/I)/(J/I)$

$R \xrightarrow{\pi_1} R/I \xrightarrow{\pi_2} (R/I)/(J/I)$

This kernel should be

$\ker \pi = \{r \in R | \pi_1(r) \in \ker \pi_2\}$

and till this point I understand but then it continues by writing

$\ker \pi =\{ r \in R | r + I = j +I, j\in J\} = J$

I don't understand the part $\ldots r + I = j +I, j\in J\}$ and how this set is equal to $J$. I know that $r +I \in R/I$ but I don't understand how $j$ is involved.

答: The author is just writing out the definition of $\ker \pi_2$. After all, $\ker \pi_2$ is defined as those elements $r + I \in R/I$ such that $\pi_2(r + I) = 0$, which happens exactly if $r + I = j + I$ (in $R/I$) for some $j \in J$. Thus, the author's equation $\ker \pi = \{ r \in R \space|\space r + I = j + I, j \in J \}$ is true. Now see if you can show that $r + I = j + I \implies r \in J$.


问:
Understanding how we can make $M$ into an $R[X]$-module by setting $Xm = f(m)$ for $m \in M.$
Here is the question I want to answer:

Let $M$ be a finite $R$-module. Show that if $f \in \operatorname {End_R(M)}$ is surjective then it is also injective.

Hint: Let $R[X] \cong R^{[1]}$ and make $M$ into an $R[X]$-module by setting $Xm = f(m)$ for $m \in M.$ Show that $XM = M$ and apply Nakayama's Lemma.

My question is:

1- I do not understand why setting $Xm = f(m)$ for $m \in M$ will make $M$ into an $R[X]$-module. Could anyone clarify this to me, please?

2- Also, I do not see how I will use surjectivity in the proof that $XM = M$. Could any one elaborate this to me please?

答:
Let $R$ be a commutative ring with an $R$-module $M.$ Given any $R$-linear map $f : M \to M,$ consider the map $\cdot : R[X] \times M \to M$ defined by $r \cdot m = rm$ and $X \cdot m = f(m).$ By hypothesis that $f$ is $R$-linear, it follows that $X \cdot (m + m') = f(m + m') = f(m) + f(m') = X \cdot m + X \cdot m',$ hence $\cdot$ gives $M$ an $R[X]$-module structure that can be extended linearly to find $p(X) \cdot m$ for any element $p(X)$ of $R[X].$ (Check for yourself that the other properties work out.)

Given that $f : M \to M$ is surjective, it follows that for any element $m$ of $M,$ there exists an element $m'$ of $M$ such that $m = f(m') = X \cdot m'.$ Consquently, we have that $M \subseteq XM$ (here, we are suppressing the $\cdot$ notation for brevity). On the other hand, it is clear that $XM \subseteq M$: for any element $m$ of $M,$ we have that $X \cdot m = f(m)$ is in $M.$ We conclude that $XM = M.$

Last, we claim that $M$ is a finitely generated $R[X]$-module. By hypothesis that $M$ is a finitely generated $R$-module, there exist elements $m_1, \dots, m_n$ of $M$ such that every element $m$ of $M$ can be written as $m = r_1 m_1 + \cdots + r_n m_n$ for some elements $r_1, \dots, r_n$ of $R.$ By viewing the $r_1, \dots, r_n$ as constant polynomials of $R[X],$ it follows that $m = r_1 \cdot m_1 + \cdots + r_n \cdot m_n$ so that $M$ is a finitely generated $R[X]$-module. Consequently, by Nakayama's Lemma (applied to the finitely generated $R[X]$-module $M$ and the principal ideal $(X)$ of $R[X]$), there exists an element $p(X)$ of $(X)$ such that $(1_R + p(X)) M = 0.$ Can you finish the proof from here?


问:
If we have $T^2 = S^2$ then $T=S$
$T$ and $S$ are two linear self-adjoint and positive and bounded operators defined on complex Hilbert space $H$. show that if we have $T^2 = S^2$ then $T=S$.

I know that $T^2$ and $S^2$ should have the same characteristic function and eigenvectors. So should we conclude that the same thing for eigenvectors of $T$ and $S$? And then what can we do?

答: Every positive element in a $C^*$-algebra has a unique positive square root. Your assumption means that the positive operator $T^2$ has positive square roots $S$ and $T$. Hence, $S = T$.

Title: If we have $T^2 = S^2$ then $T=S$


问:
Proof that ${\left(\pi^\pi\right)}^{\pi^\pi}$ (and now $\pi^{\left(\pi^{\pi^\pi}\right)}$) is a noninteger.
Conor McBride asks for a fast proof that $$x = {\left(\pi^\pi\right)}^{\pi^\pi}$$ is not an integer.  It would be sufficient to calculate a very rough approximation, to a precision of less than $1,$ and show that $n < x < n+1$ for some integer $n$.  But $x$ is large enough ($56$ digits) that that method is not obviously practical without electronic computation tools.

Is there a purely theoretic proof that works with no calculation at all, or is there a calculational argument that would be practical prior to $1950$?

Edit: Let me phrase the question a little differently, as a spur to your mathematical imagination. I go back in time to $1847$ and have lunch with Carl Friedrich Gauss.  We are chatting about the developments in $20$th and $21$st-century mathematics, and because I am a horrible person, I mention that $(\pi^\pi)^{\pi^\pi}$ has been proven to be an integer. Gauss is incredulous, but I insist. After I leave, can Gauss convince himself that I am nothing but a troll?

Edit: After I looked back at McBride's original question, I saw to my dismay that he had not asked about $x= (\pi^\pi)^{\pi^\pi}$, but $$y=\pi^{\left(\pi^{\pi^\pi}\right)}$$ which is a rather different animal. The value of $x$ has $57$ decimal digits; the value of $y$ has $10^{57}$ digits.  So Alexander Walker's impressive display of direct numerical calculation will avail nothing for deciding that $y$ is not an integer, neither in $1873$ nor in $2013,$ and perhaps not in $2153$ either. ("We should attempt to destroy the aliens.") More powerful theoretical methods will be required.  I am posting an additional bounty to see if anyone can suggest anything of that sort; it need not be theory that was available in the $19$th century.

答:
I think this calculation would have been doable in 1873, as follows:

(1) Compute $\log \pi$ to $60$ digits.  For this, we begin with the expansion

$$\log \pi = \log \left(\frac{\pi}{22/7}\right)+\log 2 + \log 11-\log 7$$

and take full advantage of the fact that the logarithm tables of small integers were known to great precision.  As for the logarithm of $\pi/(22/7) \approx 1.00041$, the Taylor series for the logarithm spits out $60$ digits after only $17$ terms.  (Surprisingly, this exact analysis has been done on MSE before.)

(2) Compute $(\pi+1)\log \pi$ to $60$ digits.  This is no big deal, given our value for $\log \pi$ from (1), since $\pi$ was known (to Gauss, no less!) to at least $100$ digits back in 1844.  For reference, this value is

$$ \approx 4.7410048855785583722294291029994190930164741026691888020108672.$$

(The multiplication will of course be a pain, as it requires around $1800$ flops.  Nevertheless, this computation would likely be delegated to a lesser mathematician.  The Wikipedia article on Zacharias Dase (a temporary assistant to Gauss) puts these computations into perspective:

At age 15 [Zacharias Dase] began to travel extensively, giving exhibitions in Germany, Austria and England. Among his most impressive feats, he multiplied 79532853 × 93758479 in 54 seconds. He multiplied two 20-digit numbers in 6 minutes; two 40-digit numbers in 40 minutes; and two 100-digit numbers in 8 hours 45 minutes. The famous mathematician Carl Friedrich Gauss commented that someone skilled in calculation could have done the 100-digit calculation in about half that time with pencil and paper.

(3) Exponentiate the product, again to $60$ places. Using just the series
$$e^z = \sum_{k=0}^\infty \frac{z^k}{k!}$$
such a calculation would require $77$ terms.  For this reason, we instead calculate
$$e^{4.741} \approx 114.54869311801681310751748724665811195370661075419665168411647;$$
$$e^{0.0000048\cdots} \approx 1.0000048855904928305900123833767696556988185632721564706179420.$$
The latter approximation requires a mere $10$ terms of the exponential Taylor series to achieve $60$ digits, a commitment of around 18000 flops.  By Gauss's metric, we might expect a skilled mathematician to complete this task in just over seven hours.

The prior calculation could be done in one of two ways: directly, at a cost of another $18000$ flops (e.g. $77$ consecutive multiplications of a $4$-digit and a $60$-digit number); or by calculating $\mathrm{exp}(4)$ and $\mathrm{exp}(.741)$ independently (a slight time savings, I believe, even after the final multiplication).  Of course, now it just takes another $1800$ flops to multiply these numbers to $60$ places.

Note: In hindsight, it appears that calculating the triple product
$$e^4 e^{3/4}e^{-9/1000}$$
may expedite this most recent step.

In case you've lost track, we now know the value of
$$e^{(\pi +1)\log \pi}=\pi^{\pi+1}$$
to $60$ digits, all within a day or two of starting our calculations.

(4) Multiply $\log \pi$ and $\pi^{\pi+1}$ to $60$ digits.  This step is easy, given steps (1) and (3).  This value is
$$\approx 131.12795303153615589452803943707399841542170349230159549341360.$$
Of course, this value is also the logarithm of $(\pi^\pi)^{\pi^\pi}$, so it remains to:

(5) Exponentiate the term from (4).  Since it worked out so well in (3), we'll again split our exponential into a product of simpler terms.  Here, we luck out - since the binary expansion of $131.127953\ldots$ begins as
$$10000011.001000001100000\cdots_2,$$
the partial approximation $131+1/8$ is surprisingly accurate: to within $\approx 0.002953$.  The exponential of this remainder can be made accurate to over $60$ digits with a mere $18$ terms (i.e. $32000$ flops).

Secondly, we compute $e^{131}$ to $60$ digits, using iterated multiplication and an approximation of $e$ to $62$ digits ($205$ digits were known to William Shanks in 1871).  Since it suffices here to successively compute
$$e^2,e^4,e^8,e^{16},e^{32},e^{64},e^{128},$$
this step can be done in less than $15000$ flops.  Thirdly, we compute $e^{1/8}$ to $60$ digits using three applications of Newton's method for the square root (another $6000$ flops).  We find a value of
$$ \approx 887455172183124295874631455225434602688412866765466125005\color{red}{.16},$$
a non-integer.

All said and done, I would be surprised if this calculation took much longer than a week.  At the same time, I stress that this problem would have been barely doable in that era.  If twice the number of digits were required, for example, this work may very well have taken the better part of a year.


问:
What is the maximum number of cells we can select in a n x m square such that no 2 cells share a side?
Consider a square of having n rows and m columns of cells. What is the maximum number of cells we can select such that no 2 selected cells share a side?

From intuition I can say that the answer should be (nm)/2 rounded up.
I tried proving this by dividing the n x m square into 2 x 1 blocks. Now each block can have a max of 1 selected cell. Total such blocks is (nm)/2 rounded down. So we can select (nm)/2 rounded down cells.
Now one cell will remain if the total (i.e. nm ) is odd, and we can select that square as well. So now we will have (n*m)/2 rounded up selected cells.

My question is - Is the answer (i.e. (n*m)/2 rounded up ) correct . And if it is , then is my proof correct?

答: The answer is correct, but your proof only shows that $\lceil \frac{nm}{2} \rceil$ is an upper bound and does not establish that this bound can actually be reached. Of course you can easily fix the proof by showing that a checkerboard colouring is a solution that reaches that bound.


问:
Visualizing infinitesimals
I am having problems visualizing infinitesimals. Like in a curve, when we integrate we can adjust in it a perfect infinitely thin rectangle. But how do I visualize it? Every time I try to visualize it, I always notice some gap area or error in the curve. Also in differentiation, $ dx $ is supposed to an infinitely close step. But every time I draw a line, I can cut it into smaller line and that becomes $ dx $. So I am facing bit problema visualizing infinitesimals.

答: sometimes there is error/gap, however as the dx (infinitesimal) tends to 0 this error tends to 0 faster (quadratically, cubicly...). Yes you may be able to make dx smaller which is what i think you are saying by "cut it into smaller line" and thats correct because thats the whole idea of a Reimann Sum which is what approximates an integral as the dx basically becomes 0 (almost)</p>

Title: Visualizing infinitesimals


问:
Is there a connection between minimal surfaces and tree graphs?
For any connected graph $G=(V,E)$, we must have $|E|\geq|V|-1$. Any edges less, and the graph wouldn't be connected. It's sorta like saying a tree graph is a "local minimum" in the space of graphs, where height is determined by the edge count. This is akin to minimal surfaces, where such a surface is a local minimum in the variation of family of regular surfaces. Is there a connection between the two?

答:
Kind of. For this analogy, the comparison function between graphs would be the inclusion ; and the constraints would be connectivity-related properties.

There are lots of constraints you can have for different problems. If you want your tree to be below a particular graph, then your local minima below this are the spanning trees associated with that graph. If each edge has a cost, you can also find a global minimum (the MST, minimum spanning tree) with algorithms such as Kruskal's or Prim's.
If your constraint is just that a specified subset $S \in V$ of vertices is in the same connected component, then you're looking for a Steiner tree (this is a famous NP-complete problem).


问:
How to prove the exercise related to stopping time in stochastic process?
N is stopping time and $Y_n \in \mathbf{F}_n$, prove that $Y_N \in \mathbf{F}_N$.

答:
To prove $Y_N \in \mathcal F_N$, we need to show $\{ Y_N \le a \} \in \mathcal F_N$ for all $a \in \mathbb{R}$.  From the definition of $\mathcal F_N$, this means showing $\{N \le n \} \cap \{ Y_N \le a \} \in \mathcal F_n$ for all $n \in \mathbb{N}$.  We know $\{N \le n \} = \bigcup_{m=1}^n \{N = m\}$ so
\begin{align*}
\{N \le n \} \cap \{ Y_N \le a \} &= \bigcup_{m=1}^n( \{N = m\}\cap \{ Y_N \le a \}) = \bigcup_{m=1}^n( \{N = m\}\cap \{ Y_m \le a \})
\end{align*}

is the finite union of events in $\mathcal F_n$ and hence is in $\mathcal F_n$.  We conclude $\{ Y_N \le a \} \in \mathcal F_N$ and hence $Y_N \in \mathcal F_N$.


问:
Proof verification: two elements of a vector space are linearly dependent
I am inexperienced with math and need verification of my answer to the following question:

Let $v,w$ be elements of a vector space and assume that $v\neq0$. If $v,w$ are linearly dependent, show that there is a number $a$ such that $w=av$.

My answer/proof:

$w=av \implies v=a^{-1}w$. Since $v\neq0$ this implies that $w\neq0$ and $a^{-1}\neq0$, which implies that $a$ exists because $a\in\ \mathbb{F}$ has a multiplicative inverse $a^{-1}\in\mathbb{F}$. $($Where $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$.$)$

Am I missing anything?

答:
Linear dependence of $v,w$ means that there exist scalars $a,b$ NOT both zero such that

$av+bw=0$.

Suppose $a\ne 0$. Then $v=(-b/a)w$ as claimed. Both vectors are collinear.

Suppose $a=0$. Then $bw=0$ and so $w=0$.