答:
Alternative approach.

Label the dice $D_1, D_2, \cdots, D_{10}$.

Let $E_1$ denote the event represented by problem 1.

Let $E_2$ denote the event represented by problem 2.

Assume that for the purposes of the first problem, when the dice are rolled, $D_1,D_2$ are rolled together, $D_3,D_4,$ are rolled together, and so forth.

Assume that each pair of dice, re above paragraph are rolled.  $E_2$ will be deemed a success if exactly $5$ of the $10$ dice are a $1$.

Note that if $E_2$ does not occur, then $E_1$ can not occur.

Assume that $E_2$, succeeded.  Given that,  what is the probability that $E_1$ succeeded?

Without loss of generality (WLOG) $D_1$ was a $1$. Then the chance that $D_2$ was a $6$ is $(5/9)$.  Note that if $D_1$ had been a $6$, the chance that $D_2$ was a $1$ would similarly have been $(5/9)$.

Assuming that $D_1$ and $D_2$ have been navigated successfully, then WLOG, $D_3$ was a $1$ and the chance that $D_4$ was a $6$ is $(4/7)$.

Continuing to the end, you see that under the assumption that $E_2$ succeeded, the chance that $E_1$ then succeeded is

$$(5/9) \times (4/7) \times (3/5) \times (2/3).$$

Title: Probability question from Brilliant


问:
Find the value of this 3 digit number such that dividing by it leaves REM = 11
REM = Remainder.

Now , It is easy to think of doing it by dividing the numbers starting from 100 till whatever until you get remainder as 11. But isthere a way to solve it by formula.

Like , what I have doing is and not getting the right answer for it.

$\frac{100 + x}{13}$= y (for y it is Some value but we don’t know)and (100+x because it has to be a number either greater or or equal to 100)

Then , 100+x - 13(y) = 11.

I’m not getting how to solve further than this.

One way I tried is this but got wrong ans.

So , from here.

100 + x - 13(6) = 9 + x.

9+x = 9+x

But x gets cancel out.

答:
Actually , I already solved the answer in Q itself but did a silly mistake.

As you can see , I did write 9+x as remainder. So , 9+x = 11 gives x = 2. Therefore , 102 is ans.

Also , one thing to notice is that if at 13*6 = 91 we have remainder = 0.(91/13).

Then , 92/13 gives remainder 1. Therefore , 102 when we reach , we get remainder as 11.

ContentLicense: CC BY-SA 4.0


问:
Hamiltonian system invariant in the unit circle but not periodic orbits
I consider the a nonlinear system of the following:

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

I have to demonstrate that $x^2 + y^2 = 1$ is an invariant set, but not periodic orbits.

My attempt is that:

I first note that this system is Hamiltonian so that I can find a Hamiltonian function which is of the form:
$$
H(x,y) = x^2 y + y^3 - y
$$

I find that: if $(x,y) \in C = \{(x,y) \in \mathbb{R}^{2} | x^2 + y^2  = 1  \}$, then

$$
H(x,y) \equiv 0
$$

so $C$ is an invariant set.

But I am not really sure how to proceed to show that there exists no periodic orbits. Any hints will be much appreciated. Thanks in advance!.

答:
You have correctly shown that the unit circle is invariant.

To show that it contains no periodic orbits, consider the equilibrium points of the system.


问:
Question about long proofs?
How do people write 50-page long proofs (and longer)? So they have a target in mind, but I can't get my head around them foreseeing that these 50 pages of work will actually lead them to exactly their target.</p>

答:
When I hear extremely long proofs mentioned, the classification of finite simple groups comes to my mind. One often hears that the complete proof is some 15000 pages long! But note that this is not a single contiguous piece of work, but rather some 500 published articles by more than 100 mathematicians over a period of half a century. So this is an excellent example of a proof

that is unbearably long when starting from the basics

was started with a clear target in mind: to classify all finite simple groups

was split up into lots and lots of subresults

ContentLicense: CC BY-SA 3.0

Title: Question about long proofs?


问:
Consistency of exp(x) being its own derivative as a special 1-dimensional case in vector differentiation
Recently I learnt the more general definition for differentiation of a vector as:

If c is an interior point of S $\subset \mathbb{R}^n $ and $\mathbf{f}: S \rightarrow \mathbb{R}^m$, then f is differentiable at the point c if there exists a linear transformation $ \mathbf{M}: \mathbb{R}^n \rightarrow \mathbb{R}^m$ such that in some open ball centred on c we have
$$ \mathbf{f}(\mathbf{x}) = \mathbf{f}(\mathbf{c}) + \mathbf{\mathsf{M}}*(\mathbf{x-c}) + \boldsymbol{\eta}$$
where $\boldsymbol{\eta} = o(\mathbf{x-c})$ as $\mathbf{x} \rightarrow \mathbf{c}$, and  $\mathbf{\mathsf{M}}$ is the matrix dependent on c, $\mathbf{\mathsf{M}}:= \mathbf{\mathsf{Df}}(\mathbf{c})$. (I think also that $\mathbf{M}(\mathbf{x}) = \mathbf{\mathsf{M}}\mathbf{x}) $

And was introduced to linear transformations $\mathbf{L} :\mathbb{R}^n \rightarrow \mathbb{R}^m$ being their own derivative s.t. $\mathbf{\mathsf{DL}}=\mathbf{\mathsf{L}}$. I didn't find this result particularly hard to prove, but it doesn't seem to be consistent, with m=n=1, with the real analysis result that exp{x} is the unique function being its own derivative, since:

exp{x} is not a linear transformation of x, as neither exp{x+y}=exp{x}+exp{y} nor exp{ax}=a*exp{x} for real a is true. Hence there should be no relevant 1x1 $\mathbf{\mathsf{M}}$ matrix representative of the linear transformation, and so no derivative. Is that correct?

I think, by doing the algebra, that here $\mathbf{\mathsf{M}}=\exp{c}$ which would imply that $M(x)=\exp{x}=\exp{c}*x$ which is pretty clearly not true for most x.

The function $f(x)=ax$ is a linear transformation in $\mathbb{R}$ which would imply the uniqueness is not true. Here $\mathbf{\mathsf{M}}=a$.

My linear algebra is really rusty so I'm sure some of the things above are wrong, the application of linear transformations especially. I'd appreciate if someone could explain how exactly the exponential is consistent with the linear transformation derivative property, and also possibly give some remarks on the accuracy of my thoughts.

Thanks for your help

答:
The result is still true for $n=m=1$. There is a difference between derivatives and derivative functions. Linear maps are the only functions which are their own derivative. The exponential function is the only function which is its own derivative function.

The derivative is just a linear map. The derivative function is a function mapping each point to a linear map. In the case where $n=m=1$, we identify linear maps with numbers. The linear map $x\mapsto ax$ is identified with the number $a$. And lo! the derivative of this map is $a$ at all points, which is exactly the number with which we identified the map.

For $n\neq1$, it doesn't make sense to talk about a function being its own derivative function. The derivative is a linear map $\mathbb R^n\to\mathbb R^m$, and the derivative function is then a function $S\to\operatorname{Lin}(\mathbb R^n,\mathbb R^m)$ into the space of linear maps. Under certain conditions ($n=1$, for instance), this space can be identified with $\mathbb R^m$, and then we can say that a function is its own derivative function. But that's a different thing from saying that it is its own derivative.


问:
Selecting From a Question Pool (Combination Question)

Alice is taking her probability exam tomorrow and she will be assigned  5 questions out of a pool
consisting of  13 questions. In particular, the pool includes  5 easy, 2 medium, and 6 difficult
questions.

If 5 questions are selected at random at the beginning of the exam, find the probability that Alice
will be assigned: both of the medium questions and at least 1 difficult question.

I've been trying hard to find a solution to this question but I couldn't. Here is what I did,

(nCr(2,2) * nCr(6,1))/nCr(13, 5) + (nCr(2,2) * nCr(6,2))/nCr(13, 5) + ... + (nCr(2,2) *
nCr(6,6))/nCr(13, 5) = 0.04895

But it turned out to be wrong, what am I doing wrong? Any help would be appreciated!

答:
Total number of unrestricted selections of $5$ questions $ = \displaystyle {13 \choose 5} = 1287$

Total number of possible selections where both medium questions are selected and at least one difficult question is selected
$\displaystyle = {11 \choose 3} - {5 \choose 3} = 155$

So the desired probability $ = \displaystyle \frac{155}{1287}$.

In your working, you also need to select remaining questions from easy. So it should be $\displaystyle \small \bigg[{6 \choose 1}{5 \choose 2} + {6 \choose 2} {5 \choose 1} + {6 \choose 3}\bigg] \bigg/ {13 \choose 5}$


答:
We have:

$N\equiv 11\bmod 13\equiv -2 \bmod 13$

Now we consider the least odd prime of 2 greater than 100 which is 7 and we can write:

$2^7=128=117+11=7\times 13+11+2\times 13$

$26=2\times 13$ can be ignored, so the smallest 3 digit number can be $7\times 13+11=102$.

Title: Find the value of this 3 digit number such that dividing by it leaves REM = 11

Body:
REM = Remainder.

Now , It is easy to think of doing it by dividing the numbers starting from 100 till whatever until you get remainder as 11. But isthere a way to solve it by formula.

Like , what I have doing is and not getting the right answer for it.

$\frac{100 + x}{13}$= y (for y it is Some value but we don’t know)and (100+x because it has to be a number either greater or or equal to 100)

Then , 100+x - 13(y) = 11.

I’m not getting how to solve further than this.

One way I tried is this but got wrong ans.

So , from here.

100 + x - 13(6) = 9 + x.

9+x = 9+x

But x gets cancel out.


问:
Measure Space without any form of metric
I'm with a difficult question that I was thinking . Any help will be useful

We know that the Urysohn's Theorem states that: every Hausdorff second-countable regular space is metrizable.

There are other metrization theorems, but this is the most famous and easy example that I remember.

I'm trying to make a Measure Space in a non-metrizable space. I think in use Urysohn's theorem, but the problem with this theorem is that the converse is not true. There are examples of non-second countable topological spaces that are metrizable.

So I don't know how to proceed, there are others metrizable theorems and colloraries of the Urysohn, but I can't tink in any form that how can I make progress with this theorems

答:
Trivial remark:

If we have a finite or countable measure space $X$ with $\sigma$-algebra $\mathcal{M}$, then this can only be the Borel-algebra of a metrisable topology on $X$ iff $\mathcal{M}=\mathscr{P}(X)$.


问:
Why is the power series of $\tan(x)$ not convergent everywhere $\cos(x)$ is non-zero?
The Taylor series expansion of $\tan(x)$ centered at $0$ has a radius of convergence of $\pi/2$, which means the power series will not converge for $|x|>\pi/2$.

How can this be when you consider $\tan(x)=\sin(x)/\cos(x)$, so Taylor series of $\tan(x)$ is just the Taylor series of $\sin(x)$ divided by the power series of $\cos(x)$, both of which converge everywhere.

At some point $|x|>\pi/2$ where $\cos(x)$ is not equal to $0$, how can the power series of $\tan(x)$ diverge? It is simply the quotient of two convergent series at that point which seems to me like it shouldn't have any problems.

答:
The reason is that, if a function $f:\Bbb C\to \Bbb C$ has a power-series representation $$f(z)=\sum_{k=0}^\infty a_kx^k$$ that fails to converge at a point $z=c\in\Bbb C$, then it cannot converge at any point $z$ such that $|z|>|c|$. This is because, roughly speaking, if the terms $a_nc^n$ do not tend to zero fast enough for the series to converge as $n\to\infty$, then $a_nz^n$ will tend to zero even more slowly when $|z|>|c|$, or not even tend to zero at all, and so a fortiori will prevent the series from converging.

In the case of the tangent function, because it blows up at $\frac12\pi$, the Taylor series $\tan z=\sum_{k=0}^\infty a_kx^k$ for it will not converge at $z=\frac12\pi$ and so will not converge outside the circle $|z|=\frac12\pi$.

The borderline case is when $|z|$ equals the radius of convergence. In some instances, the series will converge (conditionally) at certain points. For example, $$\ln(1-z)=-\sum_{k=1}^\infty\frac1kz^k$$still converges at $z=-1$.


问:
Is the Riemann sphere really just the entire complex plane where you add a "single point" at infinity?
It's well known that the Riemann Sphere is defined as the complex plane plus a point at infinity.

Question: Is the point at infinity really a single point, or is it a circle that has dimension 0?

If we make the replacement $z \to \frac{1}{z}$ then we can say that the origin represents the point at infinity.  But this is really not accurate, since the origin under this mapping got deleted, and it's necessary to treat every possible angle of approach as a distinct path.  This corresponds to the fact that starting from any finite part of the complex plane, there are an infinite number of ways to move in a straight line off to infinity. [1]

The problem of treating infinity as a single point becomes apparent when we take some polynomial which is well defined everywhere in the entire complex plane, and only has a single singularity at infinity, and make the transformation

$$z \to \frac{1}{z}$$

Then what happens is our polynomial just got turned into a rational function that has poles which are all over the complex plane, and whereas the polynomial can approach infinity from any direction and it's only singularity is at the point at infinity, when we take $z \to \frac{1}{z}$ the singularity at infinity disappears and is replaced by poles which are not all at the same point.

This makes perfect sense (to me) when the point at infinity is a circle with dimension 0.  Although the concept of "a circle with dimension 0" is something I just made up, and which is not well defined, my intuition tells me that there is a topological distinction between a single point and a circle with dimension 0.  Is there any such concept as a circle with dimension 0, or is that a bad way of thinking about this?

I want a circle with dimension 0 to be the point of tangency between two spheres that only touch at a single point.  In other words, its not just a single point because it has an extra dimension to play around with, but the intersection just looks like a single point, if that makes sense.

This issue about the difference between the Riemann sphere and the standard projective geometry approach has been bothering me for over 10 years, and any help clarifying my confusion here is very much appreciated.

Thanks!

[1] Modern Anaylysis, Whittaker & Watson, 4th edition, 5.62 The "point at infinity"

答:
I have never really thought of the Riemann Sphere as anything more than a abstract tool. It helps us better understand why particular complex functions behave the way they do. Take for example, LFTs (linear fractional transformations). Now it is a well known theorem that LFTs map "circles and lines" to "circles and lines". However this feels cumbersome and there is a certain beauty lost in trying to distinguish lines from circles, or in attempting to figure which lines go to which circles or which lines and visa-versa. However, a particular clarity is found when LFTs are placed on the Riemann Sphere; and we suddenly realize that the statement was in error, really LFTs map circles to circles, that's it. There are no lines on the Riemann Sphere; lines in the complex plane are simply the special case of circles on the Riemann Sphere that happen to go through $\infty$. Now doesn't that sound nice to say, "LFTs map circles to circles".

This kind of abstract extension of a space "to complete it" is done all the time. A similar example would be Elliptic Curve Geometry. Without going into the nuances, assume for a moment that their exist curves (symmetric about the x-axis) with the property that given any two points on the curve and the tangent line between them, then there is guaranteed to exist a third point on the curve that passes through the tangent line. We can then define an algebraic "addition" on the curve, by $A+B=C$ if $C$ is the reflective point across the x-axis of the third point that intersects the tangent line between $A$ and $B$. This works everywhere, except one small caveat. What happens if $A$ and $B$ are exactly vertical from each other, then algebraically they are apparently $A$ and $-A$, opposites of each other. But there is no third point that passes through a vertical line on an Elliptic Curve. So how do we resolve that in our mind? The geometry shows there is no third point, yet logically, $A+(-A)$ should be $0$. The solution is that $0$ is really the point at $\infty$, and so on an Elliptic Curve, the abstraction that $A+\infty=\infty+A = A$ for any $A$, makes sense geometrically; thus $\infty$ is the zero of the abstract group.


问:
Flux of $v=(y_1e^{y_2}-e^{y_1},y_2e^{y_3}-e^{y_2},y_3e^{y_1}-e^{y_3})$ through a sphere

$v=(y_1e^{y_2}-e^{y_1},y_2e^{y_3}-e^{y_2},y_3e^{y_1}-e^{y_3})$

$Y = (y_1^2+y_2^2+y_3^2=r^2, y_2>=0, y_3>=0)$

Calculate flux of v through Y.

My first intuition was to use spherical coordinates so I set $\varphi(x)=(r\sin x_2\cos x_1, r\sin x_2 \sin x_1, r\cos x_2)$, with $x_1$ between $0$ and $\pi$ and $x_2$ between $0$ and $\pi/2$.

Then I wanted to use the Stokes-Ampère theorem, but I don't really know how to use it. I saw that div v = $0$ and found $\omega_v^2$. Then I found $\omega_u^1$ with $u=(0,y_3e^{y_1}-y_1e^{y_3},y_1y_2e^{y_3}-y_1e^{y_2})$.

But what now? How do I set the integral? In class we saw an example with polar coordinates and a simple u and it seemed easy with just one variable. But here I have spherical coordinates and the vector u is very confusing.

Could someone please explain how to use this theorem correctly and how do I integrate from here?

答:
I just got the correction back and this is how we were supposed to do the exercise. Note that we only saw the divergence theorem this week, so we were supposed to solve this without using it.

So, as I wrote before, the divergence of $v$ is $0$.

Since the divergence of $v$ is $0$, there exists a $u$ such that rot$u=v$. And, by Stokes-Ampère, $\Phi=\int_{\partial Y}\omega_u^1$.
We're also going to be using Green-Riemann: $\int_{\varphi (K)}d\omega = \int_{\varphi (\partial K)}\omega$

So we have:

$\omega_v^2 = (y_3e^{y_1}-e^{y_3})dy_1\wedge dy_2 - (y_2e^{y_3}-e^{y_2})dy_1\wedge dy_3+(y_1e^{y_2}-e^{y_1})dy_2\wedge dy_3$ = $d(y_3e^{y_1}dy_2)+d(y_1e^{y_2}dy_3)+d(y_2e^{y_3}dy_1)$

And so $\omega_u^1=y_2e^{y_3}dy_1+y_3e^{y_1}dy_2+y_1e^{y_2}dy_3$

We need to calculate $\Phi=\int_{\partial Y}\omega_u^1$.

How do we parametrise $\partial Y$?

$\gamma_1 = r(-\cos t, \sin t,0)$, $\gamma_2 = r(\cos t, 0, \sin t)$, $t=(0,\pi)$

And so:

$\int_{\partial Y}\omega_u^1$ = $\int_{\gamma_1}\omega_u^1 + \int_{\gamma_2}\omega_u^1$=$r^2\int_0^{\pi}((\sin t) e^0 (\sin t) +0 +0)dt+r^2\int_0^{\pi}(0+0+(\cos t) e^0 (\cos t) dt$ = $r^2\int_0^{\pi}(\cos^2 t +\sin^2 t)dt = \pi r^2$


问:
Show that a set $A$ has a fundamental neighborhood iff its basin of attraction is a neighborhood of $A$
Let $\Omega$ be a topological space, $\tau:\Omega\to\Omega$, $A\subseteq\Omega$, $\mathcal N(A):=\{N\subseteq\Omega:N\text{ is a neighborhood of }A\}$ and $$B(A):=\left\{x\in\Omega\mid\forall N\in\mathcal N(A):\exists n_0\in\mathbb N:\forall n\ge n_0:\tau^n(x)\in N\right\}.$$

$U\in\mathcal N(A)$ is called fundamental neighborhood of $A$ if $$\forall N\in\mathcal N(A):\exists n_0\in\mathbb N:\forall n\ge n_0:\tau^n(U)\subseteq V\tag1.$$

It is easy to show that if $U$ is a fundamental neighborhood of $A$, then $$U\subseteq B(A)\tag2$$ and hence $$B(A)\in\mathcal N(A).$$ Are we able to show the converse, i.e. does $B(A)\in\mathcal N(A)$ imply that $A$ has a fundamental neighborhood?

EDIT

I'm not sure if it is relevant or not, but please note that in the other direction (i.e. the direction where we assume that $A$ has a fundamental neighborhood $U$) we can show more than $(2)$, namely $$B(A)=\bigcup_{n\in\mathbb N_0}\tau^{-n}(U).\tag3$$

"$\subseteq$": Let $x\in B(A)$. Since $U\in\mathcal N(A)$, there is a $n_0\in\mathbb N$ with $\tau^n(x)\in U$ for all $n\ge n_0$ and hence it even holds $$x\in\bigcap_{n\ge n_0}\tau^{-n}(U)\tag4.$$

"$\supseteq$": Let $x$ belong to the right-hand side of $(3)$. Then there is a $n_1\in\mathbb N_0$ with $\tau^{n_1}(x)\in U$. Let $N\in\mathcal N(A)$. By $(1)$, there is a $n_2\in\mathbb N$ with $\tau^n(U)\subseteq N$ for all $n\ge n_2$. Let $n_0:=n_1+n_2$. Then it holds $$\tau^n(x)=\tau^{\overbrace{n-n_1}^{\ge\:n_2}}(\underbrace{\tau^{n_1}(x)}_{\in\:U})\;\;\;\text{for all }n\ge n_0\tag5.$$

答: You need stronger assumptions on the attractor. Take e.g. a circle homeomorphism that fixes precisely one point $p$. All other points then moves, say to the 'right'. Then $A=\{p\}$  is an attractor (with $B(A)=S^1$) but has no neighborhood which contracts uniformly towards $A$. The reason is that if $x_k$ converges to $p$ from the right and $N$ is a small neighborhood of $p$ then the time $n_k=n_0(x_k,N)$ it takes for $\tau^n(x_k)\in N$ for all $n\geq n_k$  will diverge with $k$.


问:
How do I know if this limit even converges?
So, there is this question which I found quite new according to my experience, and it is as follows.

Assuming that interchange of limit and integration is permissible, evaluate the following: $$\lim_{n \to \infty} {\int_0^1 \frac{nx^{n -1}}{1 + x}dx}, \quad 0 < x < 1$$

Now, I first tried integration by parts, and it doesn't give me more than an expression which is also nearly impossible to evaluate (maybe I'm doing it wrong, though) -

$$\lim_{n \to \infty} \left|\frac{1}{1+x}\int nx^{n - 1}dx  +  \int\frac{x^n}{(1 + x)^2}dx\right|_0^1$$

I'm not sure how to proceed from there, so I tried this instead:

$$\lim_{n \to \infty} {\int_0^1 {nx^{n - 1}(1 + x)^{-1}}dx}$$

Expanding $(1 + x)^{-1}$ using Taylor Series:

$$\lim_{n \to \infty} {\int_0^1 {nx^{n - 1}(1 - x + x^2 - x^3 + x^4 - x^5 \ldots)}dx}$$
$$ = \lim_{n \to \infty} {\int_0^1 {(nx^{n - 1} - nx^n + nx^{n + 1} - nx^{n + 2} + nx^{n + 3} - n^{x + 4} \ldots)}dx}$$
$$\lim_{n \to \infty} \left(1 - \frac{n}{n+1} + \frac{n}{n+2} - \frac{n}{n +3} + \frac{n}{n +4} - \frac{n}{n + 5}\ldots\right)$$

I think that diverges. Then, I tried putting the expression in Desmos and increased the $n$ value slowly upto $500$, $1000$ and similar big numbers - which all returned values around $0.5$.

The question was an multiple-choice based and there were both of them listed as options.

I'm not sure who is correct, the calculator, or me.

It would be a huge help if someone looks into this. Thank you in advance.

答:
Here's another method in the same spirit as a couple comments.  We trade off the use of integration by parts in favor of just using monotonicity of the integral.

Fix any $y$ such that $0 < y < 1$ and split the integral into two as
$$\int_0^1 \frac{nx^{n-1}}{1+x}\,dx = \int_0^y \frac{nx^{n-1}}{1+x}\,dx + \int_y^1\frac{nx^{n-1}}{1+x}\,dx$$

Then estimate each integral independently:

On $0 \leq x \leq y < 1$, we have $0 \leq \frac{nx^{n-1}}{1+x} \leq nx^{n-1}$, so $$0 \leq \int_0^y \frac{nx^{n-1}}{1+x}\,dx \leq \int_0^y nx^{n-1}\,dx = y^n$$ and since $y^n \to 0$ we have $$\lim_{n\to\infty} \int_0^y \frac{nx^{n-1}}{1+x}\,dx = 0$$

On $0 < y \leq x \leq 1$, we have $\frac{nx^{n-1}}{2} \leq \frac{nx^{n-1}}{1+x} \leq \frac{nx^{n-1}}{1+y}$, so $$\frac{1}{2}(1 - y^n) = \int_y^1\frac{nx^{n-1}}{2}\,dx \leq \int_y^1\frac{nx^{n-1}}{1+x}\,dx \leq \int_y^1\frac{nx^{n-1}}{1+y}\,dx = \frac{1 - y^n}{1+y}.$$  As once again, $y^n \to 0$, we have $$\frac{1}{2} \leq \liminf_{n\to\infty} \int_y^1 \frac{nx^{n-1}}{1+x}\,dx \leq \limsup_{n\to\infty} \int_y^1 \frac{nx^{n-1}}{1+x}\,dx \leq \frac{1}{1+y}$$

Putting it back in terms of the original integral, we've shown $$\frac{1}{2} \leq \liminf_{n\to\infty} \int_0^1 \frac{nx^{n-1}}{1+x}\,dx \leq \limsup_{n\to\infty} \int_0^1 \frac{nx^{n-1}}{1+x}\,dx \leq \frac{1}{1+y}$$
which now only involves $0 < y < 1$ on the upper bound, and letting $y \to 1^{-}$ shows $$\lim_{n\to\infty} \int_0^1 \frac{nx^{n-1}}{1+x}\,dx = \frac{1}{2}$$

The other reason I mention this method is that it generalizes nicely.  We've used monotonicity of $\frac{1}{1+x}$ to keep everything clean, but even this is actually unnecessary.  The same argument may be made with any function $f$ which is continuous on $[0,1]$ (or even just $f \in L^{\infty}([0,1])$ which is left-continuous at $1$) to give $\int_0^1 nx^{n-1}f(x)\,dx \to f(1)$.


问:
Where do $p$-adic numbers and $p$-Sylow theory both appear?
Both $p$-adic numbers and $p$-Sylow theory are by design "arithmetic" ways of "localizing," so it stands to reason they might be in cahoots in certain contexts. Are they? </p>

答:
In the context of abelian torsion groups, Sylow subgroups correspond to the summands occurring in the torsion decomposition as per structure theorem. In particular, as additive groups,

$$\frac{\bf Q}{\bf Z}\cong \bigoplus_p \frac{{\bf Q}_p}{{\bf Z}_p}.$$

The Prufer $p$-groups ${\bf Z}(p^\infty)\cong{\bf Q}_p/{\bf Z}_p$ and the $p$-adic integers ${\bf Z}_p$ are Pontryagin duals of each other. This decomposition is the number field analogue of partial fraction decomposition present in function fields, which I elaborate over at this answer, illustrating this is a global phenomenon.

The connection between nilpotence and Sylow subgroups allows zeta function techniques to study  finitely-generated torsion-free nilpotent (i.e. $\frak T$-)groups, in particular lattice-of-subgroup structure and asymptotic analysis of subgroup growth, via local Euler products. The $p$-adic numbers come into play when local zeta integrals are used. See image, source being the first reference below.

Zeta Functions of Groups and Rings - du Sautoy & Woodward

Lectures on Profinite Topics in Group Theory - Klopsch, Nikolov & Voll

Here is a connection I have not seen mentioned anywhere. Let $W_{p^n}$ be the wreath power $$W_{p^n}=\underbrace{C_p\wr C_p\wr\cdots\wr C_p}_n.$$

The Sylow $p$-subgroups of $S_n$, where $n=a_kp^k+\cdots+a_1p+a_0$ is $n$'s $p$-adic expansion, are

$$a_kW_{p^k}\oplus\cdots\oplus a_1 W_{p}\oplus a_0 W_1,$$

up to isomorphism, where $W_1$ is the trivial group and $$mG:=\underbrace{G\oplus G\oplus\cdots\oplus G}_m.$$ This can be proven by computing the order of the above subgroup in $S_n$. Note that $W_{p^n}$ is also the automorphism group of the $p$-ary rooted tree of depth $n$. Incidentally, the topology on ${\bf Z}_p$ is that of an infinite-depth rooted $p$-ary tree (see Pictures of Ultrametric Spaces by Holly).

As an aside, in Tree Representations of Galois Groups, Boston conjecturally notes that $p$-adic Galois representations into matrix groups convey very little information whereas $p$-adic Galois reps into tree automorphism groups should convey a lot, according to Hausdorff dimension.

Since isometries of ${\bf Z}_p$ correspond to automorphisms of the underlying tree topology, the isometry group is given by ${\sf Iso}({\bf Z}_p)\cong\varprojlim W_{p^n}$. The torsion subset is a subgroup isomorphic to $\varinjlim W_{p^n}$. The Sylow $p$-subgroup of ${\sf Sym}({\bf Z})\ge\varinjlim S_n$ is also seen to be $\varinjlim W_{p^n}$. Its topological closure (in terms of pointwise convergence of functions) is isomorphic to $\varprojlim W_{p^n}$. (See direct / inverse limits.) (At least I'm pretty sure about that last claim. Still trying to pin down my intuition into a proof.)


问:
Every day when you open the shop, there is an $\text{Exp}(\frac{1}{10})$ random waiting time, in minutes, before the first customer arrives.
You own an ice-cream shop. Every day when you open the shop, there is an $\text{Exp}(\frac{1}{10})$ random waiting time, in minutes, before the first customer arrives.

When the waiting time for the first customer exceeds 20 minutes, it is a bad day. Approximate the probability that you will have between 50 and 60 bad days in a year (365 days).

When the waiting time for the first customer is shorter than 3 seconds (i.e., 0.05 minutes), it is a great day. Approximate the probability that you will have exactly 3 great days in a year (365 days).

My Attempt

The probability of waiting over 20 minutes is
$$\mathbb{P}(X\geq20)=\int^\infty_{20}0.1e^{-0.1x}dx=[e^{-0.1x}]^\infty_{20}=e^{-2}\approx0.135$$
Then since $np(1-p)=365(0.135)(1-0.135)=42.7>10$, we can use the normal distribution to apprimate the probability. The expectation for an exponential distribution is $\frac{1}{\lambda}=\frac{1}{1/10}=10$. Thus
$$\mathbb{P}(50\leq x\leq60)=\mathbb{P}(a\leq\frac{x-\mathbb{E}[x]}{\sqrt{\text{Var}(x)}}\leq b)$$
$$=\mathbb{P}(\frac{50-10}{\sqrt{\text{Var}(x)}}\leq\frac{x-\mathbb{E}[X]}{\sqrt{\text{Var}(x)}}\leq\frac{60-10}{\sqrt{\text{Var}(x)}})\approx \Phi(\frac{50}{\sqrt{\text{Var}(x)}})-\Phi(\frac{40}{\sqrt{\text{Var}(x)}})$$
The part that I am having trouble with is how to find the variance for the exponential distribution. Can I get some pointers?

Edit
I think I know how to find the variance now. I got Var[X]=E[$X^2$]-E[X]=200-10=190.
So
$$\Phi(\frac{50}{\sqrt{190}})-\Phi(\frac{40}{\sqrt{190}})\approx0.00176$$

My concern now is that this probability is way too small. I don't think that it is reasonable. Can I get pointers on where I went wrong? Thank you

For this one, since 3 seconds is so small, we can consider this event rare, so we can use poisson approximation. First we find the probability of great day
$$\mathbb{P}(X\leq0.05)=1-[e^{-0.1}x]^\infty_{0.05}\approx0.00498$$
We know that the expectation is equal to the parameter, $\lambda$, of the poisson distribution. Then we can find use the formula
$$e^{-\lambda}\frac{\lambda^k}{k!}$$

Is this the right approach?

答:
The first is wrong.

After calculating $P(X>20)=e^{-2}$ your annual distribution is a binomial $Y\sim B(365;e^{-2})$ with mean $\mu=365\cdot e^{-2}$ and variance $\sigma^2=365\cdot e^{-2}\cdot(1-e^{-2})$

Yes, you can approximate is as a gaussian but the correct formula is

$$P(50\leq Y\leq 60)=\Phi\left(\frac{60.5-\mu}{\sigma}\right)-\Phi\left(\frac{49.5-\mu}{\sigma}\right)=\Phi(1.699)-\Phi(0.016)\approx 0.449$$

The second can be exactly calculated with the binomial

$$P(Y=3)=\binom{365}{3}\left[1-e^{-1/200}\right]^3\cdot\left[e^{-1/200}\right]^{362}\approx 0.1632$$

If you want to approximate this result with the poisson you can. Observe that the probability to have a great day is $e^{-1/200}$ which means that you expect to have

$$365\times e^{-1/200}\approx 1.82$$

great days per year.

Thus the probability to have exactly 3 great days in a year is

$$P(Y=3)=\frac{e^{-1.82}\cdot 1.82^3}{3!}\approx 0.1628$$

which is very close to the exact result obtained with the binomial


问:
Let $f(x) = x^4 + 2x^2 - 6$ and let E be the splitting field of $f(x)$ over $\mathbb{Q}$.
Let $f(x) = x^4 + 2x^2 - 6$ and let E be the splitting field of $f(x)$ over $\mathbb{Q}$.

(a) Find roots $\alpha$ and $\beta$ of $f(x)$ such that $E = \mathbb{Q}(\alpha, \beta)$. What are the other roots of $f(x)$ ?

(b) Compute $[\mathbb{Q}((\alpha + \beta)^2) : \mathbb{Q}]$.

(c) Compute $[\mathbb{Q}(\alpha + \beta) : \mathbb{Q}]$.

$\textbf{My Attempt:}$

(a) Notice that $f(x) = x^4 + 2x^2 - 6 = (x-\sqrt{-1+\sqrt{7}})(x+\sqrt{-1+\sqrt{7}})(x-\sqrt{-1-\sqrt{7}})(x+\sqrt{-1-\sqrt{7}})$.

Then, Case 1: $\alpha = \sqrt{-1+\sqrt{7}}$ and $\beta = \sqrt{-1-\sqrt{7}}$;

or Case 2: $\alpha = -\sqrt{-1+\sqrt{7}}$ and $\beta = -\sqrt{-1-\sqrt{7}}$.

Which in both of the cases $\alpha$ cannot be represent in terms of $\beta$ and $\beta$ cannot be represent in terms of $\alpha$.

So, $E = \mathbb{Q}(\sqrt{-1+\sqrt{7}}, \sqrt{-1-\sqrt{7}}) = \mathbb{Q}(-\sqrt{-1+\sqrt{7}}, -\sqrt{-1-\sqrt{7}})$. $\textbf{But how do I prove this is true ?}$

(b) for case 1: if $\alpha = \sqrt{-1+\sqrt{7}}$ and $\beta = \sqrt{-1-\sqrt{7}}$

Then, $(\alpha + \beta)^2 = (\sqrt{-1+\sqrt{7}} + \sqrt{-1-\sqrt{7}})^2 = -2 + 2i\sqrt{6}$.

Notice that $-2$ and $2$ are in $\mathbb{Q}$. So, $\mathbb{Q}((\alpha + \beta)^2) = \mathbb{Q}(-2 + 2i\sqrt{6}) = \mathbb{Q}(i\sqrt{6})$.

Let $c = i\sqrt{6} \implies c^2 = -6 \implies c^2 + 6 = 0$.

Also, by Eisenstein’s criterion we know that $x^2 + 6$ irreducible over $\mathbb{Q}$. And $x^2 + 6$ is monic polynomial.

So, the minimal polynomial of $c$ over $\mathbb{Q}$ is $m_{c, \mathbb{Q}}(x) = x^2 + 6$.

Hence, $[\mathbb{Q}((\alpha + \beta)^2) : \mathbb{Q}] = [\mathbb{Q}(i\sqrt{6}) : \mathbb{Q}] = $ degree of $m_{c, \mathbb{Q}}(x)$ which is equal to $2$.

(c) Since, $\mathbb{Q}(\alpha + \beta) = \mathbb{Q}(\alpha, \beta)$. $\textbf{But how do I prove this is true ?}$

Then, $[\mathbb{Q}(\alpha + \beta) : \mathbb{Q}] = [\mathbb{Q}(\alpha,\beta) : \mathbb{Q}] = [\mathbb{Q}(\alpha,\beta) : \mathbb{Q}(\alpha)] \times [\mathbb{Q}(\alpha) : \mathbb{Q}]$

$= [\mathbb{Q}(\alpha,\beta) : \mathbb{Q}(\beta)] \times [\mathbb{Q}(\beta) : \mathbb{Q}] = 4 \times 4 = 8$.

By Tower Law.

$\textbf{Are these correct ?}$

$\textbf{Also, how do I prove that}$ $E = \mathbb{Q}(\sqrt{-1+\sqrt{7}}, \sqrt{-1-\sqrt{7}}) = \mathbb{Q}(-\sqrt{-1+\sqrt{7}}, -\sqrt{-1-\sqrt{7}})$

$\textbf{and}$ $\mathbb{Q}(\alpha + \beta) = \mathbb{Q}(\alpha, \beta)$ $\textbf{are true ?}$

答:
Let me abbreviate your four roots to $r,s,t,u$. Note $s=-r$ and $u=-t$, so any field containing $r$ automatically contains $s$, and any field containing $t$ automatically contains $u$. So, extending the rationals by $r$ and $t$ gives you all four of the roots and hence gives you the splitting field $E$.

Now, let $x=r+t$. You have shown $x^2=-2+2i\sqrt6$. So, $(x^2+2)^2=-24$, so $x^4+4x^2+28=0$. If you can show that this polynomial is irreducible over the rationals, then you have proved that $x$ has degree four over the rationals, which answers (c).


问:
How Could I calculate Probability Of Probability Between 2 Price
In x Axis, I have 2 Price that have independent probability of occur.

1- First Price 50: Probability Of Price become Above that is is 40% and below that is 60%.

2- Second Price 110: Probability Of Price become Above that is is 20% and below that is 80%.

How Could I Calculate Probability of price these price toghether:

below 50, Between 50 and 110 and above 110?

答:
Welcome to MSE.

You should use the fact that for every event $A$, the complementary event $A^C$ has probability $$\mathbb{P}(A^C)=1-\mathbb{P}(A) . $$

Noting that "price below 50" is the complementary of "price above 50", we get
$$\mathbb{P}(\textrm{"price below 50"}) = 1-\mathbb{P}(\textrm{"price above 50\"}) = 1-0.4 = 0.6 $$

In the same way, since "prince above 110" is the complementary of "price below 110", we have
$$\mathbb{P}(\textrm{"price above 110"}) = 1-\mathbb{P}(\textrm{"price below 110"}) = 1-0.8 = 0.2 $$

Finally, the probability of all space must be 1, so
$$ \mathbb{P}(\textrm{"price below 50"})+\mathbb{P}(\textrm{"price between 50 and 110"})+\mathbb{P}(\textrm{"price above 110"}) = 1 ,$$
and we can calculate
$$
\begin{aligned}
\mathbb{P}(\textrm{"price between 50 and 110"}) &= 1 - \mathbb{P}(\textrm{"price below 50"}) - \mathbb{P}(\textrm{"price above 110"}) \\
&= 1 - 0.6 - 0.2 = 0.2 .
\end{aligned}$$


问:
integrate function with change of variable
Find the primitive of $\;\displaystyle \int x^2 \sqrt{x+1}$ $dx$

So (...)

$u = x + 1 \quad \iff \quad u - 1 = x$

$u' = 1 \quad \iff \quad \frac{du}{dx} = 1 \rightarrow \;du = dx$

$$\int(u - 1)^2 . u^\frac12 \; du \;\;= \;\; {{(u-1)^3}\over3} \cdot {u^{3/2}\over{3/2}} + C \;\; =\;\; {1\over3} x^3\cdot{2\over3}(x+1)^{3/2} + C$$

Is this correct?

答:
The substitutions will work fine...but your evaluation is problematic.

Most problematic is the fact that you are integrating each factor of a product in the integrand, and expressing this as the product of integrated factors: which you cannot do, unless you are using, say, integration by parts, which proceeds much differently: . I.e. $$\int [f(x)\cdot g(x)]\,dx \;\neq \;\int f(x) \,dx \cdot \int g(x)\,dx$$

So, let's start from the point after which we've substituted:

$$\int \underbrace{(u - 1)^2}_{\text{expand}} . u^{1/2} \; du \; = \;\int \underbrace{(u^2 - 2u + 1)u^{1/2}}_{\text{distribute}} \,\;du = \int \left(u^{5/2} - 2u^{3/2} + u^{1/2}\right) \,du$$

Now integrate, and then back-substitute.


问:
Golden Exponent? Tetration
I read somebody say “golden exponent $ x^x=x+1 $” now I didn’t understand what he meant but it really fascinated me thinking about some type of tetration version of the golden number. A number with a special or awesome property (not necessarily linked to the golden number). Now that I think about the equation mentioned $ x^x=x+1 $ that does look like an interesting equation for x to satisfy. Anyways, what does everybody think about whether such an analog for tetration exists or not?

答:
Consider that you look for the non trivial zero's of function $$f(x)=x^x-x-1$$
$$f'(x)=x^x (\log (x)+1)-1\qquad \text{and} \qquad f''(x)=x^{x-1}+x^x (\log (x)+1)^2$$ The first derivative cancels at $x=1$ and $f(1)=-1$ with $f''(1)=2$. So, we have a minimum and two roots (one of them being the trivial $x=0$.

If you do not want to use a root-finding method, expand $f(x)$ as a series around $x=1$ and obtain
$$f(x)=-1+\sum_{n=1}^\infty a_n(x-1)^{n+1}$$ where the first coefficients are
$$\left\{1,\frac{1}{2},\frac{1}{3},\frac{1}{12},\frac{3}{40},-\frac{1}{120},\frac{59}{2520
   },-\frac{71}{5040},\frac{131}{10080},-\frac{53}{5040},\frac{179063}{19958400},\cdots\right\}$$

Now, using series reversion, we obtain
$$x=1+\sum_{n=1}^p b_n\,t^n \qquad \text{where}\qquad t=\sqrt{f(x)+1}$$ the first $b_n$ being
$$\left\{1,-\frac{1}{4},-\frac{1}{96},\frac{1}{12},-\frac{7621}{92160},\frac{173}{2880},-
   \frac{243979}{6881280},\frac{901}{60480},\frac{33853717}{118908518400},\cdots\right\}$$

Making $f(x)=0$ that is to say $t=1$, we then have the estimate
$$x=\frac{211659504277}{118908518400}\sim 1.78002$$ while the "exact" solution given by Newton method would be $1.77678$.

Title: Golden Exponent? Tetration


答:
Once you know that the probability of having a bad day is $e^{-2} \approx 0.135335$, then the rest of the question is no longer about an exponential distribution, but rather, a binomial distribution (and its approximation by a normal distribution).  In particular, the random number $X$ of bad days in a year is a binomial distribution with $n = 365$ and $p = e^{-2}$, assuming the probability that any given day is bad is independent of the outcome of any other day, and that the probability of a bad day is the same for any day of the year.

As such, we want the probability $$\Pr[50 \le X \le 60] = \sum_{x=50}^{60} \binom{365}{x} (e^{-2})^x (1 - e^{-2})^{365 - x}. \tag{1}$$  But this is difficult to calculate without a computer.  So we use the normal approximation to the binomial, by fitting a normal distribution whose mean $\mu$ equals the binomial mean $np$, and whose variance $\sigma^2$ equals the binomial variance $np(1-p)$.  Thus, we let $$Y \sim \operatorname{Normal}(\mu = 49.3974, \sigma^2 = 42.7122).$$  Then the desired probability is approximately
$$\Pr[50 \le Y \le 60] \approx \Pr\left[ \frac{50 - 49.3974}{\sqrt{42.7122}} \le \frac{Y - \mu}{\sigma} \le \frac{60 - 49.3974}{\sqrt{42.7122}}\right] = \Pr[0.0922081 \le Z \le 1.62232] \approx 0.947633 - 0.536734 = 0.410899.$$

However, the exact probability, using Equation $(1)$ above, is $0.43861306037477480884\ldots$.  Why the discrepancy?  The reason is that we needed to use continuity correction in the normal approximation, to account for half of the probability mass that we are missing.  The corrected approximation is:

$$\Pr[50 \le X \le 60] \approx \Pr[49.5 \le Y \le 60.5] \approx \Pr[0.0157023 \le Z \le 1.69883] \approx 0.44906,$$ which has about half the error of the uncorrected approximation.

For the second part of the question, the same type of calculation applies, except now the exact binomial probability is tractable because the number of good days is exactly $3$.  The probability of a good day is $$\Pr[T \le 0.05] = 1 - e^{-(0.1)(0.05)} \approx 0.00498752,$$ hence $$\Pr[X = 3] = \binom{365}{3} (0.00498752)^3 (1 - 0.00498752)^{365-3} \approx 0.163204.$$  Or you could use a Poisson approximation since the event is reasonably rare.  In such a case, the rate parameter is $\lambda = np = 365(0.00498752) \approx 1.82045$, which represents the average number of good days we can expect in a year.  Then  $$\Pr[X = 3] \approx e^{-\lambda} \frac{\lambda^3}{3!} \approx 0.162844.$$  I think the exact calculation is preferable, and just as accessible for a handheld calculator.

Body:
You own an ice-cream shop. Every day when you open the shop, there is an $\text{Exp}(\frac{1}{10})$ random waiting time, in minutes, before the first customer arrives.

When the waiting time for the first customer exceeds 20 minutes, it is a bad day. Approximate the probability that you will have between 50 and 60 bad days in a year (365 days).

When the waiting time for the first customer is shorter than 3 seconds (i.e., 0.05 minutes), it is a great day. Approximate the probability that you will have exactly 3 great days in a year (365 days).

My Attempt

The probability of waiting over 20 minutes is
$$\mathbb{P}(X\geq20)=\int^\infty_{20}0.1e^{-0.1x}dx=[e^{-0.1x}]^\infty_{20}=e^{-2}\approx0.135$$
Then since $np(1-p)=365(0.135)(1-0.135)=42.7>10$, we can use the normal distribution to apprimate the probability. The expectation for an exponential distribution is $\frac{1}{\lambda}=\frac{1}{1/10}=10$. Thus
$$\mathbb{P}(50\leq x\leq60)=\mathbb{P}(a\leq\frac{x-\mathbb{E}[x]}{\sqrt{\text{Var}(x)}}\leq b)$$
$$=\mathbb{P}(\frac{50-10}{\sqrt{\text{Var}(x)}}\leq\frac{x-\mathbb{E}[X]}{\sqrt{\text{Var}(x)}}\leq\frac{60-10}{\sqrt{\text{Var}(x)}})\approx \Phi(\frac{50}{\sqrt{\text{Var}(x)}})-\Phi(\frac{40}{\sqrt{\text{Var}(x)}})$$
The part that I am having trouble with is how to find the variance for the exponential distribution. Can I get some pointers?

Edit
I think I know how to find the variance now. I got Var[X]=E[$X^2$]-E[X]=200-10=190.
So
$$\Phi(\frac{50}{\sqrt{190}})-\Phi(\frac{40}{\sqrt{190}})\approx0.00176$$

My concern now is that this probability is way too small. I don't think that it is reasonable. Can I get pointers on where I went wrong? Thank you

For this one, since 3 seconds is so small, we can consider this event rare, so we can use poisson approximation. First we find the probability of great day
$$\mathbb{P}(X\leq0.05)=1-[e^{-0.1}x]^\infty_{0.05}\approx0.00498$$
We know that the expectation is equal to the parameter, $\lambda$, of the poisson distribution. Then we can find use the formula
$$e^{-\lambda}\frac{\lambda^k}{k!}$$

Is this the right approach?


问:
Probability question from Brilliant
Which is more likely?

You roll two dice 5 times and, every time, one of the two comes up as 1 and the other as 6.

You roll 10 dice all at once. 5 come up as 1s and the other 5 come up as 6s.

The answer is 2 and I'm confused by it.

Here's my thinking. When rolling two dice the probability of getting 1 and 6 or 6 and 1 is $2\cdot(1/6\cdot 1/6) = 2/36 = 1/18.$ This means that probability of getting this 5 times in a row is $(2/36)^5$.

In the second case there are $2^5$ dice combinations of 1s and 6s where we get 5 of each. The probability of every combination is $(1/6)^{10}$ and so the probability of the second case is $2^5\cdot (1/6)^{10}=(2/36)^5$ which is exactly the same as the first case.

What's wrong with my thinking?

答:
Which has more ways of occurring?

Obtaining five ‘1’s and five ‘6’s among ten concurrent die rolls in a row, with the ‘1’s spaced apart.

Obtaining five ‘1’s and five ‘6’s among ten concurrent die rolls in a row, with no restriction on the numbers’ order of appearance.

(Whether the die rolls are successive or concurrent is immaterial to the answer.)

Title: Probability question from Brilliant


答: You have it right up to $$\int(u-1)^2 u^{1/2} du$$. However it is not in general correct that $\int (f\cdot g)(x)dx=\int f(x)dx \cdot \int g(x) dx $, that is, integration does not "distribute" over multiplication.</p>

Title: integrate function with change of variable

Body:
Find the primitive of $\;\displaystyle \int x^2 \sqrt{x+1}$ $dx$

So (...)

$u = x + 1 \quad \iff \quad u - 1 = x$

$u' = 1 \quad \iff \quad \frac{du}{dx} = 1 \rightarrow \;du = dx$

$$\int(u - 1)^2 . u^\frac12 \; du \;\;= \;\; {{(u-1)^3}\over3} \cdot {u^{3/2}\over{3/2}} + C \;\; =\;\; {1\over3} x^3\cdot{2\over3}(x+1)^{3/2} + C$$

Is this correct?


答:
Look at the type of patterns of two pictures, and it will be quite clear:

Two dice rolled 5 times each

$\fbox1\fbox6\,\fbox6\fbox1\,\fbox1\fbox6\,\fbox6\,\fbox1\,\fbox6\fbox1$

Ten dice rolled together

$\fbox1\fbox1\fbox6\fbox6\fbox6\fbox1\fbox1\fbox6\fbox6\fbox1$

The second case can generate all of the types the first one can, and then more...

Title: Probability question from Brilliant

Body:
Which is more likely?

You roll two dice 5 times and, every time, one of the two comes up as 1 and the other as 6.

You roll 10 dice all at once. 5 come up as 1s and the other 5 come up as 6s.

The answer is 2 and I'm confused by it.

Here's my thinking. When rolling two dice the probability of getting 1 and 6 or 6 and 1 is $2\cdot(1/6\cdot 1/6) = 2/36 = 1/18.$ This means that probability of getting this 5 times in a row is $(2/36)^5$.

In the second case there are $2^5$ dice combinations of 1s and 6s where we get 5 of each. The probability of every combination is $(1/6)^{10}$ and so the probability of the second case is $2^5\cdot (1/6)^{10}=(2/36)^5$ which is exactly the same as the first case.

What's wrong with my thinking?


问:
Cyclic numbers are characterized by the reciprocals of full reptend primes?
The number $142,857$ is widely known as a cyclic number, meaning consecutive multiples are cyclic permutations, i.e.

$1 × 142,857 = 142,857$

$2 × 142,857 = 285,714$

$3 × 142,857 = 428,571$

and so on.

142857 is the repeating unit of $\frac{1}{7} = 0.\overline{142857}$ and in fact, every prime for which $10$ is a primitive root will generate a cyclic number (if we allow $0$ as a first digit, for example $0588235294117647$ which is the repeating unit of $\frac{1}{17}$). These primes are called full reptend primes.

From what I've read, it seems that there is a bijection between full reptend primes and cyclic numbers: a number is cyclic if and only if it is the repeating unit for the reciprocal of a full reptend prime.

I was able to find a proof for the if part. I was wondering if anyone can provide a proof for the only if part.

Edit: It's been a while since I posed this question and I still haven't found a proof. I've added a bounty in hopes of prompting some interest.

答:
I consider Matthew L. Wright's excellent article
highly worth reading before approaching this problem.
It outlines how long division, the ring of integers modulo a prime $p$
and its group of units $U_p$ is central to understanding the problem,
and demonstrates the "if" direction: Every "full reptend prime",
an odd prime $p$ for which $10$ is a primitive root,
generates (corresponds to) a (unique) cyclic number (by the map)
$n=\frac{10^{p-1}-1}{p}=\lfloor10^{p-1}/p\rfloor$ with period/length $d=p-1$.
Some partial work in the "only if" direction follows below.

Suppose that $$n=(n_1n_2\cdots n_d)_{10}=\sum_{i=1}^{d}n_i10^{d-i}$$
is a cyclic number, i.e., for each $1<k \le d$,
the $k$th multiple $kn$ of $n$ has decimal representation
$$kn=(n_{\sigma_k(1)}n_{\sigma_k(2)}\cdots n_{\sigma_k(d)})_{10}
=\sum_{i=1}^{d}n_{\sigma_k(i)}10^{d-i},$$
where $\sigma_k\in S_d$ is a cyclic permutation (bijection) of $\{1,2,\dots,d\}$
satisfying $$\sigma_k(i)\equiv{i+t_k}\pmod{d}$$ for some shift $0\le t_k<d$;
let $t_1=0$ (since $1n=n$ fixes all the digits of $n$).
Note that the map $k\to t_k$ is a bijection from $\{1,\dots,d\}$ to $\{0,\dots,d-1\}$.
We wish to prove that
$$\frac1{d+1}=\frac{n}{10^d-1}=(0.\overline{n_1n_2\cdots n_d})_{10},$$
i.e., that $(d+1)n=10^d-1$.
Let $s=\sum_{i=1}^{d}n_i$ be the sum of the digits of $n$.
Noting that
$$ \frac{d(d+1)}{2}n = \sum_{k=1}^{d}kn
=\sum_{k=1}^{d}\sum_{i=1}^{d}n_{\sigma_k(i)}10^{d-i}
=\sum_{i=1}^{d}\sum_{k=1}^{d}n_{\sigma_k(i)}10^{d-i}
=s\sum_{i=1}^{d}10^{d-i}=\frac{10^d-1}{10-1}s,
$$
we have
$$ (d+1)n=\frac{2s}{9d}\left(10^d-1\right)
\qquad\mbox{or}\qquad
u=\frac{2s}{9d}=\frac{(d+1)n}{10^d-1}, $$
so what remains is to show that $2s=9d$, or
$$\mbox{(wanted):}%\qquad\boxed{2s=9d}\quad\mbox{or}
\quad\boxed{s=\frac92d}
\quad\mbox{or}\quad\boxed{n=\frac{10^d-1}{d+1}}
\quad\mbox{or}\quad\boxed{u=1} . $$
Note that the desired quantity for $s$ is the expected value
of the sum of $d$ equiprobable decimal digits,
and the desired quantity for $n$ implies a cyclic, repeating fraction
of length $d$ for $\frac1{d+1}$ upon performing long division.

Perhaps a thorough consideration of the material in the article mentioned above
should be enough to stop here (see also the last paragraph below).
But if we could prove it by even more elementary means, without resorting to
the structure of cyclic groups and needing to assume that $p$ is an odd prime
so that $d$ is even and $t_{1+d/2}=d/2$ (since the order of $10$ modulo $p$
is even and hence $10^{(p-1)/2}\equiv-1\pmod p$), we might continue as below.

Now the "trivial" case $n=1$ gives rise to the fraction
$0.\overline{1}=\frac19$, with $s=1$ and a succession of $d=8$ multiples,
which are not cyclic, and do not satisfy this equation
or give rise to a full reptend prime.
Therefore, since we have already assumed $n$ is a cyclic number,
we can also assume, without loss of generality, that $d>1$.
Backtracking slightly in preparation to examine things modulo $10$, we have that
$$\frac{d(d+1)}2\cdot9n=s\left(10^d-1\right)$$
$$\implies s \equiv \frac{d(d+1)}2 \cdot n \pmod{10}.$$

Also, since $dn$ has $d$ digits, we have
$$
\frac29\frac{s}{d+1}=\frac{dn}{10^d-1}\le1
\quad\implies\quad s\le\frac92(d+1),
$$
a promising upper bound which could be enough if we also had
$s\ge\frac92(d-1)$ and ${2s}\equiv{9d}\pmod{10}$.
This is as far as I have gone by elementary means,
and is enough to see where I started in 2012 with the general base $b$.

However, I suspect that the answer is pretty clear once one realizes that
the order (minimal power) $d$ of $10$ (to reach $1$) modulo an odd prime $p$
is always even, and that $10^{d/2}\equiv-1\pmod{p}$ will always hold;
this, then, implies that the second half of the sequence of cyclic numbers
will always complement the first half, adding to $9$ digit by digit,
so that for $1\le k\le\frac{d}2$ and $j=k+\frac{d}2$,
we will have $(kn)+(jn)=(9\cdots9)_{10}=10^d-1$,
where the addition is $9$ in each decimal place with no carries,
i.e., $n_{\sigma_k(i)}+n_{\sigma_j(i)}=9$ for each $1\le i\le d$.

The first few cyclic numbers and the full reptend primes generating them are given below:

```<code>7 -> 142857
17 -> 0588235294117647
19 -> 052631578947368421
23 -> 0434782608695652173913
29 -> 0344827586206896551724137931
47 -> 0212765957446808510638297872340425531914893617
59 -> 0169491525423728813559322033898305084745762711864406779661
61 -> 016393442622950819672131147540983606557377049180327868852459
97 -> 010309278350515463917525773195876288659793814432989690721649484536082474226804123711340206185567
109 -> 009174311926605504587155963302752293577981651376146788990825688073394495412844036697247706422018348623853211
113 -> 0088495575221238938053097345132743362831858407079646017699115044247787610619469026548672566371681415929203539823
131 -> 0076335877862595419847328244274809160305343511450381679389312977099236641221374045801526717557251908396946564885496183206106870229
149 -> 0067114093959731543624161073825503355704697986577181208053691275167785234899328859060402684563758389261744966442953020134228187919463087248322147651
167 -> 0059880239520958083832335329341317365269461077844311377245508982035928143712574850299401197604790419161676646706586826347305389221556886227544910179640718562874251497
179 -> 0055865921787709497206703910614525139664804469273743016759776536312849162011173184357541899441340782122905027932960893854748603351955307262569832402234636871508379888268156424581
181 -> 005524861878453038674033149171270718232044198895027624309392265193370165745856353591160220994475138121546961325966850828729281767955801104972375690607734806629834254143646408839779
193 -> 005181347150259067357512953367875647668393782383419689119170984455958549222797927461139896373056994818652849740932642487046632124352331606217616580310880829015544041450777202072538860103626943
</code>```

Body:
I consider Matthew L. Wright's excellent article
highly worth reading before approaching this problem.
It outlines how long division, the ring of integers modulo a prime $p$
and its group of units $U_p$ is central to understanding the problem,
and demonstrates the "if" direction: Every "full reptend prime",
an odd prime $p$ for which $10$ is a primitive root,
generates (corresponds to) a (unique) cyclic number (by the map)
$n=\frac{10^{p-1}-1}{p}=\lfloor10^{p-1}/p\rfloor$ with period/length $d=p-1$.
Some partial work in the "only if" direction follows below.

Suppose that $$n=(n_1n_2\cdots n_d)_{10}=\sum_{i=1}^{d}n_i10^{d-i}$$
is a cyclic number, i.e., for each $1<k \le d$,
the $k$th multiple $kn$ of $n$ has decimal representation
$$kn=(n_{\sigma_k(1)}n_{\sigma_k(2)}\cdots n_{\sigma_k(d)})_{10}
=\sum_{i=1}^{d}n_{\sigma_k(i)}10^{d-i},$$
where $\sigma_k\in S_d$ is a cyclic permutation (bijection) of $\{1,2,\dots,d\}$
satisfying $$\sigma_k(i)\equiv{i+t_k}\pmod{d}$$ for some shift $0\le t_k<d$;
let $t_1=0$ (since $1n=n$ fixes all the digits of $n$).
Note that the map $k\to t_k$ is a bijection from $\{1,\dots,d\}$ to $\{0,\dots,d-1\}$.
We wish to prove that
$$\frac1{d+1}=\frac{n}{10^d-1}=(0.\overline{n_1n_2\cdots n_d})_{10},$$
i.e., that $(d+1)n=10^d-1$.
Let $s=\sum_{i=1}^{d}n_i$ be the sum of the digits of $n$.
Noting that
$$ \frac{d(d+1)}{2}n = \sum_{k=1}^{d}kn
=\sum_{k=1}^{d}\sum_{i=1}^{d}n_{\sigma_k(i)}10^{d-i}
=\sum_{i=1}^{d}\sum_{k=1}^{d}n_{\sigma_k(i)}10^{d-i}
=s\sum_{i=1}^{d}10^{d-i}=\frac{10^d-1}{10-1}s,
$$
we have
$$ (d+1)n=\frac{2s}{9d}\left(10^d-1\right)
\qquad\mbox{or}\qquad
u=\frac{2s}{9d}=\frac{(d+1)n}{10^d-1}, $$
so what remains is to show that $2s=9d$, or
$$\mbox{(wanted):}%\qquad\boxed{2s=9d}\quad\mbox{or}
\quad\boxed{s=\frac92d}
\quad\mbox{or}\quad\boxed{n=\frac{10^d-1}{d+1}}
\quad\mbox{or}\quad\boxed{u=1} . $$
Note that the desired quantity for $s$ is the expected value
of the sum of $d$ equiprobable decimal digits,
and the desired quantity for $n$ implies a cyclic, repeating fraction
of length $d$ for $\frac1{d+1}$ upon performing long division.

Perhaps a thorough consideration of the material in the article mentioned above
should be enough to stop here (see also the last paragraph below).
But if we could prove it by even more elementary means, without resorting to
the structure of cyclic groups and needing to assume that $p$ is an odd prime
so that $d$ is even and $t_{1+d/2}=d/2$ (since the order of $10$ modulo $p$
is even and hence $10^{(p-1)/2}\equiv-1\pmod p$), we might continue as below.

Now the "trivial" case $n=1$ gives rise to the fraction
$0.\overline{1}=\frac19$, with $s=1$ and a succession of $d=8$ multiples,
which are not cyclic, and do not satisfy this equation
or give rise to a full reptend prime.
Therefore, since we have already assumed $n$ is a cyclic number,
we can also assume, without loss of generality, that $d>1$.
Backtracking slightly in preparation to examine things modulo $10$, we have that
$$\frac{d(d+1)}2\cdot9n=s\left(10^d-1\right)$$
$$\implies s \equiv \frac{d(d+1)}2 \cdot n \pmod{10}.$$

Also, since $dn$ has $d$ digits, we have
$$
\frac29\frac{s}{d+1}=\frac{dn}{10^d-1}\le1
\quad\implies\quad s\le\frac92(d+1),
$$
a promising upper bound which could be enough if we also had
$s\ge\frac92(d-1)$ and ${2s}\equiv{9d}\pmod{10}$.
This is as far as I have gone by elementary means,
and is enough to see where I started in 2012 with the general base $b$.

However, I suspect that the answer is pretty clear once one realizes that
the order (minimal power) $d$ of $10$ (to reach $1$) modulo an odd prime $p$
is always even, and that $10^{d/2}\equiv-1\pmod{p}$ will always hold;
this, then, implies that the second half of the sequence of cyclic numbers
will always complement the first half, adding to $9$ digit by digit,
so that for $1\le k\le\frac{d}2$ and $j=k+\frac{d}2$,
we will have $(kn)+(jn)=(9\cdots9)_{10}=10^d-1$,
where the addition is $9$ in each decimal place with no carries,
i.e., $n_{\sigma_k(i)}+n_{\sigma_j(i)}=9$ for each $1\le i\le d$.

The first few cyclic numbers and the full reptend primes generating them are given below:

7 -> 142857
17 -> 0588235294117647
19 -> 052631578947368421
23 -> 0434782608695652173913
29 -> 0344827586206896551724137931
47 -> 0212765957446808510638297872340425531914893617
59 -> 0169491525423728813559322033898305084745762711864406779661
61 -> 016393442622950819672131147540983606557377049180327868852459
97 -> 010309278350515463917525773195876288659793814432989690721649484536082474226804123711340206185567
109 -> 009174311926605504587155963302752293577981651376146788990825688073394495412844036697247706422018348623853211
113 -> 0088495575221238938053097345132743362831858407079646017699115044247787610619469026548672566371681415929203539823
131 -> 0076335877862595419847328244274809160305343511450381679389312977099236641221374045801526717557251908396946564885496183206106870229
149 -> 0067114093959731543624161073825503355704697986577181208053691275167785234899328859060402684563758389261744966442953020134228187919463087248322147651
167 -> 0059880239520958083832335329341317365269461077844311377245508982035928143712574850299401197604790419161676646706586826347305389221556886227544910179640718562874251497
179 -> 0055865921787709497206703910614525139664804469273743016759776536312849162011173184357541899441340782122905027932960893854748603351955307262569832402234636871508379888268156424581
181 -> 005524861878453038674033149171270718232044198895027624309392265193370165745856353591160220994475138121546961325966850828729281767955801104972375690607734806629834254143646408839779
193 -> 005181347150259067357512953367875647668393782383419689119170984455958549222797927461139896373056994818652849740932642487046632124352331606217616580310880829015544041450777202072538860103626943


问:
Basis for vector space with complex vectors

Find a basis for $V = \operatorname{span}\{v_1,v_2,v_3\}$, where  $v_1=(-i,-i,i), v_2=(5,1,0), v_3=(4,0,1)$.

This is asking to find a basis for the column space of $$\begin{pmatrix}-i&5&4\\ -i&1&0\\ i&0&1\end{pmatrix}.$$

I have that $$\begin{pmatrix}-i&5&4\\ -i&1&0\\ i&0&1\end{pmatrix}\begin{pmatrix}x_1\\ x_2\\ x_3\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix} \implies x_1=ix_3,x_2=-x_3.$$ So any vectors that would form a basis for $V$ would be of the form $(x_1,x_2,x_3)=(ix_3,-x_3,x_3)?$ I'm a bit confused how can I construct the basis from here?

答:
You solved for the kernel (=nullspace) of the given matrix, and found that it's the span of $(i,-1,1)$, which has dimension $1$.

By the rank-nullity theorem we know that the rank of the matrix must be $2$, which means that any two independent vectors in the column space form a basis of the column space.

So, just take any $2$ columns of the matrix, it's easy to see that they are not parallel, hence are linearly independent.

Specifically, $v_1,v_2$ is such a basis.

You can also note that $iv_1=(1,1,-1)$ and so $v_3=v_2-i\,v_1$.


问:
Category with endomorphisms only
How is called a category with endomorphisms only?

How is called a subcategory got from an other category by removing all morphisms except of endomorphisms?

答:
Every category in which every arrow is an endomorphism is a coproduct (in the category of categories) of monoids (a monoid is a category with just one object). So a category with all morphisms endormophisms is a coproduct of monoids. I'm not aware of any specific terminology for it.

Clearly the category of all coproducts of monoids admits an inclusion functor into the category of all categories. The construction you describe (of removing all non-endo morphisms) describes a right adjoint to this inclusion functor.

Title: Category with endomorphisms only


答:
Substitute $x=t^{1/n}$ to obtain
$$
\int_0^1 {\frac{{nx^{n - 1} }}{{1 + x}}dx}  = \int_0^1 {\frac{{dt}}{{1 + t^{1/n} }}}
$$
Now, since
$$
0 \le \frac{1}{{1 + t^{1/n} }} \le 1,
$$
we can apply the dominated convergence theorem to deduce
$$
\mathop {\lim }\limits_{n \to  + \infty } \int_0^1 {\frac{{nx^{n - 1} }}{{1 + x}}dx}  = \mathop {\lim }\limits_{n \to  + \infty } \int_0^1 {\frac{{dt}}{{1 + t^{1/n} }}}  = \int_0^1 {\mathop {\lim }\limits_{n \to  + \infty } \frac{{dt}}{{1 + t^{1/n} }}}  = \int_0^1 {\frac{{dt}}{2}}  = \frac{1}{2}.
$$

Title: How do I know if this limit even converges?

Body:
So, there is this question which I found quite new according to my experience, and it is as follows.

Assuming that interchange of limit and integration is permissible, evaluate the following: $$\lim_{n \to \infty} {\int_0^1 \frac{nx^{n -1}}{1 + x}dx}, \quad 0 < x < 1$$

Now, I first tried integration by parts, and it doesn't give me more than an expression which is also nearly impossible to evaluate (maybe I'm doing it wrong, though) -

$$\lim_{n \to \infty} \left|\frac{1}{1+x}\int nx^{n - 1}dx  +  \int\frac{x^n}{(1 + x)^2}dx\right|_0^1$$

I'm not sure how to proceed from there, so I tried this instead:

$$\lim_{n \to \infty} {\int_0^1 {nx^{n - 1}(1 + x)^{-1}}dx}$$

Expanding $(1 + x)^{-1}$ using Taylor Series:

$$\lim_{n \to \infty} {\int_0^1 {nx^{n - 1}(1 - x + x^2 - x^3 + x^4 - x^5 \ldots)}dx}$$
$$ = \lim_{n \to \infty} {\int_0^1 {(nx^{n - 1} - nx^n + nx^{n + 1} - nx^{n + 2} + nx^{n + 3} - n^{x + 4} \ldots)}dx}$$
$$\lim_{n \to \infty} \left(1 - \frac{n}{n+1} + \frac{n}{n+2} - \frac{n}{n +3} + \frac{n}{n +4} - \frac{n}{n + 5}\ldots\right)$$

I think that diverges. Then, I tried putting the expression in Desmos and increased the $n$ value slowly upto $500$, $1000$ and similar big numbers - which all returned values around $0.5$.

The question was an multiple-choice based and there were both of them listed as options.

I'm not sure who is correct, the calculator, or me.

It would be a huge help if someone looks into this. Thank you in advance.


问:
why $\lambda(S^1) \le\lim_{\epsilon\to 0}\lambda (S_\epsilon) $?
Note : Im posting this question because user is not active for past $3-4$ months

find the lebesgue measure of the unit circle?

My attempt : I found the answer here

Here is the outline of the given answer

For a very nice region, the Lebesgue measure in the two-dimensional space measures the area of the region.

Consider the region consisting on the points $S_\epsilon$ between the circle of radius $1-\epsilon$ and $1+\epsilon$. The area of this region is

$$\begin{align*}
S_\epsilon&=\pi\cdot (1+\epsilon)^2 - \pi\cdot (1-\epsilon)^2\\
&=\pi\cdot [(1+\epsilon)+(1-\epsilon]\cdot [(1+\epsilon)-(1-\epsilon)]\\
&=4\pi\cdot \epsilon
\end{align*}$$
Since the circle of radius $1$ (which is usually denoted by $S^1$) is contained in each $S_\epsilon$, we know that

$$0\leq \lambda(S^1)\leq \lim_{\epsilon\to 0}\lambda (S_\epsilon)=\lim_{\epsilon\to 0}4\pi\cdot \epsilon=0.$$
Thus, $\lambda(S^1)=0$.

My confusion : why   $\lambda(S^1) \le\lim_{\epsilon\to 0}\lambda (S_\epsilon) $?

My thinking :  Here  $\epsilon < 1$.Generally we  take $\epsilon=\frac{1}{2}$ in most of the book

Now we have  circle  with radius $1$ and circle with radius $1/2$ .

obviously circle  with radius $1$  will be bigger than the circle  with radius $1/2$

$\implies$  circle with radius $1$ will have more measure than circle with radius $1/2$

So $$\lambda(S^1) \ge \lambda (S_\epsilon) $$

答: Note that $S_\varepsilon = \{ x \in \mathbb{R}^2 : 1 - \varepsilon \leq ||x|| \leq 1 +\varepsilon \}$ so if $x \in S^1$ then $||x||=1$, therefore $x \in S_\varepsilon$. Now, remember that given a sequence $(x_n) \subset \mathbb{R}$ and $c$ a real number, if  $c \leq x_n$ then $c \leq \displaystyle \lim_{n \to \infty} x_n$.


问:
probability of$P(A∣\lnot B)$?
If we have the probabilities of $P(A)$ and $P(A∣B)$, how can we calculate the probability of $P(A∣\lnot B)$?

The question statement is as follows:

A disease affects 6.5% of the population. There is however, an inheritance factor. If one’s father has the disease, the probability that the child will get the disease is 0.13. What is the probability a child will get the disease when the father does not have the disease?

How do we do this?

答:
You can do the following
\begin{align*}
P(A)&=P(A|B)P(B)+P(A|\lnot B)P(\lnot B)\\
&=P(A|B)P(B)+P(A|\lnot B)(1-P(B))
\end{align*}
So that
\begin{align*}
P(A|\lnot B) &= \frac{P(A)-P(A|B)P(B)}{1-P(B)}
\end{align*}
The problem is that from $P(A)$ and $P(A|B)$ only we cannot determine $P(B)$.

Title: probability of$P(A∣\lnot B)$?


问:
Counting 11 letter words with subword "FRED" - check my answer
I'm preparing for an exam and just want to make sure I am getting things correctly.

How many 11-letter 'words' constructed from the 26 letters of the english alphabet contain the subword "FRED" Where 'words' are just any string of 11 letter words (does not have to be a word in the dictionary)

a subword is a word inside another word. So an 11 letter word must have the word "FRED" in it, so FREDAABBCDEF is allowed, or XYZFREDABBC

My answer:
FRED contains 4 letters, so:

Choose the 4 positions to put FRED in: $c(11,4)$

Choose the remaining 7 letters (can be any repeatable letters): $p(26,7)$

So we have $c(11,4) * p(26,7)$

答:
You double-count words. This is a very frequent error in this sort of problems, where you fix some positions to have a specific value, and choose freely the values for the other positions. The problem is that those other positions may also have the specific value, and so you'll double count the solution.

Here is a simple example: Count all binary words that contain at least one occurrence of 1. There are 3 of them, of course: 11,01,10 (but not 00). However, your method yields 4: You choose a place in which to place "1" (2 choices: either first or last letter) and then you choose the value of the other position (2 choices: 0 or 1). It is easy to see that you count "11" twice.

Usually such problems are solved using the Inclusion-Exclusion principle.


答:
No, it isn't. If you take the derivative of the product applying Leibniz's rule it won't give you the original integrand. The error is that you integrated both parts of the functions instead of the function as a whole.

The integral is easier if you make the change of variable:

$$u=\sqrt{1+x}$$

Then:

$$du=\frac{dx}{2\sqrt{1+x}}$$

So the integral becomes:

$$2\int u^2(-1+u^2)^2 du$$

And now it's an inmmediate integral.


问:
Bayes estimate of upper limit of uniform distribution with exponential prior
Let $X_{1}, . . . , X_{n} > 0$ be a random sample from $U(0, \theta)$. Suppose $\theta$ has the prior $\pi(θ) = e^{-\theta} ; \theta > 0$. Find the Bayes estimate of $\frac{1}{\theta}$ with respect to squared error loss function.

Here is my approach:

Let $D = \{X_{1}, . . . , X_{n}\}$

First, we find the expression for the likelihood $p(D|\theta)$

Now $X_{(n)}$ cannot be greater than $\theta$. So, likelihood is $0$ for $\theta < X_{(n)}$

\begin{equation}
    p(D|\theta) =
    \begin{cases}
      0, & \theta < X_{(n)} \\
      1/\theta^{n}, & \text{otherwise}
    \end{cases}
  \end{equation}

The posterior is proportional to prior times likelihood

\begin{equation}
    p(\theta | D) \propto
    \begin{cases}
      0, &  \theta < X_{(n)} \\
      \theta^{-n} e^{-\theta}, & \text{otherwise}
    \end{cases}
  \end{equation}

Let $\frac{1}{\theta} = \beta$
\begin{equation}
    p(\beta | D) \propto
    \begin{cases}
      0, &  \beta > X_{(n)} \\
      \theta^{-n} e^{-\theta}, & 0 < \beta \leq X_{(n)}
    \end{cases}
  \end{equation}
Bayes estimate of $\beta$,
\begin{equation}
\hat{\beta} = \mathbb{E}_{p(\beta| D)}[\beta] = \int_{0}^{X_{(n)}} \beta p(\beta|D) d\beta = \int_{0}^{X_{(n)}} \theta^{-(n + 1)}e^{-\theta} d\beta =  \int_{X_{(n)}}^{\infty} \theta^{-(n + 3)}e^{-\theta} d\theta
\end{equation}

First, I don't know if what I have done till now is correct or not. Second, I have no idea how to solve this integral. Third, is it correct to just find the Bayes estimate of $\theta$ and invert it to get Bayes estimate of $1/\theta$?

答:
First you have to normalize you posterior which results

$$\pi(\theta|\mathbf{x})=\frac{1}{\Gamma[1-n,x_{(n)}]}\theta^{-n}e^{-\theta}\cdot\mathbb{1}_{(x_{(n)};+\infty)}(\theta)$$

Then, given that the loss function is quadratic, Bayes estimation is the posterior expectation, that is

$$\mathbb{E}\left[\frac{1}{\theta};\mathbf{x}\right]=\frac{1}{\Gamma[1-n,x_{(n)}]}\int_{x_{(n)}}^{+\infty}\theta^{-(n+1)}e^{-\theta} d\theta=\frac{\Gamma[-n,x_{(n)}]}{\Gamma[1-n,x_{(n)}]}$$


问:
Let $n$ and $k$ be positive integers. Show that $\gcd(n, nk + 1) = 1$
This is what I've begun with:

So lets say $x=n,y=nk+1,x,y,n\in\Bbb Z$
Lets say we have a integer, d that divides both x and y. So, $d\vert x$ and $d\vert y$ so if there is a gcd, then d must also divide y-x. $d\vert(y-x)$ but we said $x=n,y=nk+1$. Then, $d\vert(nk+1-n)$=$d\vert (n(k-1)+1)$.
But I have no idea if I'm heading in the right direction? Any tips?

答:
Let $p\mid n$ and $p\mid nk+1$. Then $n=pq$ and $nk+1=pr$ for some $q,r\in\mathbb Z$. Therefore \begin{align*}
nk+1=kpq+1=pr&\implies p(r-kq)=1\\
&\implies p\mid 1.
\end{align*}


问:
Finding the upper and lower bounds, positive and negative each if applicable, for a complex function
Given $|z^4 -8 + 3|, |z|=2 $

I need to find the positive upper and lower bounds of the above expression.

My attempt:

By triangle inequality, we have

$||z^4|+|-8z|+3|\geq|z^{4} -8z + 3| \leq  ||z^4 -8z| +3|???||z^4|+|3-8z|| \geq ||z^4|-|8z|+3|\geq||z^4+3|-|8z|| $

(I used ??? since I don't know which is larger than which, and I believe i got all the combinations possible)

At the left most, I believe that will give me the upper bound so $|2^4|+|-8(2)|+3| = 35$

At the right most, $||2^4 +3| -|8(2)| = 3|$

BUT, and here is my question, given that $z = r*e^{i\theta}, |e^{i\theta}| = 1$

assuming my triangle inequality is true, we know $|8z| = |8r*e^{i\theta}|=|8r|$, however $|z^4+3|$ is not necessarily equal to $|r^4 +3|$, I wonder if $|z^4+3| = |8z|$ such that $||z^4+3|-|8z||= 0$ ?

Alternatively if $|z^4 - 8z| = -3$ is possible??

答:
Given two complex numbers $a$ and $b$, then the triangle inequality is $$||a|-|b|| \le |a -b| \le |a|+|b|$$

Hence, if we want simple bounds, assuming $|z|=2$, then we can do the following, $$13= |16-3| = ||-8z|-|3|| \le |-8z+3| \le |-8z|+|3| = 16+3 = 19$$ Hence, $|-8z+3|$ is bounded between $13$ and $19$ (notice that it could be $16$). Thus, we have $$0\le |16-|-8z+3||=||z^4|-|-8z+3||\le |z^4-8z+3| \le |z^4|+|8z|+3 = 35$$ So, you have lower bound $0$ and upper bound $35$.

If you want tighter bounds, you could solve the absolute value explicitly. That is, let $z=x+yi$, then assuming $|z|=2$ we have that $y^2 = 4 - x^2$, so we perform substitution, $$|(x+yi)^4-8(x+yi)+3| $$ $$= |(x^4-6x^2y^2+y^4-8x+3)+(4x^3y-4xy^3-8y)i|$$ $$=\sqrt{(x^4-6x^2y^2+y^4-8x+3)^2+(4x^3y-4xy^3-8y)^2}=\sqrt{48x^4-256x^3-192x^2+720x+617}$$ we can explicitly take the derivative of the polynomial ($192x^3-768x^2-384x+720$) in the square root and find the absolute maximum and absolute minimum, where $x$ is between $-2$ and $2$. Doing so we find the absolute minimum is $1.79964$ and is found with $x=-1.0785$ ($z=-1.0785+1.68429i$), likewise the absolute maximum is $35$ and is found with $x=-2$ ($z=-2+0i$). So we have exact bounds, $$1.79964 \le |z^4-8z+3| \le 35$$ moreover equality is achieved at specific $z=-1.0785+1.68429i$ and $z=-2$.


问:
Determine the equations of the tangents to the curve $y=2x^2+3$ that pass through point $(2,3)$
I am unsure of how to answer this question. All I have found is the slope of the tangents: $y'= 4x$.

答:
Figure 1 Showing the problem (Figure 1 is just for general reference)

You have a 2nd order polynomial $y = f(x)$ which we know is a parabola. You have a point that lies outside of the parabola $P\,(x_P,y_P)$. There are two points on the parabola $(a\,\,and\,\,b)$ who's tangent lines pass through point $P$. The tangent line that passes through points $a$ and $P$ is defined by $y=a(x) = f'(x_a)x + b_a$ and the tangent line that passes through points $b$ and $P$ is defined by $y=b(x) = f'(x_b)x + b_b$. First we need to find the slope of $f(x)$ then solve for $x_a$ and $x_b$.

Find the slope of $f(x)$ by taking the derivative
$$\begin{align}y=f(x)&=2x^2+3\\\\f'(x)&=4x\end{align}$$,

Solve for $x_a$ and $x_b$. Looking at Figure 1 both points $a\,(x_a,y_a)\,\, and\,\,b\,(x_b, y_b)$ have been defined as $[x, f(x)]$, this is done so we can find a general solution using the Slope Formula
$$\begin{align}m&=\frac{y_2-y_1}{x_2-x_1}\\\\Making\,\,substitutions\,\,&with\,\,our\,\,function,\,\, derivative,\,\,and\,\,point\\\\f'(x)&=\frac{f(x)-y_P}{x-x_P}\end{align}$$
The above states that derivative at $x$ is equal to the slope between the point defined by $f(x)$ and our known point $P$. Now we will substitute our known values into the slope formula and solve for x.
$$\begin{align}x_P&=2\\y_p&=3\\\\4x&=\frac{2x^2+3-3}{x-2}\\\\0&=\frac{2x^2+3-3}{x-2}-4x\\\\&=2x^2+3-3-4x(x-2)\\\\&=2x^2-0-4x^2+8x\\\\&\Rightarrow 2x^2-8x+0\,\,\,(Multiplied\,\,both\,\,sides\,\,by\,\,-1)\end{align}$$
We now have a Quadratic Equation where $a=2$, $b=-8$, and $c=0$. We can solve for x using the quadratic formula.
$$\begin{align}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\&=\frac{8\pm8}{4}\\\\&\Rightarrow x_a=0,\,\,x_b=4\end{align}$$

Now we can solve for the slope at points $a$ and $b$.
$$\begin{align}m_a&=f'(x_a)\\&=4(0)\\&=0\\\\m_b&=f'(x_b)\\&=4(4)\\&=16\end{align}$$

Now we will solve for $b_a$ and $b_b$ in the tangent line equations using the values we found for $x$. We will evaluate the tangent line equations at point $P$.
$$
  \begin{align}
    x_a&=0\\
    x_b&=4\\
    x_P&=2\\
    y_P&=3\\\\
    y_P=a(x_P)&=m_ax+b_a\\\\
    3&=(0)(2)+b_a\\\\
    &=0+b_a\\\\
    b_a&=3\\\\
    y_P=b(x_P)&=m_bx+b_b\\\\3&=(16)(2)+b_b\\\\&=32+b_b\\\\b_b&=-29\end{align}$$

Putting it all together... We have solved for the slope and y-intercept for the lines tangent to points $a$ and $b$ which pass through point $P$. Now we will substitute the values into our tangent equations.

$$\begin{align}
  y&=mx+b\\\\\\
  &For\,\,the\,\,tangent\,\,line\,\,at\,\,a\\
  a(x)&=m_ax+b_a\\\\
  &=(0)x+3\\\\
  &=3\\\\\\
  &For\,\,the\,\,tangent\,\,line\,\,at\,\,b\\
  b(x)&=m_bx+b_b\\\\
  &=16x-29\end{align}$$

Check that $f(x_a) = a(x_a)$ and $f(x_b) = b(x_b)$
$$\begin{align}f(0)&=a(0)\\2(0)^2+3&=3\\\\f(4)&=b(4)\\2(4)^2+3&=16(4)-29\end{align}$$


问:
Can any non-singular real number matrix be diagonalized without swapping any rows or columns?
According to Nering's Linear Algebra and Matrix Theory, there are three types of elementary row operations on matrices:

I. Multiply a row by a non-zero scalar

II. Add one row to another

III. Interchange two rows

There are three analogous elementary column operation.

The associated elementary matrices are, for example example:

$$I.\left(\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & a & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{array}\right)$$

$$II.\left(
\begin{array}{cccc}
 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 \\
 0 & 1 & 1 & 0 \\
 0 & 0 & 0 & 1 \\
\end{array}
\right)$$

$$III.\left(
\begin{array}{cccc}
 0 & 0 & 1 & 0 \\
 0 & 1 & 0 & 0 \\
 1 & 0 & 0 & 0 \\
 0 & 0 & 0 & 1 \\
\end{array}
\right)$$

Nering gives:

Theorem II-6.1 Any non-singular matrix can written as a product of elementary matrices.

As mention in Are the statement and proof of this theorem regarding the linear transformation of a contented set satisfactory? CH Edwards, Theorem IV 5.1, CH Edwards asserts that a standard theorem of linear algebra says a non-singular matrix can be expressed as a product of two types of elementary matrices.  These elementary matrices are of type I and II given by Nering.

In his book General Theory of Relativity, Dirac says that the determinant $g$ of the flat space-time metric is negative for oblique axes because it is $g=-1$ for the Minkowski metric.

With oblique axes $g$ must still be negative, because the oblique axes can be obtained from the orthogonal ones by a continuous process, resulting in $g$ varying continuously, and $g$ cannot pass through the value zero.

I take that as intuitively obvious.  But, if a set of axes (basis vectors) can be continuously transformed to and from an orthogonal set without changing the sign of the matrix whose columns are the basis vectors, can't the same thing be accomplished using elementary operations?

答:
You can swap rows and columns using only the first two types of operations. For simplicity let us consider the vector $(a,b)$. Here are the steps:

Add the second component to the first: $(a+b,b)$

Multiply the first component by $(-1)$: $(-a-b,b)$

Add the first component to the second: $(-a-b,-a)$

Multiply the second component by $(-1)$: $(-a-b,a)$

Add the second component to the first: $(-b, a)$

Multiply the first component by $(-1)$: $(b, a)$


问:
Computing integral of absolute value
Compute $ \int_{0}^{\pi} |$sin$(kx)| \,dx \quad (\forall k \in \mathbb{Z}$).

My question: Can I say $\int_{0}^{\pi} |$sin$(kx)| \,dx = \int_{0}^{\pi} $sin$(kx) \,dx$, because sin$(x)$ is positive on $[0;\pi]$?

答:
No. While $\sin(x)$ is positive on $[0, \pi]$, $\sin(kx)$ is not positive on $[0, \pi]$ (equivalently, $\sin(x)$ is not positive on $[0, k\pi]$) for all $k$. Your logic is fine for $\int_0^\pi |\sin(x)|\;\mathrm{d}x$, but this is not the integral in question.

Instead, note that $|\sin(x)|$ has period $\pi$, and so $|\sin(kx)|$ has period $\pi/k$ (where we assume $k > 0$ for now). Thus, $|\sin(kx)|$ achieves $k$ full periods in the interval $[0, \pi]$, so
$$\int_0^\pi |\sin(kx)| \; \mathrm{d}x = k\int_0^{\pi/k} |\sin(kx)| \; \mathrm{d}x = k\int_0^{\pi/k} \sin(kx) \; \mathrm{d}x = 2.$$
For negative $k$, note that $|\sin(-kx)| = |\sin(kx)|$, so we can simply replace $k$ with $|k|$ and get the same result. For $k = 0$, direct computation shows the integral is $0$.

Title: Computing integral of absolute value


问:
Question on cofactors
If the row sums of a symmetric matrix of size 4 by 4 are all $0$, then why are all the cofactors of the matrix equal?

Thanks in advance for any helpful answers.

答:
There is no reason to consider a $4\times 4$ matrix in particular. When $n=1$, there is only one cofactor so nothing to prove. The result is easy to figure out for $2\times 2$ matrices, and that's maybe a good starting point.

Then one realizes that more generally, as observed by user1551, it suffices that $A$ be an $n\times n$ matrix over an arbitrary field such that the row sums and the column sums be null (the latter being equivalent to the former for symmetric matrices).

In short: we use the adjugate formula $A\cdot\mbox{adj}(A)=(\det A)I_n$ to prove that the rows and the columns of $C=\mbox{adj}(A)^T$, the matrix of cofactors, are all in the span of  $(1,\ldots,1)$, which is the nullspace of $A$ and $A^T$ in the only nontrivial case.

Proof: the row condition means that the sum of the columns of $A$ is null. Whence the columns the columns are linearly dependent and $\mbox{rank} A\leq n-1$. In particular, $\det A=0$.
Let $C$ be the matrix of the cofactors of $A$. Then the transpose $C^T$ is the adjugate of $A$ and, as is well-know $AC^T=(\det A)I_n$. Therefore
$$
AC^T=0\quad\iff \quad \sum_{k=1}^na_{ik}c_{jk}=0\quad \forall 1\leq i,j\leq n.
$$

If $\mbox{rank} A\leq n-2$, then all the cofactors are zero and the result follows. So we assume that $\mbox{rank}\, A=n-1$. That is, the nullspace of $A$ has dimension $1$. But we already know that $v=(1,\ldots,1)$ is in $\ker A$ by assumption. So $\ker A$ is the one-dimensional span of $v$. Note that $\ker A^T$ also has dimension $1$ and is spanned by $v$ as-well by assumption.

Now looking at the above, we see that it means that every row $R_j$ of $C$ is in the nullspace of $A$, whence $R_j=r_jv=(r_j,\ldots,r_j)$.

Since the adjugate matrix of $A^T$ is the transpose of the adjugate matrix of $A$, we get $A^TC=0$ (or simply by taking the tranpose of $C^TA=0$). From that, we deduce that the columns $C_j$ of $C$ are of the form $C_j=c_jv=(c_j,\ldots,c_j)$.

Starting from $c=c_{1j}$ for every $j$ (first row), we see that $c_{ij}=c$ ($j$th column) for every $i$. That is, all the coefficients of the cofactor matrix are equal as desired. QED.

Title: Question on cofactors


问:
Does $\int_0^{2\pi} \sqrt{(1-a\cos x)^2 + b} \mathrm{d}x$ evaluate in terms of elliptic integrals?
Consider the integral

$$\int_0^{2\pi} \mathrm{d}x \sqrt{(1-a\cos x)^2 + b}$$

with $a,b>0$. I believe I can somehow evaluate this integral by incomplete elliptic integrals. This would be obvious, for example, if $a=1$. But, in the general case, this integral appears rather challenging. Mathematica, and SymPy have not been useful. And the Weierstrass substitution returns polynomials of 4th degree.

答:
Here is a step-by-step solution where each step is not too hard.

Step 1: Instead of doing a Weierstrass substitution, substitute $t=\cos x$ and take factors out to get
$$I=2a\int_{-1}^1\frac{\sqrt{(t-1/a)^2+b/a^2}}{\sqrt{1-t^2}}\,dt$$
The roots of the upper quadratic are always complex since $a,b>0$: $c=\frac{1+\sqrt{-b}}a$ and $c^*$.
$$I=2a\int_{-1}^1\frac{\sqrt{(t-c)(t-c^*)}}{\sqrt{1-t^2}}\,dt$$
Step 2: Perform a linear fractional transformation that keeps the integrand's poles at $\pm1$ but shifts the zeros at $c$ and $c^*$ to the imaginary axis, so as to leave only even powers of the polynomial under root. This is the most involved part. Define$\newcommand{Re}{\operatorname{Re}}$
$$A=\frac{1+|c|^2+|c^2-1|}{2\Re(c)}>1$$
$$A_1=A^2-2A\Re(c)+|c|^2>0\qquad A_2=A^2|c|^2-2A\Re(c)+1>0\qquad B=\frac{A_2}{A_1}>0$$
The actual substitution is $u=\frac{At+1}{t+A}$. After taking out more factors
$$I=2a\sqrt{A_1(A^2-1)}\int_{-1}^1\frac1{(u+A)^2}\sqrt{\frac{u^2+B}{1-u^2}}\,du$$
Define $g=2a\sqrt{A_1(A^2-1)}$ and move on.

Step 3: Multiply top and bottom by $(u-A)^2\sqrt{u^2+B}$ to get
$$I=g\int_{-1}^1\frac{(u-A)^2(u^2+B)}{(u^2-A^2)^2}\frac1{\sqrt{(u^2+B)(1-u^2)}}\,du$$
$$=g\left(\int_{-1}^1\frac{(u^2+A^2)(u^2+B)}{(u^2-A^2)^2}\frac1{\sqrt{(u^2+B)(1-u^2)}}\,du-A\int_{-1}^1\frac{(2u)(u^2+B)}{(u^2-A^2)^2}\frac1{\sqrt{(u^2+B)(1-u^2)}}\,du\right)$$
The two integrands are even and odd about zero respectively, so
$$I=2g\int_0^1\frac{(u^2+A^2)(u^2+B)}{(u^2-A^2)^2}\frac1{\sqrt{(u^2+B)(1-u^2)}}\,du$$
We perform a partial fraction decomposition of the rational part of the integrand:
$$I=2g\int_0^1\left(1-\frac{3A^2+B}{A^2-u^2}+\frac{2A^2(A^2+B)}{(A^2-u^2)^2}\right)\frac1{\sqrt{(u^2+B)(1-u^2)}}\,du$$
Step 4: Finally we use elliptic integrals (all arguments here follow Mathematica/mpmath conventions). Byrd and Friedman 213.11 gives$\newcommand{sn}{\operatorname{sn}}$
$$I=\frac{2g}{\sqrt{1+B}}\left(V_0-\frac{3A^2+B}{A^2-1}V_1+\frac{2A^2(A^2+B)}{(A^2-1)^2}V_2\right)$$
where
$$V_k=\int_0^{K(m)}\frac1{(1-n\sn^2u)^k}\,du\qquad m=\frac1{B+1}\qquad n=\frac1{1-A^2}$$
Step 5: B&F 336.00, .01, .02 gives solutions for the $V_k$.
$$V_0=K(m)\qquad V_1=\Pi(n,m)$$
$$V_2=\frac{nE(m) + (m-n)K(m) + (2nm+2n-n^2-3m)\Pi(n,m)}{2(n-1)(m-n)}$$
Putting everything together and simplifying a lot we get the final answer.
$$I=4a\sqrt{-n(A_1+A_2)}(E(m)-K(m)+(1-n)\Pi(n,m))$$


问:
Prove some member of the sequence $7, 77, 777, 7777, \dots$ is divisible by $2019$.

Prove that some member of the sequence $7, 77, 777, 7777, \dots$ is divisible by $2019$.

So far I have figured that as $2019$ is divisible by $3$, then if one of the terms of the sequence
$$
a_{n} = 7\left(\frac{10^{n}-1}{9}\right)
$$
is divisible by $2019$ it is also divisible by $3$. Hence the number of digits in the solution will be a multiple of $3$, i.e. $777, 777777, 777777777, \dots$

I'm not too sure where to go from here any help would be appreciated.

答:
Inspired by a book titled "A walk through combinatorics" by Miklos Bona on page 2 and page 3.

I prove that an even stronger statement is true, in fact, one of the first 2019 elements of the sequence is divisible by 2019.

I start my proof by assuming that its contrary is true. It means that none of the first 2019 elements is divisible by 2019. When being divided by 2019, their remainders must be in a closed set of $[1,2018]$.

As there are 2018 possible remainders for 2019 elements, there are at least two elements with the same remainder.

Let $A_x$ and $A_y$ (with $x<y$) be the $x$-th and $y$-th elements with the same remainder.

For certain integers $\kappa_x$, $\kappa_y$, and $r$,  I can represent $A_x$, $A_y$, and $A_y-A_x$ as follows.
\begin{align}
    A_x &= 2019\kappa_x +r\\
    A_y &= 2019\kappa_y +r\\
    A_y-A_x&= 2019 (\kappa_y-\kappa_x)  \\
   \end{align}
It means that $A_y-A_x$ is divisible by 2019.  It is not the end of proof.

The first $(y-x)$ digits of $A_y-A_x$ are all 7. They are followed by $x$ zeros. Therefore, $A_y-A_x$ can be written as $A_y-A_x=\underbrace{777\cdots777}_{(y-x) \,\text{digits}}\times 10^x$.

As $10^x$ is not divisible by 2019 then  $\underbrace{777\cdots777}_{(y-x) \,\text{digits}}$, which is the $(y-x)$-th element of the given sequence, must be divisible by 2019. It is contradictory to my assumption in the beginning.

The proof by contradiction is complete. Some elements of the given sequence are divisible by 2019.


问:
Solving a Polynomial Equation by Factoring.
$n^3+12n^2+48n+64$
I know the sum of two cubes formula, $(a+b)(a^2-ab+b^2)$. I'm not sure how to apply it here? Any help would be appreciated.

Body:
You were going in the correct direction, as you said:

$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$

If we apply the same here:
$^3+12^2+48+64 = ^3+ 4^3 + 12^2+48 = (n+4)(n^2 - 4n + 16) + 12n(n+4)$

If we take (n+4) common, we get:

$(n+4)(n^2 - 4n + 16 + 12n) = (n+4)(n^2 + 8n + 16) = (n+4)(n+4)^2 = (n+4)^3$

Cheers


问:
What is the area of the shaded part of the circle?
In the diagram below, line segment $AT$ is a diameter of the circle with center $O$. What is the area of the shaded part of the circle?

$AT= 16$.

Half of the circles area is equal to $100.48$, on the other half of the circle there is a triangle spitting it up, The triangle is a right triangle with interior angle measures of $30, 60$, and $90$. There is one side known and that side is the diameter which is equal to $16$, it is the hypotenuse of the triangle.

答: Sides of a 30-60-90 triangle of hypotenuse $16$ are $8$ and $8 \sqrt{3}$.  The area of that triangle is then $(1/2) (8)(8 \sqrt{3}) = 32 \sqrt{3}$.  The shaded area is then the area of the circle minus the area of that triangle, or $64 \pi - 32 \sqrt{3}$.</p>


问:
Proving that the set of all integers cannot be a vector space whatever the fields are
Let Z be the set of all integers. Prove that there doesn't exist a field F and a way to define
a scalar multiplication on Z over F such that Z is a vector space over F (the vector
addition is the usual addition)</p>

答:
Assume there is a field $F$ such that $\mathbb Z/F$ is a vector space.

Let $e$ be the field's multiplicative identity and $\theta$ its additive identity (i.e. its zero).

Denote the elements of $\mathbb Z$ by their integer representation.

Apply the standard vector space principles and integer addition.....

$e.1 = 1$

$1 + 1 = 2 = e.1 + e.1 = (e + e).1$

So, there is some element $\alpha = e + e \in F$ that satisfies $\alpha.1 = 2$, and $\alpha \ne \theta $ because $\theta.1 = 0 $

Then $\alpha $ has a field inverse $\beta = e/\alpha$

By closure then $\beta.1 \in \mathbb Z$ and is therefore an integer.

Also, $ 1 = e. 1 = \alpha.\beta. 1 = (e + e).\beta.1 = e.\beta.1 + e.\beta.1 = \beta.e.1 + \beta.e.1 = \beta.1 + \beta.1$

So, $\beta.1$ must be an integer, say $p$, which satisfies $p + p = 1$

This is a contradiction, so there is no such field.


问:
What is the ring $\mathbb{Z}[a]$ for a constant number $a$?
I know that the notation $\mathbb{Z}[x]$ is the notation for the ring of polynomials with coefficients in $\mathbb{Z}$, however I am confused with the notation $\mathbb{Z}[a]$ where $a$ is just a number, e.g. $a=1/3$.

Is this saying that one of the roots of the polynomials is $1/3$, i.e. $f(x)=(1-1/3)\cdot g(x)$, where $g(x)$ is in $\mathbb{Z}[x]$?

答:
You can think of $\Bbb Z[\alpha]$ as the evaluation of every $f\in\Bbb Z[X]$ at $\alpha$, that is, $\Bbb Z[\alpha]=\{f(\alpha)\mid f(X)\in\Bbb Z[X]\}$, even though $\alpha$ is not necessarily an element of the coefficient ring.

In your case, the elements of $\Bbb Z[1/3]$ are expressions like $a_0+\frac{a_1}3+\frac{a_2}{3^2}+\dots+\frac{a_n}{3^n}$, for some $n\in\Bbb N$.

Another example would be $\Bbb Z[\sqrt3]=\{a+b\sqrt3\mid a,b\in\Bbb Z\}$. Notice that even powers of $\sqrt3$ are integers, and odd powers are the product of an integer and $\sqrt3$, thus we only need two terms to write these expressions.

Same goes for another classic example, the Gaussian integers $\Bbb Z[i]=\{a+ib\mid a,b\in\Bbb Z\}$.


问:
$\lim_{n\to\infty} \sqrt[n]{\sum_{k =0}^n \binom{n}{k}^t} .$
I'm trying to find the following limit for $t\geq1$

$$\lim_{n\to\infty} \sqrt[n]{\sum_{k =0}^n \binom{n}{k}^t}.$$

I suspect this can be solved using generating functions but did not make any headway.

Looking at this thread, Combinatorial proof of summation of $\sum\limits_{k = 0}^n {n \choose k}^2= {2n \choose n}$.
<s>I'm guessing the answer to be $t$.</s> As pointed out by Gary, the limit is $4$ when $t = 2$ and the appropriate guess would be $2^t$.

答:
Assume that $n$ is even (we can deal with odd cases as well, but it's a little messier), writing $n=2m$. Then, since for all $0\leq k\leq 2m$ we have
$\binom{2m}{k}\leq \binom{2m}{m}$, we get
$$
\binom{2m}{m}^t\leq \sum_{k=0}^{2m} \binom{2m}{k}^t \leq (2m+1)\binom{2m}{m}^t \tag{1}
$$
and, using standard estimates for the Binomial coefficient (Stirling's inequality) we have
$$
\binom{2m}{m} \operatorname*{\sim}_{m\to\infty} \frac{C}{\sqrt{m}}\cdot 2^{2m} \tag{2}
$$
for some constant $C>0$. Combining (1) and (2) and taking the $2m$-th root, we get by the squeeze theorem that, as $m\to\infty$, the limit is $\boxed{2^t}$.


问:
Proving a factorization of ideals in a Dedekind Domain
Let $R=\mathbb{Z}[\sqrt{-13}]$. Let $p$ be a prime integer, $p\neq 2,13$ and suppose that $p$ divides an integer of the form $a^2+13b^2$, where $a$ and $b$ are in $\mathbb{Z}$ and are coprime. Let $\mathfrak{P}$ be the ideal generated in $R$ by $p$ and $a+b\sqrt{-13}$ and $\overline{\mathfrak{P}}$ the ideal generated by $p$ and $a-b\sqrt{-13}$:
$$\mathfrak{P}=(p,a+b\sqrt{-13})\qquad;\qquad\overline{\mathfrak{P}}=(p,a-b\sqrt{-13}).$$

I want to show that $\mathfrak{P}\cdot\overline{\mathfrak{P}}=pR$.

I have already proved one inclusion: certainly $\mathfrak{P}\cdot\overline{\mathfrak{P}}$ can be generated by $p^2,p(a\pm\sqrt{-13}),a^2+13b^2$ which are all $p$-multiples, hence $\mathfrak{P}\cdot\overline{\mathfrak{P}}\subseteq pR$.

For the reverse inclusion, I am tring to express $p$ in terms of product of elements of the two ideals, but with no results up to now. Any help would be appreciated, thanks.

答:
Let $\ w = a\!+\!b\sqrt{-13}.\:$ $\ \color{#c00}{p\mid ww' = a^2\!+\!13b^2}\,$ so $\,(p,w)(p,w')=(p)\,$ by the Lemma below, since $\,\color{#0a0}{(p,w\!+w') = (p,2a) = 1},\,$ else $\,\color{#c00}{p\mid a\Rightarrow p\mid b},\,$ contra $(a,b)=1,\,$ using $\,p\ne 2,13.\ \ $ QED

$\begin{eqnarray}{\bf Lemma}\,\ \ \color{#c00}{p\mid ww'},\ \color{#0a0}{(p,w\!+\!w')=1}\,\Rightarrow\, (p,w)(p,w')\! &=\,& (p^2,pw,pw',ww') \\ &=\,&\  \ p\ (p,w,w',\color{#c00}{ww'/p}) \\ &=\,& (p)\ \ by\ \ (p,w,w')\supset \color{#0a0}{(p,w\!+\!w') = 1}\end{eqnarray}$

Remark $\ $ It is very helpful to become adept with the above form of ideal (and gcd) calculation since it is ubiquitous in number theory, from elementary results in $\Bbb Z$ such as Euler's four number theorem (or Riesz/Schreier refinement), to analogs of the above in higher-degree number rings, such as the classical Kummer-Dedekind Criterion for ideal factorization of rational primes.


答:
The graph of $|sin(kx)|$ and $sin(kx)$ will look something like this(for simplicity).Upward graph is of $|sin(kx)|$

On Introducing $(kx)$ instead of $x$ in sine function, the graph will cut $x$-axis $(k-1)$ times more than usual graph of $sinx$ in interval $[0,\pi]$ and that to symmetrically. After introducing modulus you have to take image of graph below $x$-axis about $x$-axis itself that will make graph similar to above.

So you have find area under graph in interval $[0,\pi]$.Notice that area under each small branches are equal.

$$\int_{0}^{\pi}|sin(kx)|dx=\int_{0}^{\frac{\pi}{k}}|sin(kx)|dx+\int_{\frac{\pi}{k}}^{\frac{2\pi}{k}}|sin(kx)|dx+\int_{ \frac{2\pi}{k}}
 ^{\frac{3\pi}{k}}|sin(kx)|dx+.....+\int_{ \frac{(k-1)\pi}{k}}^{\frac{k\pi}{k}}|sin(kx)|dx$$

Since value of all integrals are equal(Area under each small branches are equal)

$$\int_{0}^{\pi}|sin(kx)|dx=k\int_{0}^{\frac{\pi}{k}}|sin(kx)|dx$$

Also In interval $\Big[0,\frac{\pi}{k}\Big]$ ,$sin(kx)$ is positive as it is first branch of graph. Similarly in next brach i.e$\Big[\frac{\pi}{k},\frac{2\pi}{k}\Big]$ $sin(kx)$ is negative. So sign of $sin(kx)$ changes alternatively from $+ $ to $-$.

$$\int_{0}^{\pi}|sin(kx)|dx=k\int_{0}^{\frac{\pi}{k}}sin(kx)dx$$

This gives $$\int_{0}^{\pi}|sin(kx)|dx=2$$

Title: Computing integral of absolute value

Body:
Compute $ \int_{0}^{\pi} |$sin$(kx)| \,dx \quad (\forall k \in \mathbb{Z}$).

My question: Can I say $\int_{0}^{\pi} |$sin$(kx)| \,dx = \int_{0}^{\pi} $sin$(kx) \,dx$, because sin$(x)$ is positive on $[0;\pi]$?


问:
Find $a$ such that the piecewise function $f $ is injective and $a$ such that $f$ is surjective
Let $f(x)= \begin{cases}
      ax+2, & x\leq 1 \\
       \
      x+2a, & x\gt 1
   \end{cases}
$ where $a\in\mathbb{R}$

Find $a$ such that:

$f$ is injective

$f$ is surjective

EDIT: I need $a$ so $f$ is injective first and then surjective, not both at the same time.

I usually study injectivity and surjectivity with the first derivative, but in this case I do not think i could use that because i would have to assume the function is differentiable at $x=1$. So I tried taking 3 cases for injectivity:

$f(x) = f(y)$ where $x\leq1$ and $y\leq1$

$f(x) = f(y)$ where $x\gt1$ and $y\gt1$

$f(x) = f(y)$ where $x\leq1$ and $y\gt1$

The first 2 cases were easy to solve but i am stuck at the third case as I have too many unknowns and I do not know what to do.

For surjectivity, I tried taking limits at $+\infty$, $1$ and $-\infty$ to figure out the Image of $f$, but aside from limit to $+\infty$ where i get $+\infty$, I still have $a$ in the result for the other limits.

How should i solve this problem ?

Thanks in advance !

答:
For injectivity and surjectivity, I advise finding the range of $f$ when restricted to $(-\infty, 1]$ and when $f$ is restricted to $(1, \infty)$.

The union of these two ranges produces the range of $f$. So, $f$ is surjective if and only if the union of these ranges is $\Bbb{R}$.

If these ranges intersect, then the function is not injective (since there is some function value attained both in $(-\infty, 1]$ and $(1, \infty)$, contradicting injectivity). Moreover, $f$ is injective on both subdomains and these ranges fail to intersect if and only if $f$ is injective over $\Bbb{R}$.

So, let's consider these two ranges. Let $R_1 = f(-\infty, 1]$ and $R_2 = f(1, \infty)$. Considering $R_1$ first, we need to consider certain possibilities for $a$. If $a = 0$, then $f(x) = 2$ for all $x \le 1$, so $R_1 = \{2\}$. If $a > 0$, then $f$ is increasing on $(-\infty, 1]$, and tends to $-\infty$ as $x \to -\infty$, so $R_1 = (-\infty, a + 2]$. Finally, if $a < 0$, then $f$ is decreasing, tending to $\infty$ as $x \to -\infty$, giving us $R_1 = [a + 2, \infty)$. In summary,
$$R_1 = \begin{cases} (-\infty, a + 2] & \text{if } a > 0 \\ \{2\} & \text{if } a = 0 \\ [a + 2, \infty) & \text{if } a < 0. \end{cases}$$

On the other hand, $f$ is always increasing on $(1, \infty)$, with $f(x) \to \infty$ as $x \to \infty$, so
$$R_2 = (1 + 2a, \infty).$$

Where is $f$ surjective? Certainly, if $a \le 0$, then both $R_1$ and $R_2$ are intervals with a lower bound, thus they cannot cover all of $\Bbb{R}$. So, we are considering $a > 0$ from the get-go. In order for the two intervals $(-\infty, a + 2]$ and $[1 + 2a, \infty)$ to cover $\Bbb{R}$, we need $a + 2 \ge 1 + 2a \iff a \le 1$. So, $f$ is surjective if and only if $0 < a \le 1$.

Now, $f$ is always injective on $(1, \infty)$, and $f$ is injective on $(-\infty, 1]$ if and only if $a \neq 0$. Note that if $a < 0$, then both intervals tend are unbounded above, and thus non-trivially intersect. This implies $f$ is not injective if $a \le 0$.

If $a > 0$, then we need the two intervals not to intersect, which occurs precisely when $a + 2 \le 1 + 2a \iff a \ge 1$.

Thus, in conclusion, $f$ is injective when $a \in [1, \infty)$ and is surjective when $a \in (0, 1]$. It is bijective precisely when $a = 1$.


问:
The relation $[X,Y]=fX+gY$ to be established for vector fields $X$ and $Y$ on a smooth manifold $M$.
Why would smooth functions f and g exist so that $[X,Y]=fX+gY$ for vector fields $X,Y$ on a manifold $M$ of dimension $n$?
One starts to think of expanding the Lie bracket $[X,Y]$ in terms of the local expression of the vector fields $X=\sum\limits_{j=1}^n x^j(p)\frac{\partial}{\partial x_j}$, but how properly the functions $f$ and $g$ come out when
$$[X,Y](fg)=X(Y(fg))-Y(X(fg))= \cdots \quad
\text{is used}$$ Perhaps not the right approach, please advise me.

答:
As higlighted by @Yuval , the expression you are trying to prove is not true in general. In particular, we can say that it is true if and only if (by definition) the tangent distribution generated by the vector fields $X$ and $Y$ is involutive.

By distribution we mean the union, over all the points $m\in M$, of subspaces of the tangent spaces $T_mM$. In particular, in your case you have the distribution $\mathcal{D} := \cup_{m\in M}span\{X|_m,Y|_m\}$.

A simple example of non-involutive distribution on $M=\mathbb{R}^3$ is
$$ X=\partial_x,\quad Y = x\partial_z + \partial_y, $$
since if you compute $[X,Y]$ you get $\partial_z = [X,Y]$, which can not be expressed as a linear combination of the original fields $X$ and $Y$, and hence it does not belong to the tangent distribution generated by them.

However, in many books or online you can find many more examples.