答:
Let's start with a theorem:

Suppose $f : [a, b] \to \Bbb{R}$ is continuous and differentiable on $(a, b)$. Further, let $c, d \in [a, b]$ such that $f(c) = \min f$ and $f(d) = \max f$. Then $c = a$, $c = b$, or $f'(c) = 0$ (and similarly for $d$).

In other words, the global maximum/minimum must either occur at a stationary point, or at an endpoint of the interval.

In this case, we have a function defined on the interval $[5, 8]$. You have already computed the stationary point $x = 6.5$ (beware, your derivative is missing factors of $1/2$, but they would have cancelled out when solving $f'(x) = 0$ anyway). So, our maximum and minimum must occur at $5$, at $8$, or at $6.5$. So, we just need to figure out which has the largest function value, and which has the least function value.

We have $f(6.5) = \sqrt{6}$ as you said, and $f(5) = f(8) = \sqrt{3}$. So, we have an absolute minimum of $\sqrt{3}$ and an absolute maximum of $\sqrt{6}$.

That said, even though you found the maximum instead of the minimum, $\sqrt{6}$ is still correct. It means that there is one and only one value of $y$ such that $\sqrt{y - 5} + \sqrt{8 - y} \ge \sqrt{6}$, but one and only one value of $y$ is still a solution! Anything larger than $\sqrt{6}$, and there'd be no solution.

The theorem applies to maximising/minimising a function on a closed and bounded interval. If you have open ends, or infinite ends, you can compute the limit(s) of the function as it approaches these end(s). So, instead of comparing the function value at the endpoint, compare the limit(s) at these end(s) to the function value at the stationary points.

(Note that I used neither the first nor second derivative tests to determine global maxima/minima. These tests are for testing local maxima/minima. While every global maximum/minimum is also local, these tests are useless for determining global maximma/minima.)


问:
The probabilities of components A, B, C, D and E working are 0.7, 0.7, 0.8, 0.8 and 0.8, respectively. What is the probability that:

An electrical circuit has 5 components labelled A, B, C, D and E. The probabilities of components A, B, C, D and E working are 0.7, 0.7, 0.8, 0.8 and 0.8, respectively. What is the probability that:

(a) The system will be working.

(b) Component A will not be working given that the system is working.

(c) Component A will not be working given that the system is not working.

I solved but I'm not sure it is correct.

(a)
W is event that the system will be working
\begin{align}
 & P(W) \\
={}& P(A \cap B \cup C \cap D \cap E)\\
={}& P(A \cap B) + P(C \cap D \cap E) - P(A \cap B \cap C \cap D \cap E)\\
={}& 0.7^2 + 0.8^3 - 0.7^2 0.8^3\\
={}& 0.75112
\end{align}

(b)
\begin{align}
 & P(\bar{A}|W) \\
={}& \frac{P(\bar{A} \cap W)}{P(W)}\\
={}& \frac{P(\bar{A}BCDE \cup \bar{A} \bar{B}CDE)}{P(W)}\\
={}& \frac{0.3\times0.7\times0.8^3 + 0.3^2\times0.8^3}{0.75112}\\
={}& \frac{1920}{9389}
\end{align}

Is my answer for (a) and (b) correct?

How to solve (c)?

答:
The figure gives $P(C)=0.8$ but the question says $P(C)=0.7$.

Your answers are correct according to the figure but there are some typos: in $(a)$ it should be $P(A\cap B\cup C\cap D\cap E)$ as taking union of probabilities does not make sense. In $(b)$, judging from your calculation and assuming $XY$ is shorthand for $X\cap Y$, you probably meant to have $P(\bar ABCDE\color{red}\cup\color{blue}{\bar A\bar B}CDE)$ in the numerator. Note that $\overline{AB}\ne\bar A\bar B$.

ContentLicense: CC BY-SA 4.0


问:
Maximum value of $k$ so that the inequality $\sqrt{y-5}+\sqrt{8-y} \ge k$ should have answer

Suppose $k$ be a real number so that the inequality
$\sqrt{y-5}+\sqrt{8-y} \ge k$ should have answer. what is the maximum
possible value of $k$?

$1)\sqrt{6}-1\quad\quad2)\frac12+\sqrt{3}\quad\quad3)\sqrt6\quad\quad4)2\sqrt2\quad\quad5)\sqrt6+1$

So I translated the question into finding the minimum value of $f(y)$ where $f(y)=\sqrt{y-5}+\sqrt{8-y}\quad$.(because for $k=\min(f(y))$ the inequality certainly has answer.) therefore we should solve  $f'(y)=0$:

$$f'(y)=\frac{1}{\sqrt{y-5}}-\frac{1}{\sqrt{8-y}}=0$$
$$y-5=8-y\rightarrow\quad y=6.5$$

And $f(6.5)=2\sqrt{\frac32}=\sqrt{6}$. so the answer is third choice.

Is my answer right? I know to find maximum or minimum of a function we take derivative of it and equate it to $0$. so here I don't know why $f(6.5)$ is minimum. can you please explain it to me?

EDIT:

I just noticed that $\sqrt6$ is not the minimum because $f(5)=f(8)=\sqrt3<\sqrt6$ maybe it is maximum! so I conclude the answer to this problem should be $\sqrt3$ but I don't have this option in the choices

答:
$k$ is the maximum of the function $\sqrt{y-5}+\sqrt{8-y}$ on $[5,8]$.

Using Cauchy-Schwarz you get immediately

$$\sqrt{y-5}+\sqrt{8-y} \leq \sqrt 2 \sqrt{y-5+8-y} = \sqrt 6$$

Equality is reached for $y-5 = 8-y \Leftrightarrow y =\frac{13}{2}$.

So, $\boxed{k=\sqrt 6}$.


答:
Since $\sqrt x$ is concave, via Jensen's inequality we have
$$k \le \sqrt{y-5}+\sqrt{8-y} \le 2 \sqrt{\frac{(y-5)+(8-y)}{2}} = \sqrt 6$$

Title: Maximum value of $k$ so that the inequality $\sqrt{y-5}+\sqrt{8-y} \ge k$ should have answer

Body:
Suppose $k$ be a real number so that the inequality
$\sqrt{y-5}+\sqrt{8-y} \ge k$ should have answer. what is the maximum
possible value of $k$?

$1)\sqrt{6}-1\quad\quad2)\frac12+\sqrt{3}\quad\quad3)\sqrt6\quad\quad4)2\sqrt2\quad\quad5)\sqrt6+1$

So I translated the question into finding the minimum value of $f(y)$ where $f(y)=\sqrt{y-5}+\sqrt{8-y}\quad$.(because for $k=\min(f(y))$ the inequality certainly has answer.) therefore we should solve  $f'(y)=0$:

$$f'(y)=\frac{1}{\sqrt{y-5}}-\frac{1}{\sqrt{8-y}}=0$$
$$y-5=8-y\rightarrow\quad y=6.5$$

And $f(6.5)=2\sqrt{\frac32}=\sqrt{6}$. so the answer is third choice.

Is my answer right? I know to find maximum or minimum of a function we take derivative of it and equate it to $0$. so here I don't know why $f(6.5)$ is minimum. can you please explain it to me?

EDIT:

I just noticed that $\sqrt6$ is not the minimum because $f(5)=f(8)=\sqrt3<\sqrt6$ maybe it is maximum! so I conclude the answer to this problem should be $\sqrt3$ but I don't have this option in the choices


问:
Find $\lim_{n \rightarrow \infty} \int_0 ^{\infty} \left(1+ \frac{x}{n}\right)^n e^{-ex} dx$
Find $$\lim_{n \rightarrow \infty} \int_0 ^{\infty} \left(1+ \frac{x}{n}\right)^n e^{-ex} dx.$$

I know that $\lim_{n \rightarrow \infty}  \left(1+ \frac{x}{n}\right)^n = e^x.$

Also, I showed that $ \left(1+ \frac{x}{n}\right)^n \leq e^x$ for all $x \geq -n$.

So I'm thinking about the Lebesgue Dominated Convergence Theorem. But I'm not sure how to deal with the $e^{-ex}$ part.

答:
$$\int \left(1+ \frac{x}{n}\right)^n e^{-ex}\, dx= {n^{-n}}\int(x+n)^ne^{-ex}\, dx$$
$$\int (x+a)^b e^{-c x}\, dx=-e^{a c} c^{-b-1} \Gamma (b+1,c (x+a))$$
$$\int_0^\infty (x+a)^b e^{-c x}\, dx=e^{a c} c^{-b-1} \Gamma (b+1,a c)\quad \text{if}\quad \Re(c)>0\land \Re(a)>0$$ All of that makes
$$I_n=\int_0^\infty  \left(1+ \frac{x}{n}\right)^n e^{-ex}\, dx=e^{(e-1) n-1}\, n^{-n} \,\Gamma (n+1,e n)$$

Have a look here [at $§8.11(iii)$] for $\lambda=e$.


问:
How to sum the series $\sum_{n=4}^{\infty}\frac{1}{(2n-7)(2n-6)}$
I have been tinkering with telescoping sum and I've encountered by chance this series:

$$\sum_{n=4}^{+\infty}\dfrac{1}{(2n-7)(2n-6)}=\log(2)$$

I use Wolfram to sum this series because I am not sure if it is a telescoping sum. I wonder how can you sum this series? By converting the infinite sum to an improper integral?

I integrate this function:

$$\int_{4}^{\infty}\dfrac{1}{(2n-7)(2n-6)}dx=\dfrac{\log(2)}{2}$$

I am new to improper integration and numerical infinite series. I thought the sum of the series and the result of the integral should match, but they don't.

It seems that I miss something very basic and crucial. I am learning myself Cal 3 so I may miss some crucial theorems. What are conditions for the sum of the series and the limit of the integral to be the same?

Come to think of it, summation is just a discrete version of integration, so the values naturally cannot be the same.

Is there a way to sum this series and obtain its sum? This doesn't seem to be a telescopic series in my humble opinion.

答:
$$\frac{1}{(2n-6)(2n-7)}=\frac{1}{2n-7}-\frac{1}{ 2n-6  }$$
so your series is the Taylor series for $\log(1+x)=\sum_{k \ge 1} (-1)^{k-1} x^k/k$ evaluated at $x=1$.


问:
Where is the mistake in a point closest to cross section of two planes and a line problem
The problem:

Find the point that lies on the line that is the intersection of two planes:

plane one: $$3x-2y+z = 2$$

plane two: $$x+y+2z = 4$$

The point also  needs to be the same distance away from points $A(1,-1,-1),B(1,-1,3)$

My try:
First I found the direction of the line that is the intersection of two planes. Here I used cross product between $(3,-2,1)$ and $(1,1,2)$, which are the normals of the two planes. I got $s = (-1,-1,1)$ That is the direction of the intersection line.

What I did next was I solved the system of equations: $$3x-2y+z-2 = 0\\x+y+2z-4 = 0$$

I got: $z = 7y +2$ and $x = \frac{-5y}{3}$

Here I chose $y$ to be equal to $1$ $\implies y = 1 \implies x = \frac{-5}{3} \implies z = 9$

So the equation of the intersection line is: $(1,\frac{-5y}{3}, 9) + t(-1,-1,1) = (x,y,z)$

Then I wanted to find the plane where all the points that have the same distance from $A(1,-1,-1),B(1,-1,3)$ are located. it is trivial this plane's normal will be the direction vector $AB$, thus the normal is $n=(0,0,4)$

Then I wanted to position the plane($O$...origin) : $$OA + \frac{1}{2}AB = (1,-1,-1) +\frac{1}{2}(1,-1,3) = (\frac{3}{2},\frac{-3}{2},\frac{1}{2}) = \text{\{This is the point to position the plane\}}$$

So the equation of this plane is: $4z = 4*\frac{1}{2}=2$

I got $z_0$ form the last coordinate from the point.

Then I Solved for intersection between the first line(intersection between plane one and two) and this last plane. I got the wrong answer: $x = \frac{19}{2}, y = \frac{41}{6}, z = \frac{1}{2}$ I got this results by inserting parametrically defined $z$ into the plane equation. My answer is wrong.

The correct solution is: $T(1,1,1)$

Where did I go wrong?

答:
When you got  $z = 7y +2$ and $x = \frac{-5y}{3}$ (which, I believe, isn't correct), that describe the line of intersection right there. Finding the cross product of the normals is needlessly complicated.

Given the points $A(1,-1,-1),B(1,-1,3)$, you can find the plane of equidistance through the methods discussed here: Equation of a plane equidistant from two points $A$ and $B$?

Taking $N = B-A=(0,0,4)$ and $(A+B)/2= (1,-1,1)$, we get $0x+0y+4z=4$, which gives $z=1$. Now you have the simultaneous equations

$$3x-2y+z = 2$$
$$x+y+2z = 4$$
$$z = 1$$


问:
Are there counter intuitive interpretations of ZF or ZFC?
Are there interpretations of ZF or ZFC that are non standard in the sense that $\epsilon$ is interpreted in a counter intuitive way that intuitively has nothing to do with "belongs to" or "is part of" (apart from formally modelling the axioms of ZFC)?

答:
This is very late to the party, but I think it's worth noting that there are in fact specific obstacles to finding "natural" nonstandard models of set theories, coming from computability theory. To start with, we have:

There is no computable model of $\mathsf{ZF}$.

(The argument here works for other set theories like $\mathsf{NFU}$ as well, it's quite general.)

That is, there is no computable binary relation $R$ on $\mathbb{N}$ such that $(\mathbb{N},R)\models\mathsf{ZF}$. And things get even worse if we demand "nice" models: there is no hyperarithmetic model of $\mathsf{ZF}$ with no "false natural numbers" (= $\omega$-model), and the situation is even nastier for fully well-founded models (well-founded models are guaranteed to be $\omega$-models, but not conversely).

If we drop the axiom of infinity, things get better per Olivier Begassat's answer above. However, they don't get much better: Tennenbaum's theorem winds up implying that there are no other computable models of $\mathsf{ZFC}$-$\mathsf{Inf}$. (Interestingly, $\mathsf{ZF}$-$\mathsf{Inf}$ is a more complicated beast - see here.)

So while the downward Lowenheim-Skolem theorem tells us that if $\mathsf{ZFC}$ is consistent then it has a very "unintended" model (namely a countable one), we aren't going to find any particularly simple models.

ContentLicense: CC BY-SA 3.0


问:
Numbers of the kind $0.aaa\ldots =\frac{1}{aaa\ldots a}$
What are the integers $a$ such that
$$
0.\underbrace{aaa\ldots}_{\infty\text{ times}} =\frac{1}{\underbrace{aaa\ldots}_{k\text{ times}}}
$$

eg.
$$
0.333\ldots=1/3\\
$$
while
$$
0.1616\ldots\ne1/16\ne1/1616\ne1/\underbrace{161616\ldots}_{\text{any }k}
$$

I was able to reduce $a$ to
$$
a=\frac{10^d-1}{\sqrt{10^{k\,d}-1}}
$$

where
$d$ is the number of digits in $a$ in base $10$. If this is correct, is $k=1,d=1$ the only solution for integral $a$?

答:
I think a complete answer should cover all bases (no pun intended). We will have:

$$a = \frac {b^d-1}{\sqrt{b^{kd}-1}}$$

for any base $b$. However for $k\ge 2$:

$$\sqrt{b^{kd}-1}\ge \sqrt{b^{2d}-1} = \sqrt{(b^d-1)(b^d+1)} \ge b^d-1$$

forcing $a<1$. Hence we must have $k=1$, with $a = \sqrt{b^d-1}$, or $b^d - a^2 = 1$.

By Catalan's conjecture (or Mihăilescu's theorem, or if you are interested, this $1+a^2 = b^d$ is a special case proven by V. A. Lebesgue using Gaussian integers), if $d \ge 2$ there are no solutions. This forces $d = 1$, and we have only the uninteresting case:

$$1+a^2 = b$$

so our choice of $b$ must be one more than some square. For base $10$, $10 = 1+3^2$, so $0.\dot3 = \dfrac13$.


答:
The long and winding road. We will prove the result, only using that $H_N = \log N +\gamma + o(1)$, where $H_N$ is the Harmonic sum and $\gamma$ the Euler constant.

First, perform a change of index to get something more reasonable:
$$
\sum_{n=4}^{\infty}\frac{1}{(2n-7)(2n-6)}
= \sum_{k=0}^{\infty}\frac{1}{(2k+1)(2k+2)}
$$
Here, you may recognize that, for $k\geq 0$,
$$
\frac{1}{(2k+1)(2k+2)} = \frac{1}{2k+1}-\frac{1}{2k+2}
$$
and now you can use the following:
$$
\sum_{k=0}^{n}\left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)
= \sum_{k=0}^{n}\left(\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}\right)
= \sum_{k=1}^{2n+2}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{k}
= H_{2n+2}-H_n
$$
Using now the fact that $H_{N} = \log N + \gamma +o(1)$, you get
$$
\sum_{k=0}^{n}\left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)
= H_{2n+2}-H_n
 = \log(2n+2) - \log n + o(1) = \log 2 + o(1)
$$
since $\log(2n+2) = \log 2 + \log (n+1) = \log 2 + \log n + o(1)$.

Taking $n\to \infty$, you get
$$
\sum_{k=4}^{\infty}\frac{1}{(2k-7)(2k-6)}
= \sum_{k=0}^{\infty}\frac{1}{(2k+1)(2k+2)} = \lim_{n\to\infty}\sum_{k=0}^{n}\left(\frac{1}{2k+1}-\frac{1}{2k+2}\right) = \log 2
$$
as desired.

Title: How to sum the series $\sum_{n=4}^{\infty}\frac{1}{(2n-7)(2n-6)}$

Body:
I have been tinkering with telescoping sum and I've encountered by chance this series:

$$\sum_{n=4}^{+\infty}\dfrac{1}{(2n-7)(2n-6)}=\log(2)$$

I use Wolfram to sum this series because I am not sure if it is a telescoping sum. I wonder how can you sum this series? By converting the infinite sum to an improper integral?

I integrate this function:

$$\int_{4}^{\infty}\dfrac{1}{(2n-7)(2n-6)}dx=\dfrac{\log(2)}{2}$$

I am new to improper integration and numerical infinite series. I thought the sum of the series and the result of the integral should match, but they don't.

It seems that I miss something very basic and crucial. I am learning myself Cal 3 so I may miss some crucial theorems. What are conditions for the sum of the series and the limit of the integral to be the same?

Come to think of it, summation is just a discrete version of integration, so the values naturally cannot be the same.

Is there a way to sum this series and obtain its sum? This doesn't seem to be a telescopic series in my humble opinion.


答:
You can use partial fractions. $\frac{1}{(2n-7)(2n-6)} = \frac{A}{2n-7} + \frac{B}{2n-6}$
If you combine you get A(2n-6) + B(2n-7) = 1, which must be true for all n>3.
When n = 4, 2A + B = 1
when n = 5, 4A + 3B = 1
Solving, we find A = 1 and B = -1, so
$\frac{1}{(2n-7)(2n-6)} = \frac{1}{2n-7} - \frac{1}{2n-6}$
which is an alternating series, in fact the alternating harmonic series.  I have memorized the fact that this series sums to ln(2), and I can prove it converges by the Alternating Series Test, but I don't recall offhand how to prove that it has that particular limit. In general, only telescoping series and geometric series can be summed exactly without advanced techniques that are not taught in calculus classes.


问:
show $(1+\frac{x}{n})^n \leq e^x$
I want to show that $(1+\frac{x}{n})^n \leq e^x$ for all $x \geq -n$, $n\in \mathbb{N}$.

I already showed that positive case ($x>0$) and zero case $(x=0)$.

Not sure how to approach the negative case, because in the case $x>0$ I used that $x^n$ is an increasing function on $(0, \infty)$. But I can't do the same when $x$ is negative.

答:
Assuming you know that
\begin{align}
\lim_{n\rightarrow \infty}\left( 1+ \frac{x}{n}\right)^n = e^x,
\end{align}
then it suffices to prove that
\begin{align}
\left( 1+ \frac{x}{n}\right)^n \le \left( 1+ \frac{x}{n+1}\right)^{n+1}
\end{align}
for all $n \in \mathbb{N}$.

By Bernoulli's inequality, we have that
\begin{align}
\frac{\left( 1+ \frac{x}{n+1}\right)^{n+1}}{\left( 1+ \frac{x}{n}\right)^{n+1}} =&\ \left( 1+\frac{-\frac{x}{n(n+1)}}{1+\frac{x}{n}}\right)^{n+1} = \left( 1-\frac{x}{n(n+1)+(n+1)x}\right)^{n+1} \\
 \ge&\ 1-(n+1)\left(\frac{x}{n(n+1)+(n+1)x} \right) = 1-\frac{x}{n+x} = \frac{1}{1+\frac{x}{n}}.
\end{align}
Multiplying both side by $(1+x/n)^{n+1}$ yields the desired inequality.

Remark: This is a standard proof given in a first course of analysis without assuming any knowledge of derivatives.

Title: show $(1+\frac{x}{n})^n \leq e^x$


问:
Abstract Geometry?
Are there similar terms in other areas for the idea the "angle" conveys in geometry ? I find that functions for abstract things such as pressure,electrical currents ( nothing geometric there ) on AC circuits are described by the sin function ( argument is an angle, there fore an geometric entity ) .Do we really have that abstract things are necessarily described by geometric entities ?   I understand that we can relate time to angle with the concept of angular frequency, but it seems too unintuitive and unnecessary to have to relate a sin-like variation generally ( a variation that decelerates smoothly towards the peaks and accelerates smoothly towards the center ) with geometry and angles.

Maybe the concept of angle we know is a simply an example of a greater idea ( like a  a measure of difference between instances of the same thing ) applied to geometry or am i tripping ?

答:
You tagged this question as differential equations, and that gives part of the answer. Many physical objects behave like little springs near equilibrium (like a pendulum almost at rest, a man falling through a hole to the center of the earth, a wobbly top, etc.). One term for this is the harmonic oscillator. Such things can often be described by second order linear differential equations, whose solutions are either polynomial, exponential, involve sin and cos, or a mix of the three.

The not way that the solutions get oscillation, then, is if sin or cos are involved. Thus, sin and cos are, to a very good approximation, the most common kind of wave-like variation in the universe.

Title: Abstract Geometry?


问:
Notation for Range for a Function
I was given a question where I was asked to determine if a function is one-to-one and/or onto. It was given like this:

$\mathbb{Z}^+\to\mathbb{Z}^+,f\left(x\right)=\left(\frac{x}{5}\right)-4$

In this case, what would the $\mathbb{Z}^+\to\mathbb{Z}^+$ mean? Does the first $\mathbb{Z}^+$ signal the range for $x$ and the second $\mathbb{Z}^+$ signal the range for $f\left(x\right)$?

答:
When we say a function is defined by $f : A \to B$, we mean it takes elements of $A$ as inputs, and maps them to some subset of $B$. Sometimes $B$ is precisely equal to the range of $f$, and sometimes $B$ is larger.

In your case, the notation $f : \Bbb Z^+ \to \Bbb Z^+$ means that $f$ takes in positive integers, and maps them to some subset of the positive integers. Though unless something significant is meant by the parentheses in your question, this is also problematic - $1 \in \Bbb Z^+$ but $f(1) < 0$ if we assume they're just normal parentheses, and thus $f(1) \not \in \Bbb Z^+$. Every output needs to be housed in $\Bbb Z^+$ or a subset thereof.

Title: Notation for Range for a Function


问:
Proving $\tan 3^{\circ}$ is irrational
I want to show $\tan 3^{\circ}$ is irrational I have looked at other similar questions but they all have different patterns to  approach trignometric values like $\tan3$ so what I am thinking of is assuming $tan 3$ is rational that is to say $\tan 3 = a/b$ so this means $\sin 3^{\circ} / \cos 3^{\circ} = a/b$ so we have $\sin 3^{\circ}b= \cos 3^{\circ} a$ so from here I want to find a contradiction but I need help.

答:
Assume $\tan(3^{\circ})$ is rational that is $\exists a,b\in\mathbb{N}$, $b\neq0$ and $\gcd(a,b)=1$ such that $\displaystyle\tan(3^{\circ})=\frac{a}{b}$. Therefore :
$$
\tan(6^{\circ})=\frac{2\tan(3^{\circ})}{1-\tan^{2}(3^{\circ})}
=\displaystyle\frac{\displaystyle\frac{2a}{b}}{\displaystyle\frac{b^{2}-a^{2}}{b^{2}}}
=\frac{2ab}{b^{2}-a}\in\mathbb{Q}
$$
Repeat this until you observe that $\tan(12^{\circ})\in\mathbb{Q}\implies\tan(24^{\circ})\in\mathbb{Q}$. However notice that :
$$
\tan(30^{\circ})=\tan(24^{\circ}+6^{\circ})=\frac{\tan(24^{\circ})+\tan(6^{\circ})}{1-\tan(24^{\circ})\tan(6^{\circ})}
$$
Since $\tan(24^{\circ}) ,\tan(6^{\circ})\in\mathbb{Q}$, then $\tan(30^{\circ})\in\mathbb{Q}$. However :
$$
\tan(30^{\circ})=\frac{\sqrt{3}}{3}\notin\mathbb{Q}
$$
Contradiction is thus achieved. Hence, $\tan(3^{\circ})\notin\mathbb{Q}$

Title: Proving $\tan 3^{\circ}$ is irrational


问:
Proposition 17.2 in Loring Tu's An Introduction to manifolds
Proposition 17.2 is showed as follows in Loring Tu's An Introduction to manifolds:

Proposition 17.2. If $f:M\longrightarrow\mathbb{R}$ is a $C^\infty$ funciton, then for $p\in M$ and $X_p\in T_pM$,
$$\left.f_*(X_p)=(df)_p(X_p)\frac{d}{dt}\right|_{f(p)}.$$
Proof. Since $f_*(X_p)\in T_{f(p)}\mathbb{R}$, there is a real number $a$ such that
$$\left.f_*(X_p)=a\frac{d}{dt}\right|_{f(p)}.\tag{17.1}$$
To evaluate $a$, apply both sides of $(17.1)$ to $x$:
$$a=f_*(X_p)(t)=X_p(t\circ f)=X_pf=(df)_p(X_p).\qquad\square$$

I cannot understand why $X_p(t\circ f)=X_p f$. Can anyone clarify this for me?

答: Here, $t\colon \Bbb R \to \Bbb R$ is the global coordinate function on $\Bbb R$. The first $\Bbb R$ plays the role of the "manifold", the second $\Bbb R$ plays the role of the Euclidean space modelling the manifold. Compare: isn't a chart $(U,\varphi)$ for a manifold $M$ a homeomorphism $\varphi\colon U \subseteq M \to \varphi[U]\subseteq \Bbb R^n$? Here, $U = M  = \Bbb R$, $\varphi$ is called "$t$", and $n=1$. So given $p \in M$, we have that $f(p) \in \Bbb R$ (the manifold $\Bbb R$). What is the coordinate of this point $f(p)$ in the manifold $\Bbb R$? Apply the chart $t$ to get $t(f(p)) \in \Bbb R$ (the Euclidean space $\Bbb R$). What makes this example so particular (and caused your confusion) is that $t(f(p)) = f(p)$ (numerically).


问:
Infinite series and its upper and lower limit.
I am learning analysis on my own and I am puzzled with the following question.

Consider the series $$\frac{1}{2}+\frac{1}{3}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^3}+\cdots$$ Indicate whether this series converges or diverges.

The following is what I was able to get.

First I noticed that $$a_n = \frac{1}{2^n}+\frac{1}{3^n}$$ would be a simple and nice form of this sequence, so according to the ratio test, $$\lim_{n \to \infty} \frac{1}{6} \left( \frac{3+2(\frac{2}{3})^n}{1+(\frac{2}{3})^n}\right) = \frac{1}{2}$$ therefore the series must converge.

However, I later understood the concept of upper an lower limits a little, so taking $$ a_n =
\begin{cases}
\frac{1}{2^n},  & \text{if $n$ is odd} \\
\frac{1}{3^{n-1}}, & \text{if $n$ is even}  \\
\end{cases}$$ I figured out the following $$\lim_{n \to \infty} \frac{1}{2}\left(\frac{2}{3}\right)^n = 0$$ when $n$ is odd and $$\lim_{n \to \infty} \frac{1}{2}\left(\frac{3}{2}\right)^n = +\infty$$ when $n$ is even.

Through this experience I realized and really understood how a subsequence of a sequence may have multiple limits, and the supremum and the infimum of those limits give us the upper and lower limits, respectively. But the book does something different than I did, and I would like to have the following questions answered.

1),$\quad$ From my understanding, the ratio test tells us that $\Sigma a_n$ diverges when the upper limit is greater than 1. Clearly one of the limits goes to $+\infty$, so why cannot we conclude that the series diverge ?

2),$\quad$  I learned that by choice of the subsequence, the limit may or mayn't be different. In the case that it is different, how do we know what is the supremum and infimum? It would be clear that if they were $\pm \infty$ but in my case, why is it guaranteed that there is no other subsequence that has a negative limit ? Is there an algebraic way to guarantee them ? Or do we always have to conceptually understand all of the subsequential limits.

3),$\quad$  The book says the ratio test is inconclusive, which turns me back to question 1), but it used the ratio test to determine that the series converges. According to my even-odd wise definition of $a_n$, the "larger limit" (I am purposefully not saying supremum because I don't know how it's guaranteed that it is the supremum) using the ratio test can be found as $$\lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \sqrt[n]{\frac{1}{2^n}} = {1 \over 2}$$ But the book says $$\lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \sqrt[2n]{\frac{1}{2^n}} = {1 \over \sqrt{2}}.$$ Where did the $2n$ come from ?

Can someone help me out ?

Edit:Thanks for helping me out everyone! I really appreciate it.

答: The ratio test says that IF a certain limit is less than $1$ then a certain series converges and IF that limit is more than $1$ then that series diverges.  The ratio test does not say whether the series converges or diverges in cases in which that limit does not exist (nor when it's equal to $1$).  In this case, the limit in the ratio test does not exist, so the ratio test (in that form, at least) doesn't tell you whether the series converges or not.  Thus the ratio test does not give you any conclusion that can contradict any other conclusion about whether the series converges.</p>


问:
Jensen inequality about mapping
In $X \in L_{p}(\Omega,\mathcal{F},P).$
If $p \in [1,+\infty]$, then why the map $x \mapsto |x|^p$ is convex, and we can apply Jensen's Inequality?

答:
This version of Holder's inequality says:

$$a_1^{\lambda_1}b_1^{\lambda_2}+a_2^{\lambda_1}b_2^{\lambda_2}\le(a_1+a_2)^{\lambda_1}(b_1+b_2)^{\lambda_2}$$

where $\lambda_1+\lambda_2=1$. Let $\lambda_1+\lambda_2=1$.

$$\begin{split}|tx+(1-t)y|&\le t|x|+(1-t)|y|\text{ triangle inequality}\\
&=t^{\lambda_1+\lambda_2}|x|^{\lambda_2/\lambda_2}+(1-t)^{\lambda_1+\lambda_2}|y|^{\lambda_2/\lambda_2}\text{exponents are 1}\\
&=t^{\lambda_1}t^{\lambda_2}(|x|^{1/\lambda_2})^{\lambda_2}+(1-t)^{\lambda_1}(1-t)^{\lambda_2}(|y|^{1/\lambda_2})^{\lambda_2}\text{rewrite some terms}\\
&=t^{\lambda_1}(t|x|^{1/\lambda_2})^{\lambda_2}+(1-t)^{\lambda_1}((1-t)|y|^{1/\lambda_2})^{\lambda_2}\text{group terms}\\
&\le \underbrace{(t+1-t)^{\lambda_1}}_{=1}(t|x|^{1/\lambda_2}+(1-t)|y|^{1/\lambda_2})^{\lambda_2}\text{Holder's inequality}\end{split}$$

with $a_1=t_1, b_1=1-t, a_2=t|x|^{1/\lambda_2}, b_2=(1-t)|y|^{1/\lambda_2}$. Let $\frac 1 {\lambda_2}=p,p\in[1,\infty)$. Then

$$\begin{split}|tx+(1-t)y|^{1/\lambda_2}&\le t|x|^{1/\lambda_2}+(1-t)|y|^{1/\lambda_2}\\
|tx+(1-t)y|^{p}&\le t|x|^{p}+(1-t)|y|^{p}\end{split}$$

Hence $f(x)=|x|^p$ is convex. Since it is convex, you can apply Jensens inequality to it.

Title: Jensen inequality about mapping


问:
Expanding exponents as opposed to solving logarithmically provides different answers. $10^{2t-3} = 7$
The solution should be simplified from the above

$$10^{2t-3} = 7$$

by recognizing the equivalent logarithmic format

$$\log(7) = 2t-3$$

which solves for $t$ as

$$\frac{\log(7)+3}{2} = t$$

Where I'm having problems is I attempted to work the problem from the other direction (just what stood out to me at first) by expanding

$$10^{2t-3} = 10^2 \cdot 10^t \cdot 10^{-3}$$

I think that might be where my problem is, because my next steps solve as

$$100 \cdot 10^t \cdot \frac{1}{1000} = 7$$

$$\frac{100 \cdot 10^t}{1000} = 7$$

$$\frac{1 \cdot 10^t}{10} = 7$$

$$10^t=70$$

Then I apply the equivalent logarithmic format as

$$\log(70) = t$$

and after calculating, it is very clear to me that

$$\log(70) \not = t = \frac{\log(7)+3}{2}$$

Can anyone help me understand what I'm doing wrong here?

答: $10^{2t}$ does not equal $10^{2} * 10^{t}$.  That would be $10^{t+2}$ instead.


Title: Expanding exponents as opposed to solving logarithmically provides different answers. $10^{2t-3} = 7$

Body:
The solution should be simplified from the above

$$10^{2t-3} = 7$$

by recognizing the equivalent logarithmic format

$$\log(7) = 2t-3$$

which solves for $t$ as

$$\frac{\log(7)+3}{2} = t$$

Where I'm having problems is I attempted to work the problem from the other direction (just what stood out to me at first) by expanding

$$10^{2t-3} = 10^2 \cdot 10^t \cdot 10^{-3}$$

I think that might be where my problem is, because my next steps solve as

$$100 \cdot 10^t \cdot \frac{1}{1000} = 7$$

$$\frac{100 \cdot 10^t}{1000} = 7$$

$$\frac{1 \cdot 10^t}{10} = 7$$

$$10^t=70$$

Then I apply the equivalent logarithmic format as

$$\log(70) = t$$

and after calculating, it is very clear to me that

$$\log(70) \not = t = \frac{\log(7)+3}{2}$$

Can anyone help me understand what I'm doing wrong here?

Body: $$10^2\cdot 10^t \cdot 10^{-3} = 10^{2+t-3} = 10^{t-1} \neq 10^{2t-3}$$


问:
Characterizing McDuff Factors in Terms of Central Sequences
Let $M$ be a II$_1$ factor von Neumann algebra with two-norm $\|\cdot\|_2$ induced by its canonical normal, faithful trace. It is said to be a McDuff Factor provided it tensorily absorbs the hyperfinite II$_1$ factor $R$; that is, $M \cong M \overline{\otimes} R$.

A sequence $(x_n)_{n \in \Bbb{N}} \in \ell^{\infty}(\Bbb{N},M)$ is said to be central if $$\lim_{n \to \infty}\|xx_n - x_n x\|_2 = 0$$ for any $x \in M$.Two central sequences $(x_n)_{n \in \Bbb{N}}$ and $(y_n)_{n \in \Bbb{N}}$ are said to be equivalent if $$\lim_{n \to \infty} \|x_n - y_n\|_2 =0$$ Finally, a central sequence is said to be trivial if it is equivalent to a scalar sequence.

I've also seen that $M$ is a McDuff factor is equivalent to the existence of pairs of non-commuting non-trivial central sequences in $M$. But I don't know exactly what that means. Would someone help me parse this?

答: Non-commuting central sequences means exactly what you call non-equivalent central sequences. McDuff proved that a II$_1$-factor $M$ is isomorphic to $M\otimes R$, with $R$ the hyperfinite II$_1$-factor, if and only if there exist non-equivalent central sequences in $M$.


问:
Finding the set of generators of the ideal $I = (2, 1 + \sqrt{-5})$ of $R = \mathbb Z[\sqrt{-5}]$
If I am given this ideal $I = (2, 1 + \sqrt{-5})$ clearly in the question, does that means that the set of generators of it is $\{ 2, 1 + \sqrt{-5} \}$? if so, is this the only set of generators of it in $R$? is this set linearly independent?

答: Yes, those are generators of the ideal. Given any set of generators of an ideal, you can always multiply the generators by units and keep the same ideal, so there are certainly other sets of generators (note that the only units in $R$ are $\pm 1$). A more interesting question and one that is often asked is whether the given ideal is principal, which is I believe what you are getting at in your linear independence question. To answer that more precisely however, you need to clarify what you mean by linearly independent? Over what set of scalars?


问:
True or false statements odd integers
One is false one is true write the negation for the false one and prove both of them

a) for all integers a, there exists an integer b so that a+b=odd

b) there exists an integer b such that for all integers a, a+b=odd

I said (a) is false and (b) is true. I wrote the negation for (a) then I split it in two cases. Case 1: b is even. Let b=2k and let a=0 then a+b=2k which is even.
Case 2: b is odd. Let b=2k+1 and let a=-1 then a+b=2k

Then for (b) I set up two cases again. Case 1: a is even. A=2k and b=1 then a+b=2k+1 which is odd
Case 2: a is odd, let a=2k+1 and b=0 then a+b=2k+1 which is odd

Input in whether I'm right or not would be great thanks

答:
Here the first true,because if $a=2k$( is even) choose $b=1$ if $a=2k+1$(odd) choose $b=2$ to complete proving the first fact.

In your proof i guess you have missed out the critical point "there exists".

Title: True or false statements odd integers


问:
Taking Limits of 2 function
I am taking the log of both numerator and denominator and then trying to find out the limits through L'Hospital's Rule. I am taking log because it makes it easier to differentiate the 2 functions. But I am not sure if this is a valid way to solve the problem. Can anyone clarify this?

$$
\lim_{n\to\infty} \dfrac{f(n)}{g(n)} = \lim_{n\to\infty} \dfrac{\log_2 f(n)}{\log_2 g(n)}
$$

答:
Well, there's a slight mistake in your part. Let us take,
$$y=\lim_{n\to\infty}\frac{f\left(n\right)}{g\left(n\right)}$$
Then, if we take log on both sides, we'll get,
$$\log{y}=\log\lim_{n\to\infty}\frac{f\left(n\right)}{g\left(n\right)}=\lim_{n\to\infty}\left(\log\frac{f\left(n\right)}{g\left(n\right)}\right)=\lim_{n\to\infty}\left(\log{f\left(n\right)}-\log{g\left(n\right)}\right)$$

Title: Taking Limits of 2 function


问:
Drake, Seven Axioms of the Algebra of Events
On page 3 of Drake Fundamentals of Applied Probability he lists The Seven Axioms of the Algebra of Events.

$1. A \cup B = B \cup A \\
2. A \cup (B \cup C) = (A \cup B) \cup C \\
3. A\cap( B \cup C) = A \cap B \cup A \cap C \\
4. (A')' = A \\
5. (A\cap B)' = A' \cup B'\\
6. A\cap A' = \phi \\
7.A \cap U = A  $

I have two questions.

First while Drake states that his choice is not unique is there a reason that he selected these seven and is seven the minimum possible number of axioms with which the Algebra of events can be expressed.

Second he further states that any other relation in the algebra of events can be proved directly from these seven axioms with no additional information and in fact in the chapter problem section asks the student to do so. Does the author mean that all the other set relations in the Algebra of Events can be proved by direct manipulation of the seven axioms with out resorting to the method of $x \in A$

For example, proofs of $A \cup A' = U$ are given in this posting
Proof of union of a set and its complement is equivalent to a universe

but these proofs use the method of $x \in A$ and not direct manipulation of Axiomatic statements as required by Drake. How would I prove $A \cup A' = U$ or $A \cup A = A$ using only direct manipulation of Drake's Seven Axioms of the Algebra of Events?

答:
What Drake calls the Algebra of Events is usually called Boolean Algebra. There are many ways to axiomatize Boolean Algebra, and Drake has given one of them.

First while Drake states that his choice is not unique, is there a reason that he selected these seven...

I don't see any reason to prefer this axiomatization to others. It's a little inconvenient since, as you noted, it's tricky to see why some basic equations follow from these axioms.

...and is seven the minimum possible number of axioms with which the Algebra of events can be expressed?

Wikipedia has a page on Minimal axioms for Boolean Algebra, but the axiomatizations discussed there work in a context with fewer basic operations (e.g., viewing $A\cap B$ as an abbreviation for $(A'\cup B')'$, or just using the Sheffer stroke). I don't know what the minimum number of equations is that axiomatizes Boolean Algebra with all its usual operations.

Second he further states that any other relation in the algebra of events can be proved directly from these seven axioms with no additional information...

Here's the idea of how to prove such a thing: If some equation is not provable from the axioms of Boolean Algebra, then there is an abstract Boolean algebra in which the equation does not hold - for example, the free Boolean algebra on the variables appearing in the equation. Now the Stone representation theorem says that every abstract Boolean algebra is isomorphic to an algebra of subsets of some set $U$. So there would be a counterexample to the equation obtained by substituting subsets of $U$ for the variables in the equation. Thus, if some  equation holds whenever the variables are interpreted by events (subsets of a set $U$), then the equation can be proved from the axioms of Boolean Algebra.

Now if you want to be sure that Drake's list of axioms fully axiomatize the Algebra of Events, instead of checking that every equation that holds of events follows from Drake's axioms (there are infinitely many such equations!), you just need to check it for each axiom in some known axiomatization of Boolean Algebra (like this one).


问:
Are the conjugacy classes normal subgroups in $S_n$?
I am working on proving that if $C$ is a conjugacy class in $S_n$ then $C^g =C$ for all $g \in S_n$.

I went over similar questions which only prove that a normal subgroup is contained in union of conjugacy classes, but not claiming a conjugacy class is normal subgroup.

答:
We don’t talk about “normality” except for subgroups. You may ask whether a set is “closed under conjugation”, and conjugacy classes certainly are: let $C$ be a conjugacy class, and let $x\in G$. To show $C^x\subseteq C$, let $g\in C$. Then $x^{-1}gx$ is conjugate to $g$, hence it lies in the conjugacy class of $g$... that is, in $C$. Thus, $x^{-1}Cx\subseteq C$. As this holds for all $x\in G$, it follows that $C^g=C$ for all $g\in G$.

Note that this holds for any group, not just $S_n$>


问:
Finding Joint PDF of Two Non-Independent Continuous Random Variables
I'm in the process of reviewing some stats using A First Course in Probability by Sheldon Ross. For the chapter on Joint Distributions, it shows how to obtain the Joint PDF given two independent continuous random variables. However, if the variables weren't independent, how would I go about obtaining the joint PDF of the two variables? Is there a systematic way of going about it similar to when the variables are independent?

So for example, if $f(x)$ and $f(y)$ is the PDF of two continuous independent random variables, I can find their joint PDF $f_{x,y}(x,y)$ by simply multiplying $f(x)$ and $f(y)$. However, how will I find $f_{x,y}(x,y)$ if $X$ and $Y$ were not independent?

Thanks!

答:
You wouldn't be able to find their joint pdf $f_{X,Y}$ given just their individual pdfs if they are not independent. You would need at least a conditional pdf or the joint pdf itself to know more about the relationship of their distributions. The joint pdf is related to the conditional pdf by
$$\begin{split}f_{X|Y}(x|y)&=\frac{f_{X,Y}(x,y)}{f_Y(y)}\\\text{or} f_{Y|X}(y|x)&=\frac{f_{X,Y}(x,y)}{f_X(x)}\end{split}$$

If the variables are independent
$$\begin{split}\frac{f_{X,Y}(x,y)}{f_Y(y)}&=f_{X|Y}(x|y)\\
&=f_X(x)\end{split}$$

which is why you can directly multiply them together.


问:
prove $ F(x)=\int_0^\infty {\sin(tx)\over(t+1)\sqrt t} dt \in C^\infty(\mathbb R^*) $
prove that :

$$ F(x)=\int_0^\infty {\sin(tx)\over(t+1)\sqrt t} \, dt \in C^\infty(\mathbb R^*) $$

i end up proving that $F(x)\in C^ \infty(\mathbb R^{*+})$ not $\mathbb R^*$ , and i studied the case with :

$F(x)= \sqrt x G(x)$ where $G(x)=\int_0^\infty {\sin(t)\over(t+x)\sqrt t} \, dt$ ($\sqrt x , G(x) \in C^\infty(R^{*+}))$

is my answer wrong ? , wich is the correct interval for $C^\infty$ ?

答:
Assume $x > 0$ and write

$$ F(x) = \sqrt{x} G(x), \quad \text{where} \quad G(x) = \int_{0}^{\infty} \frac{\sin t}{(t + x)\sqrt{t}} \, dt. $$

The proof is divided into two steps, where we first improve the decaying behavior of the integral representation of $G(x)$ and then we prove that $G(x)$ is infinitely differentiable.

Step 1. We introduce the function

$$ A(x) = \int_{0}^{x} \frac{\sin t}{\sqrt{t}} \, dt = \frac{1-\cos x}{\sqrt{x}} + \frac{1}{2} \int_{0}^{x} \frac{1-\cos t}{t^{3/2}} \, dt. $$

Clearly $A(x)$ converges as $x \to \infty$. In particular, $A(x)$ is uniformly bounded on $[0, \infty)$. Then we have

\begin{align*}
G(x)
 = \int_{0}^{\infty} \frac{\sin t}{(t + x)\sqrt{t}} \, dt
&= \int_{0}^{\infty} \frac{A'(t)}{t + x} \, dt
 = \left[ \frac{A(t)}{t+x} \right]_{0}^{\infty} + \int_{0}^{\infty} \frac{A(t)}{(t + x)^{2}} \, dt \\
&= \int_{0}^{\infty} \frac{A(t)}{(t + x)^{2}} \, dt.
\end{align*}

Step 2. Now we introduce the notation

$$ G_{k}(x) = (-1)^{k} (k+1)! \int_{0}^{\infty} \frac{A(t)}{(t + x)^{k+2}} \, dt, $$

so that $G(x) = G_{0}(x)$. We claim that $G_{k}'(x) = G_{k+1}(x)$. Once this is proved, it follows from the induction that $G(x)$ is infinitely differentiable. Indeed, by the Mean Value Theorem,

$$ \frac{1}{h} \left( \frac{1}{(t + x + h)^{k+2}} - \frac{1}{(t + x)^{k+2}} \right) = -\frac{k+2}{(t + x + \theta h)^{k+3}} $$

for some $0 \leq \theta \leq 1$, with the choice of $\theta$ depending on $t$, $x$ and $h$. Thus for $h$ satisfying $0 < |h| \ll x$, we have

\begin{align*}
&\left| \frac{G_{k}(x+h) - G_{k}(x)}{h} - G_{k+1}(x) \right| \\
&= (k+1)! \left| \int_{0}^{\infty} A(t) \left\{ \frac{1}{h} \left( \frac{1}{(t + x + h)^{k+2}} - \frac{1}{(t + x)^{k+2}} \right) + \frac{k+2}{(t+x)^{k+3}} \right\} \, dt \right| \\
&= (k+2)! \left| \int_{0}^{\infty} A(t) \left( \frac{1}{(t+x)^{k+3}}-\frac{1}{(t + x + \theta h)^{k+3}} \right) \, dt \right|
\end{align*}

Since the integrand of the last integral is dominated by the integrable function

$$ |A(t)| \left( \frac{1}{(t+x)^{k+3}} + \frac{1}{(t + x - |h|)^{k+3}} \right), $$

we can apply the dominated convergence theorem to conclude that

$$ \lim_{h\to 0} \frac{G_{k}(x+h) - G_{k}(x)}{h} = G_{k+1}(x) $$

as desired. Therefore the proof is complete.


问:
Proof that square root of a sequence converges to the square root of the limit of the sequence.
I've seen other posts around here regarding this issue as well, so I apologise for the repetition. But I need help regarding a specific part of the proof that I don't see covered elsewhere. I was doing a question and it had some guiding sub-questions which I completed, but I cannot see how these sub-questions contribute to the final part of the proof using the epsilon-delta definition of the limit of sequences. Here's what I was asked to show in the sub-questions, which I managed to:
$$
If \ lim_{\to\infty}a_n=L,then\ when \ n>N_1:N_1\in\mathbb{Z}^+, a_n<\frac{L}{4} \ and \\
|\sqrt{a_n}-\sqrt{L}|<\frac{2|a_n-L|}{3\sqrt{L}}
$$
Now, I want to show that $|\sqrt{a_n}-\sqrt{L}|<\epsilon$ using these 2 inequalities. I tried
$$
|\sqrt{a_n}-\sqrt{L}|<\frac{2|a_n-L|}{3\sqrt{L}}<\frac{2}{3\sqrt{L}}*\frac{3L}{4}=\frac{\sqrt{L}}{2}<\frac{3L}{4}=\epsilon
$$
But I know this is wrong because this epsilon is the one I selected for the sequence ${a_n}$ to get the first inequality, and I must prove $\forall\epsilon$. How do the 2 inequalities apply to the proof, and do I need to still use the same $N_1$ for the proof next, or do I have another $N_2$ to create another special epsilon? Thank you very much for your help.

答:
I suppose your $L $ is positive.

You did not make full use of the fact that $a_n \to L$ and that is why you got stuck. Choose $N_1$ such that $n >N_1$ implies $|a_n-L| <\epsilon \sqrt L$. Then you get $|\sqrt a_n -\sqrt L|=\frac {|a_n-L|} {\sqrt a_n+\sqrt L}\leq \frac {|a_n-L|} {\sqrt L}<\epsilon$ for $n >N_1$.

The result is also true for $L=0$ and the proof is much simpler in that case.


问:
Computation of density function of $Y=\frac1X-X$ where $X\stackrel{\mathrm d}= U[0,1]$

Calculate the density function of $Y=\frac1X-X$, where $X\stackrel{\mathrm d}= U[0,1]$.

I am confused about the technique to deal with the problem like this when after the transformation, I can not express the density of $Y$ in terms of $X$

答:
$1/X-X$ is a decreasing function of $X$ so $Y\in(0,\infty)$ for $X\in(0,1)$.

$$P(Y\le y)=\begin{cases}0,&y\le0\\P(1-X^2\le yX),&y>0\end{cases}$$

Now $X^2+yX-1$ is an upward opening parabola with zeroes at  $\frac{-y\pm\sqrt{y^2+4}}2$, and thus $$\begin{align*}P(X^2+yX-1\ge0)&=\underbrace{P\left(X\le\frac{-y-\sqrt{y^2+4}}2\right)}_{0}+P\left(X\ge \frac{-y+\sqrt{y^2+4}}2\right)\\&=1-P\left(X\le\frac{-y+\sqrt{y^2+4}}2\right)\\&=1-\min\left\{1,\frac{-y+\sqrt{y^2+4}}2\right\}\end{align*}$$

Can you simplify this?

 Rationalizing$$\frac{-y+\sqrt{y^2+4}}2\left[\frac{\sqrt{y^2+4}+y}{\sqrt{y^2+4}+y}\right]=\frac2{\sqrt{y^2+4}+y}$$Call it $g(y)$. It is decreasing for $y>0$ so $g(y)<g(0)=1$, giving $\min\{1,g(y)\}=g(y)$.


问:
Degree of Gauss map equal to half the Euler characteristic and Poincaré-Hopf
The Poincaré-Hopf theorem states that for a smooth compact $m$-manifold $M$ without boundary and a vector field $X\in\operatorname{Vect}(M)$ of $M$ with only isolated zeroes we have the equality
$$\sum_{\substack{p\in M\\X(p)=0}}\iota(p,X)=\chi(M)$$
where $\iota(p,X)$ denotes the index of $X$ at $p$ and $\chi(M)$ denotes the Euler characteristic of $M$.

Let $m$ be even and $M\subset\mathbb{R}^{m+1}$ be a $m$-dimensional smooth compact submanifold without boundary and denote by $\nu:M\to S^m$ its Gauss map. How can I deduce from the Poincaré-Hopf theorem that the Brouwer degree of $\nu$ is equal to half the Euler characteristic of $M$ i.e.
$$\deg(\nu)=\frac{1}{2}\chi(M)?$$

After many repeated (unsuccessful) tries I was hoping hat someone else might shed some light onto this...

答:
I like Ryan Budney's proof very much. For posterity, I will add my own, different proof. I will use the Jordan-Brouwer separation theorem (see p. 89 of the book by Guillemin & Pollack), the Poincaré-Hopf theorem for manifolds with boundary (which assumes that the vector field is outward pointing at the boundary), and the fact that the Euler characteristic of any odd-dimensional smooth closed manifold vanishes. (In the orientable case, the latter claim follows from Poincaré duality and is exercise 18-9 in Lee's Intro. to Smooth Manifolds, 2nd ed.; the nonorientable case follows from the orientable case, consideration of the orientation double cover, and the fact that the Euler characteristic of a degree $n<\infty$ covering space $E$ of a compact manifold $B$ with boundary satisfies $\chi(E) = n\cdot \chi(B)$. Alternatively, see Cor. 3.37 of Algebraic Topology by Hatcher.)

Proof: Let $M\subset \Bbb{R}^{m+1}$ be a smooth closed hypersurface. By the Jordan-Brouwer separation theorem, $M=\partial N$ is the boundary of some compact, smooth, codimension-$0$ manifold $N$ with boundary $\partial N$. Let $X$ be a smooth vector field on $N$ which points strictly outward at $\partial N$ and has isolated zeros in $\text{int}(N)$. By the Poincaré-Hopf theorem, the Euler characteristic $\chi(N)$ is equal to the sum of the indices of these zeros.
On the other hand, the degree of the Gauss map $M\to S^m$ is also equal to the sum of these indices. (To see this, let $N'$ be $N$ minus the union of small open balls centered at the zeros of $X$; the Gauss map $\frac{X}{|X|}:\partial N'\to S^m$ admits a smooth extension to all of $N'$ and therefore has degree zero, as shown in Guillemin & Pollack.) Hence

$\chi(N) = \text{degree of the Gauss map }M\to S^m.$    (1)

Now, in general the following Euler characteristic formula holds for smooth closed manifolds $N$ with nonempty boundary:

$\chi(DN) = 2\chi(N)-\chi(\partial N),$ (2)

where $DN$ is the (compact, boundaryless) double of $N$ obtained by pasting two copies of $N$ together along their boundaries and smoothing the result.
When $\dim(N)=\dim(DN)$ is odd, $\chi(DN)=0$ as mentioned in the preface.

Therefore, in our case when $\dim(M) = \dim(\partial N)$ is even, the left-hand side of (2) vanishes to yield, when combined with (1),

$\frac{1}{2}\chi(M)= \text{degree of the Gauss map }M\to S^m$

when $\dim(M)$ is even.

This proves the desired result.

Bonus: From (1) and (2) we also obtain the following general formula, valid for arbitrary $\dim(M) = \dim(\partial N)$:

$\chi(N)=\frac{1}{2}[\chi(M)+\chi(DN)]= \text{degree of the Gauss map }M\to S^m$

where again $M = \partial N$ and the existence of $N$ is guaranteed by the Jordan-Brouwer separation theorem. Hence if $\dim(M)=\dim(\partial N)$ is odd, then $\chi(M)=0$ so that

$\chi(N)=\frac{1}{2}\chi(DN)= \text{degree of the Gauss map }M\to S^m$

when $\dim(M)$ is odd.


问:
homotopic to a constant map
Prove that any continuous map $f\colon RP^3\to S^1$ is homotopic to a constant map?

How to prove it? Thanks in advance.

答:
Since $\pi_1(\Bbb RP^3)=\Bbb Z_2$ and $\pi_1(\Bbb S^1)=\Bbb Z$ the induced group homomorphism $f_*:\pi_1(\Bbb RP^3)\to \pi_1(\Bbb S^1)$ is the zero homomorphism.

Hence, by the arbitrary map lifting lemma we can lift the map $f$ w.r.t the covering $\Bbb R\ni t\longmapsto \exp(2\pi it)\in \Bbb S^1$. That is there is a map $\widetilde f:\Bbb RP^3\to \Bbb R$ with $\exp\big(2\pi i\widetilde f(x)\big)=f(x)$.

Note that to lift a map, we must ensure that the domain of the map is locally path-connected and connected. But here we are dealing with connected manifolds, so we do not need to worry.

Now, consider the homotopy $H:\Bbb RP^3\times [0,1]\ni (x,t)\longmapsto \exp\big(2\pi i (1-t)\widetilde f(x)\big)\in \Bbb S^1$.

Title: homotopic to a constant map


问:
Spheres in different dimension are not homotopy equivalent
Is there a way to prove that $\textbf{S}^n$ and $\textbf{S}^m$ are not homotopy equivalent if $n\neq m$ without using the machinery of homology or higher homotopy groups?</p>

答:
Suppose $S^{n} \simeq S^{n+d}$. Notice that $S^{n}$ is a subspace of $S^{n+d}$. So there exists a deformation retraction $H:S^{n+d} \times I \to S^n$ such that $H((x_1, ..., x_{n+1}, x_{n+2}, ..., x_{n+d+1}),0)=(x_1, ..., x_{n+1}, x_{n+2}, ..., x_{n+d+1})$ and $H((x_1, ..., x_{n+1}, x_{n+2}, ..., x_{n+d+1}),0)=(x_1, ..., x_{n+1}, 0, ..., 0)$.

Restrict $H$ to $n+1$th and $n+2$ coordinate to get $F$. Notice $F$ is still continuous since $H$ is. $F$ defines a deformation retraction from a circle to a an interval. $F((x_{n+1}, x_{n+2}),0)=(x_{n+1}, x_{n+2})$ and $F((x_{n+1}, x_{n+2}),1)=(x_{n+1}, 0)$. But this is a contradiction and hence $S^{n}$ cannot be homotopy equivalent to $S^{n+d}$ as we assumed. $\blacksquare$


问:
How to find a centroid of a quarter circle?

Find the centroid of the quarter circle $x^2+y^2 \le  10$, $y\ge |x|$ assuming the density $ {\displaystyle \delta }(x,y) = 1$. give your answer as point coordinates in the form (X,Y)

So I first convert to polar coordinates:

$$rsin(\theta)^2+rcos(\theta)^2=10$$

Then I know it's a quarter circle that's bounded by $y=|x|$

So $\theta = \frac{\pi}{4}$ to $\frac{3\pi}{4}$

So now I integrate:

$$\int _{\frac{\pi }{4}}^{\frac{3\pi }{4}}\int _0^{2.5}\:\:\:rsin\left(\theta \right)^2+rcos\left(\theta \right)^2drd\theta $$

But this is just one integral. How do I get two? What am I missing?

Thank you.

edit:

答:
Another (related$ ^* $) method for locating the centroid of a region with constant density is to apply Pappus' (Second) Centroid Theorem, which states that the volume of a solid of revolution is equal to the area of the region revolved about the rotation axis times the circumference of the circle swept out by the centroid.

Revolving the quarter-circle of radius $ \ \sqrt{10} \ $ about one of its straight edges (radii) generates a hemisphere of that radius.  We then have

$$ \ V_{hemi} \ = \ \frac{2 \pi}{3} · ( \sqrt{10} )^3 \ \ = \ \ A_{qtr} \ · \ 2 \pi  \bar{x} \ \ = \ \ \frac{\pi}{4} · (\sqrt{10})^2 \ · \ 2 \pi \bar{x}  $$

$$ \Rightarrow \ \ \bar{x} \ \ = \ \ \frac{4}{3 \pi} \ · \ \sqrt{10} \ \ . $$

With the "center" of the quarter-circle at the origin and the edges along the coordinate axes, the centroid is thus found to be at $ \ \left( \  \frac{4\sqrt{10}}{3 \pi}    \ , \  \frac{4\sqrt{10}}{3 \pi} \ \right) \  $ , by the symmetry of the region.

To connect this to the other answers given by Math Lover and David G. Stork, the distance of the centroid from the vertex of the quarter-circle is

$$ \frac{4}{3 \pi} \ · \ \sqrt{10} \ · \ \sqrt{2} \ \ = \ \   \frac{4\sqrt{20}}{3 \pi} \ \ = \ \   \frac{8 \sqrt{5}}{3 \pi} \ \ . $$

For the orientation of the quarter-circle indicated in the problem statement, the coordinates of the centroid are then $ \ \left( \  0   \ , \  \frac{8 \sqrt{5}}{3 \pi}  \ \right) \ \ . $

$$ \ $$

$^*$ This is a related method because Pappus' Centroid Theorems describe the results found from the moment integrals for centroids of surfaces and solids of revolution, used in the other answers.


问:
Evaluate $\lim_{n \rightarrow \infty } \int_0^{\pi/2} \sqrt{n \sin{\frac{x}{n}} }dx$
Evaluate $\lim_{n \rightarrow \infty }  \int_0 ^{\pi/2} \sqrt{n \sin{\frac{x}{n}}} dx$ .

I already have that $\lim_{n \rightarrow \infty} \sqrt{n \sin{\frac{x}{n}}}= \sqrt{x}$.

So I'm trying to use DCT so that $\lim_{n \rightarrow \infty} \int f_n = \int \lim_{n \rightarrow \infty} f_n$.

But I'm having a hard timer finding a bound.

Also, I'm a little concerned about the range of $x$, can we still use the DCT with the range?

Thanks

答:
You can have much more that the limit since, without any restriction (except $n>0$), $$I_n=\int\sqrt{n \sin \left(\frac{x}{n}\right)}\,dx=-2 n^{3/2} E\left(\left.\frac{1}{4} \left(\pi -\frac{2 x}{n}\right)\right|2\right)$$ So
$$J_n=\int_0^{\frac \pi 2}\sqrt{n \sin \left(\frac{x}{n}\right)}\,dx=n^{3/2} \left(\sqrt{\frac{2}{\pi }} \Gamma \left(\frac{3}{4}\right)^2-2
   E\left(\left.\frac{(n-1) \pi }{4 n}\right|2\right)\right)$$ Expanding as series
$$J_n=\frac{\pi ^{3/2}}{3 \sqrt{2}}\Bigg[1-\frac{\pi ^2}{112 n^2}+\frac{\pi ^4}{84480 n^4}+O\left(\frac{1}{n^6}\right) \Bigg]$$


问:
What does "$\;\lfloor y \rfloor\;$" mean?
Could someone clarify the what the "L" and "inverted L" brackets on the $y$ variable mean? Is it weirdly formatted absolute value brackets or something different?

答:
Let $f(x) = \left\lfloor x \right\rfloor $, then $f$ is the 'floor' function, and it can be expanded as:

$$f(x) = \begin{cases} x, & \textrm{if }x \in \mathbb{Z} \\ n, & \textrm{if $n$ is the biggest integer such that } n< x\end{cases}$$


问:
Derivation of a recursive binomial coefficient algorithm
I am working through a Haskell book and I see that the binomial coefficients formula

$$\binom{n}{k} = \frac {n!}{k! (n - k)!}$$

is expressible as recurrence relationships

\begin{align}
\binom{n}{k} &= \binom{n - 1}{k - 1} \frac{n}{k} \quad
\text{for} \; n \gt 0 \\
\binom{n}{0} &= 1
\end{align}

and as

\begin{align}
\binom{n}{k} &= \binom{n - 1}{k - 1} + \binom{n - 1}{k} \\
\binom{n}{0} &= 1
\end{align}

My question is how were these derived from the original formula above? Is there a treatment, a proof anywhere of how these were derived that I could see?

Update

Yes, the classic algebra assignment is to take the recurrence relation and simplify it back to the original as a proof that the recursive version is indeed the same as the original

$$\binom{n}{k} = \frac {n!}{k! (n - k)!} = \binom{n - 1}{k - 1} + \binom{n - 1}{k} $$

Instructive, but I would like to know how the recursive version was derived in the first place. I assume that there is was no systematic way, that it was just a guess -- or perhaps some much deeper numerical analysis issue?

答: The combinatorial interpretation makes these intuitive. See https://en.wikipedia.org/wiki/Binomial_coefficient#Recursive_formula.</p>


问:
How can I find the general solution to a second order ODE when $yy''=3(y')^2$?
How can I reduce $yy''=3(y')^2$ to first order and find its general solution by using $z=y'$ and $y''=\frac{dz}{dy}z$?

At first, I reduced it to a first order, and I know that I'm supposed to do separable parts. However, I can't seem to make that work. Any help is greatly appreciated. Thank you!

答:
\begin{align}
yy''=3(y')^2,
\end{align}
divide by $yy'$
\begin{align}
\frac{y''}{y'}=3\frac{y'}{y},
\end{align}
integrate
\begin{align}
\ln(y')=3\ln(y)+c\\
y'=cy^3.
\end{align}
Divide by $y^3$ and integrate
\begin{align}
\frac{y'}{y^3}=c\\
\frac{-1}{2y^2}=c_1x+c_2\\
y=\frac{1}{\sqrt{c_1x+c_2}}
\end{align}


问:
Cuphead: How many setups (combinations) are possible?
Cuphead is a 2D shooter game in which a player can permanently acquire up to [edit: six] different Shots (i.e., weapons) and up to six different Charms (i.e., perks or buffs).

However, a player can only equip two Shots (one active, one reserve) and one Charm at a time. Note that a player cannot have the same the Shot be both active and reserve.

How many two-Shot, one-Charm [edit: permutations] are possible?

FYI I once attended a few lectures of an undergrad Probability course. My understanding is that...

```<code>[edit: 6] possible active Shots * [edit: 5] possible reserve Shots * 6 possible Charms = [edit: 180] possible [permutations]
</code>```

Is this calculation correct? Also, assuming that a second player can have any identical two-Shot, one-Charm setup as the first player, would the possible combinations of setups for two players then be...

```<code>[edit: 180] possible setups for Player 1 * [edit: 180] possible setups for Player 2 = [edit: 32,400] possible setups for the pair
</code>```

答: It depends on whether or not you distinguish between primary and secondary weapons. For instance, if you don't care which is primary and which is secondary then there are $\binom{5}{2}=10$ ways to choose any two weapons, but if you care about which is primary and which is secondary, then there are $5 \cdot 4=20$ ways to choose the 2 weapons. Your calculation uses the latter and thus distinguishes two hands if the primary and secondary weapons are swapped, so if this is what you intended then the calculation is correct.


答:
Surely you know the theorem that if $Y=r(X)$ where r(x) is differentiable and one-to-one for $a<x<b$, and $(\alpha, \beta)$ is the image of (a, b) under the function r, and if s(y) is the inverse function of r(x), then the pdf of Y is given by

$$f_Y(y)=\begin{cases}f_X(s(y))\left|\frac{d}{dy}s(y)\right|&, \alpha<y<\beta\\
0&\text{otherwise}\end{cases}$$

First find s(y):

$$\begin{split}Y&=\frac 1 X - X\\
s(Y)=X&=\frac{-y\pm\sqrt{y^2+4}}{2}\end{split}$$

Take the derivative

$$s'(y)=-\frac 1 2 \pm \frac{y}{2\sqrt{y^2+4}}$$

It remains to determine if $$\begin{split}\left|\frac{d}{dy}s(y)\right|&=\frac 1 2-\frac {y}{2\sqrt{y^2+4}}\\
\text{or}\left|\frac{d}{dy}s(y)\right|&=\frac 1 2+\frac {y}{2\sqrt{y^2+4}}\end{split}$$

Because $Y=\frac 1 X - X$ is decreasing on [0,1] you take the minus sign as $\frac {y}{2\sqrt{y^2+4}}$ is increasing on $(0,\infty)$.

Then
$$f_Y(y)=\begin{cases}\frac 1 2-\frac {y}{2\sqrt{y^2+4}}&,0<y<\infty\\
0&,y\le0\end{cases}$$

Title: Computation of density function of $Y=\frac1X-X$ where $X\stackrel{\mathrm d}= U[0,1]$

Body:
Calculate the density function of $Y=\frac1X-X$, where $X\stackrel{\mathrm d}= U[0,1]$.

I am confused about the technique to deal with the problem like this when after the transformation, I can not express the density of $Y$ in terms of $X$


问:
The expression $i^{i^{i^2}}$ can be written $e^\alpha$ for some real number $\alpha \in [0, 2\pi)$. What is $α$?

The expression $i^{i^{i^2}}$ can be written $e^\alpha$ for some real number $\alpha \in [0, 2\pi)$. What is $α$?

I am not sure what this is asking, could I get a hint on where to start?

答: You should note that if there is no parentheses, you evaluate top down. So $2^{2^{2^2}}$ would equal $2^{16}$ rather than $256$. Thus, $i^{i^{i^2}} = i^{i^{-1}} = i^{-i} = (i^i)^{-1}$. Note that $e^{\pi i} = -1$, thus $e^{\pi i/2} = i. $ Thus, we get $(e^{\pi i/2})^i = i^i $ which becomes $e^{-\pi/2} = i^i$. We had this equation before $i^{i^{i^2}} = i^{i^{-1}} = i^{-i} = (i^i)^{-1}$, so now we can substitute in our value for $i^i$ to get $i^{i^{i^2}}   = (e^{-\pi/2})^{-1} = e^{\pi/2}$.


问:
Maximum value of $k$ so that the inequality $\sqrt{y-5}+\sqrt{8-y} \ge k$ should have answer

Suppose $k$ be a real number so that the inequality
$\sqrt{y-5}+\sqrt{8-y} \ge k$ should have answer. what is the maximum
possible value of $k$?

$1)\sqrt{6}-1\quad\quad2)\frac12+\sqrt{3}\quad\quad3)\sqrt6\quad\quad4)2\sqrt2\quad\quad5)\sqrt6+1$

So I translated the question into finding the minimum value of $f(y)$ where $f(y)=\sqrt{y-5}+\sqrt{8-y}\quad$.(because for $k=\min(f(y))$ the inequality certainly has answer.) therefore we should solve  $f'(y)=0$:

$$f'(y)=\frac{1}{\sqrt{y-5}}-\frac{1}{\sqrt{8-y}}=0$$
$$y-5=8-y\rightarrow\quad y=6.5$$

And $f(6.5)=2\sqrt{\frac32}=\sqrt{6}$. so the answer is third choice.

Is my answer right? I know to find maximum or minimum of a function we take derivative of it and equate it to $0$. so here I don't know why $f(6.5)$ is minimum. can you please explain it to me?

EDIT:

I just noticed that $\sqrt6$ is not the minimum because $f(5)=f(8)=\sqrt3<\sqrt6$ maybe it is maximum! so I conclude the answer to this problem should be $\sqrt3$ but I don't have this option in the choices

答:
We need $y-5\ge0\iff y\ge5$ and $8-y\ge0\iff y\le8$

$$5\le y\le8\iff5-\dfrac{8+5}2\le y-\dfrac{8+5}2\le8-\dfrac{8+5}2$$

WLOG $y-6.5=1.5\cos2t$  where  $0\le2t\le\pi$

$$\sqrt{y-5}+\sqrt{8-y}$$

$$=\sqrt{6.5+1.5\cos2t-5}+\sqrt{8-(6.5+1.5\cos2t)}$$

$$=\sqrt3(\cos t+\sin t)$$

Now as $\cos t,\sin t\ge0;$

$$\cos t+\sin t=\sqrt2\cos\left(t-\dfrac\pi4\right)\le\sqrt2$$


问:
Proving $\sum_{k=0} \frac{(-1)^k}{1+2k} = \frac{\pi}{4} $ taking $4 \int_0^1 \sqrt{1 - x^2} \; \text{d}x$ as the definition of $\pi$.
I'm trying to prove the Madhava-Leibniz formula for $\pi$ using one quarter of the area of the unit circle as the definition of $\frac{\pi}{4}$.

$$ \int_{0}^{1} \sqrt{1-xx}\; \text{d}x = \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} $$

I have a proof of this identity, but it uses a lot of trigonometric substitutions and makes direct use of $\pi$ as a constant.

This feels a bit circular since the left hand side is supposed to be the definition of $\frac{\pi}{4}$.

Is there a way to prove this identity directly, instead of proving that both sides are equal to $\frac{\pi}{4}$?

All integrals are over the interval $[0,1]$ in the variable $x$. Let $\theta$ refer to $\frac{\pi}{4} x$. $\theta$ is a constant multiple of $x$, I'm using it so I don't need to change the bound of integration.

$$ \sum_{k=0} \frac{(-1)^k}{1+2k} $$

Using Lemma 101.

$$ \sum_{k=0} \int_{0}^{1} (-1)^k x^{2k} \text{d}x $$
$$ \sum_{k=0} \int_{0}^{1} (-xx)^k \text{d}x $$

Exchanging the sum and integral.

$$ \int_{0}^{1} \sum_{k=0} (-xx)^k \text{d}x $$

Applying Lemma 102

$$ \int_{0}^{1} \frac{1}{1+xx} \text{d}x $$

Applying Lemma 103, with $g$ equal to $\tan(\theta)$ and $g'$ equal to $\frac{\pi}{4} \sec^2(\theta)$.

$$ \int_{0}^{1} \frac{\pi}{4} \cdot \sec^2(\theta) \cdot \frac{1}{1+\tan^2(\theta)} \text{d}x $$

$$  \frac{\pi}{4} \int_{0}^{1} \sec^2(\theta) \cdot \frac{1}{\sec^2(\theta)} \text{d}x $$

$$  \frac{\pi}{4} \int_{0}^{1} 1\; \text{d}x $$

$$ \frac{\pi}{4} $$

And for the other side.

$$ \int_0^1 \sqrt{1-xx} \; \text{d}x $$

Apply lemma 103, with $g = \sin(2\theta)$ and $g' = \frac{\pi}{2}\cos(2\theta)$.

$$ \int_0^1 \frac{\pi}{2} \cdot \cos(2 \theta) \sqrt{1-\sin^2(2 \theta)} \; \text{d}x $$

$$ \frac{\pi}{2} \int_0^1 \cos^2(2 \theta) \; \text{d}x $$

Apply lemma 104.

$$ \frac{\pi}{2} \cdot \frac{1}{2} $$

$$ \frac{\pi}{4} $$

Lemmas

Lemma 101: integral of monomials

$$ \int_0^1 x^n \text{d}x = \frac{1}{1+n} $$

Lemma 102: sum of geometric series

$$ \sum_{k=0} r^k = \frac{1}{1-r} $$

Lemma 103: Special case of change of variables for integration

Suppose $g(0)=0$ and $g(1)=1$, then the following holds.

$$ \int_0^1 f(x) \text{d}x = \int_0^1 g'(x) \cdot f'\circ g(x) \text{d}x $$

Lemma 104: Integral of $\cos^2 + \sin^2$

For all values of $x$, $\cos^2(x) + \sin^2(x) = 1$, therefore the following holds

$$ \int_0^1 \cos^2(2\theta) + \sin^2(2\theta) \text{d}x = 1 $$

However, $\cos\left(\frac{\pi}{2}-x\right) = \sin(x) $, therefore, over the interval $[0, 1]$, $\sin$ and $\cos$ take on the same values at the same offsets, but in the opposite order. Thus

$$ \int_0^1 \sin^2(2\theta) \text{d}x = \int_0^1 \cos^2(2\theta) \text{d}x $$

Thus

$$ \int_0^1 \cos^2(2\theta) \text{d}x = \frac{1}{2} $$

答:
Too long for coments.

Considering
$$S_{m,n}=\sum_{k=m}^{n} \frac{(-1)^k}{2k+1}$$
$$S_{m,n}=\frac{1}{2} \left((-1)^m \,\Phi \left(-1,1,m+\frac{1}{2}\right)+(-1)^n\, \Phi\left(-1,1,n+\frac{3}{2}\right)\right)$$ where appears  the Lerch transcendent function.

$$S_{0,n}=\frac{1}{4} \left(2 (-1)^n \Phi \left(-1,1,n+\frac{3}{2}\right)+\psi
   \left(\frac{3}{4}\right)-\psi \left(\frac{1}{4}\right)\right)$$
$$\psi
   \left(\frac{3}{4}\right)-\psi \left(\frac{1}{4}\right)=\pi$$
$$S_{0,n}=\frac{1}{2} (-1)^n \Phi \left(-1,1,n+\frac{3}{2}\right)+\frac{\pi }{4}$$
$$S_{0,n}=\frac{1}{4} (-1)^n \left(\psi \left(\frac{2 n+5}{4} \right)-\psi   \left(\frac{2 n+3}{4} \right)\right)+\frac{\pi }{4}$$ Using asymptotics
$$S_{0,n}=\frac{\pi }{4}+\frac{(-1)^n }{4n} \left(1-\frac{1}{n}+\frac{3}{4 n^2}+O\left(\frac{1}{n^3}\right)\right)$$


问:
Let $W=\{p(B):p\text{ be a polynomial with real coefiicient}\}$
Let $W=\{p(B):p\text{ be a polynomial with real coefiicient}\}$ and where $B=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}$  Then the dimension $d$ of the space $W$ satisfies

1.$4\le d\le 6$

2.$6\le d\le 9$

3.$3\le d\le 8$

4.$3\le d\le 4$

I have calculated that $B^3=I$ so $I,B,B^2$ will be the basis vebtors in $W$ so I hope $\dim=3$ but I dont know why they have put bounds, It was question in an past examination.

答:
Hint: $B^k=B^{k-3}=B^{a}$ for some $a\in\{0,1,2\}$ and for all $k>3,k\in \mathbb{N}$.

This implies that any higher power of $B$ can ultimately transformed to a power $\in \{0,1,2\}$. So your claim follows.


问:
Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2\sin\theta}{2\sin\theta-1}$
Parametric equation of curve is $ x=2\sin\theta +\cos(2\theta) $ , $y=1+\cos(2\theta)$ , for $0\leq\theta\leq\pi/2$

Show that $$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2\sin\theta}{2\sin\theta-1}$$
I drived $x$ and $y$ seperatly for $x$,

$$ x=2\cos\theta+\sin\theta\pm\sin\theta$$ and for the

derivative of $y$ i got $y=-2\sin(2\theta)$

and then I did

$$dy/dx= y=-2\sin(2\theta), \qquad   x=2\cos\theta+\sin\theta\pm \sin\theta$$

but I dont know if I did it wronge or I don't know how to continue. Any help would be much appreciated. Thank you!

答: $$\frac{dy}{dx}=\frac{dy}{d\theta}/\frac{dx}{d\theta}=\frac{-2\sin 2\theta}{2\cos \theta-2\sin 2 \theta}=\frac{-\sin \theta \cos \theta}{\cos \theta(1-2\sin \theta)}=\frac{\sin \theta}{2\sin \theta-1}.$$


问:
Every set has a free abelian group
I know the fact that every set has a free abelian group with basis $S$, (i.e. for any set $S$, free abelian group on $S$ is a group whose elements can be "uniquely" written by $\mathbb Z$-linear combinations of elements on $S$).

If the fact is true for any set $S$, I don't know how any $S$ can be a basis and $\mathbb Z$-linear combinations on $S$ can be "uniquely" expressed.

答:
Let's think about this problem from a lens of linear algebra because I think this concept can get lost in the weeds if you think about from an abstract algebra perspective. I think what you are probably stuck on is an example like this: Let $\mathcal P_2$ be the free group of integer valued polynomials (we are only allowing addition). Then the sets
$$S = \{1, x, x^2\} \text{ and } S' = \{1-x, x - x^2, x^2\}$$ both generate $\mathcal P_2$ as a free abelian group. But how can we have the same group generated by two distinct sets? The error in this is that you are imposing more mathematical structure than the theorem assumes.

In linear algebra we can define a vector space $V$ with basis $\mathcal B = \{b_1, b_2, \ldots, b_n\}$. This is called an $n$-dimensional vector space and we can define a rich theory of linear algebraic properties of $V$. But what is the basis $\mathcal B$? From our perspective, it's just a set of distinct symbols which we call a "basis" of $V$.

That's what this theorem is telling you, it's just saying that given a set of symbols $S$, you can make a free algebra (vector space) with that set as a basis. If you think of $S$ and $S'$ as merely sets of symbols rather than elements of $\mathcal P_2$, then you would get two different free groups (but they would be connected by a group isomorphism taking $1 \mapsto 1-x$, $x \mapsto x-x^2$, $x^2 \mapsto x^2$.

Hope this helps.

Title: Every set has a free abelian group


问:
Finding Joint PDF of Two Non-Independent Continuous Random Variables
I'm in the process of reviewing some stats using A First Course in Probability by Sheldon Ross. For the chapter on Joint Distributions, it shows how to obtain the Joint PDF given two independent continuous random variables. However, if the variables weren't independent, how would I go about obtaining the joint PDF of the two variables? Is there a systematic way of going about it similar to when the variables are independent?

So for example, if $f(x)$ and $f(y)$ is the PDF of two continuous independent random variables, I can find their joint PDF $f_{x,y}(x,y)$ by simply multiplying $f(x)$ and $f(y)$. However, how will I find $f_{x,y}(x,y)$ if $X$ and $Y$ were not independent?

Thanks!

答:
Sometimes the dependence structure can be derived reading the text:

Let be $X , Y$ two uniform rv's in $(0;1)$ where
it is known that $X>Y$

Reading the text you can realize that

$$0<y<x<1$$

this inequalities are represented in the following drawing by the pink triangle

Thus the joint density is uniform in the pink triangle:

$$f_{XY}(x,y)=2$$

if $(x,y) \in T$ and zero elsewhere

In a compact way:

$$f_{XY}(x,y)=2\cdot\mathbb{1}_{(0;1)}(x)\cdot\mathbb{1}_{(0;x)}(y)$$


问:
Integrate $\int_0^{\infty} e^{x-ex} dx$
Integrate $\int_0^{\infty} e^{x-ex} dx$.

I'm having a hard time integrating this...

I tried u-substitution. But at one point you get $\infty-\infty$.

According to online calculator, the answer seems to be $\frac{1}{e-1}$.

Any suggestions will be appreciated!

Title: Integrate $\int_0^{\infty} e^{x-ex} dx$

Body: $$I=\int_{0}^{\infty} e^{x-ex} dx=\int_{0}^{\infty} e^{-(e-1)x} =-\frac{e^{-(e-1)x}}{e-1}|_{0}^{\infty}=\frac{1}{e-1}.$$


问:
Infinite series and its upper and lower limit.
I am learning analysis on my own and I am puzzled with the following question.

Consider the series $$\frac{1}{2}+\frac{1}{3}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^3}+\cdots$$ Indicate whether this series converges or diverges.

The following is what I was able to get.

First I noticed that $$a_n = \frac{1}{2^n}+\frac{1}{3^n}$$ would be a simple and nice form of this sequence, so according to the ratio test, $$\lim_{n \to \infty} \frac{1}{6} \left( \frac{3+2(\frac{2}{3})^n}{1+(\frac{2}{3})^n}\right) = \frac{1}{2}$$ therefore the series must converge.

However, I later understood the concept of upper an lower limits a little, so taking $$ a_n =
\begin{cases}
\frac{1}{2^n},  & \text{if $n$ is odd} \\
\frac{1}{3^{n-1}}, & \text{if $n$ is even}  \\
\end{cases}$$ I figured out the following $$\lim_{n \to \infty} \frac{1}{2}\left(\frac{2}{3}\right)^n = 0$$ when $n$ is odd and $$\lim_{n \to \infty} \frac{1}{2}\left(\frac{3}{2}\right)^n = +\infty$$ when $n$ is even.

Through this experience I realized and really understood how a subsequence of a sequence may have multiple limits, and the supremum and the infimum of those limits give us the upper and lower limits, respectively. But the book does something different than I did, and I would like to have the following questions answered.

1),$\quad$ From my understanding, the ratio test tells us that $\Sigma a_n$ diverges when the upper limit is greater than 1. Clearly one of the limits goes to $+\infty$, so why cannot we conclude that the series diverge ?

2),$\quad$  I learned that by choice of the subsequence, the limit may or mayn't be different. In the case that it is different, how do we know what is the supremum and infimum? It would be clear that if they were $\pm \infty$ but in my case, why is it guaranteed that there is no other subsequence that has a negative limit ? Is there an algebraic way to guarantee them ? Or do we always have to conceptually understand all of the subsequential limits.

3),$\quad$  The book says the ratio test is inconclusive, which turns me back to question 1), but it used the ratio test to determine that the series converges. According to my even-odd wise definition of $a_n$, the "larger limit" (I am purposefully not saying supremum because I don't know how it's guaranteed that it is the supremum) using the ratio test can be found as $$\lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \sqrt[n]{\frac{1}{2^n}} = {1 \over 2}$$ But the book says $$\lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \sqrt[2n]{\frac{1}{2^n}} = {1 \over \sqrt{2}}.$$ Where did the $2n$ come from ?

Can someone help me out ?

Edit:Thanks for helping me out everyone! I really appreciate it.

答:
1) ... is divergent when the lower limit of $a_{n+1}$ to $a_n$ is greater than $1$. If it, indeed, says "the upper" there, it is a misprint.

2) I cannot understand what you are asking. Can you restate the question in a clearer way?

3) Formally $1/2^n$ stands in the $2n-1$-th position in the original series, so it should really be $\lim_{n\to\infty}\sqrt[2n-1]{1/2^n}=\frac{1}{\sqrt{2}}$. Of course, if you remove half of the terms, you change the indexation and, as a consequence, the quantity in the root test.


答:
Let $y=(1-e)x$, then we have
$$
\int_0^{\infty}e^{x-ex}\mathrm{d}x = \frac{1}{1-e}\int_0^{-\infty}e^{y}\mathrm{d}y = \frac{1}{e-1}.
$$

Title: Integrate $\int_0^{\infty} e^{x-ex} dx$

Body:
Integrate $\int_0^{\infty} e^{x-ex} dx$.

I'm having a hard time integrating this...

I tried u-substitution. But at one point you get $\infty-\infty$.

According to online calculator, the answer seems to be $\frac{1}{e-1}$.

Any suggestions will be appreciated!


问:
The cross product of two sets
Just reading about the cross product. It says here:
$$A\times B= \{(x,y)|x\in A, y\in B\}$$
Does this mean that the result of $A\times B$ is a new set of ordered pairs? If so, what happens if both A and B share a common member? What would the result be, since an ordered pair cannot include duplicates?

Edit: Quoting the free textbook I'm looking at online,

If we are given that A is a set and no other information about A, then there is no
  ordering to the elements of A. Thus, we cannot speak of “the second element of the set A”
  unless we have specified an ordering of the elements of A. If we wish to regard A as ordered in some way, then we specify this fact explicitly: “The elements of A are ordered a, b, c,” or “A = (a, b, c).” The latter notation replaces the braces with parentheses and designates that A is ordered, left to right, as indicated. We call this an ordered set. An ordered set is also called a linear order. Various other names are also used: list, vector, string, word — all with no repeated elements. Of course, you’ve seen repeated elements in vectors, for example the point in the plane at the coordinates (1,1). That’s fine, it’s just not an ordered set.

Why is it okay to specify a set S = {a, b, c, a} where the element a has been repeated,
  but it is not okay to have repeated elements in an ordering of S?   When we say S  =
  {a, b, c, a}, we know that S contains just the three elements a, b and c.  If we were to talk
  about the ordered set (a, b, c, a) it would not make sense because it would say that the
  element a is in two places at once: the first position and the last position.

答:
No,This is no the real multiplication of sets.The real multiplication of sets is:
f(x)   multiplicity of x
A*B={x+y:f(x)f(y)|x:f(x)∈A,y:f(y)∈B}
when f(x)=f(y)=1
AB={x+y|x∈A,y∈B}
why?because:
P({a,b})={0,{a},{b},{a,b}}
P({c,d})={0,{c},{d},{c,d}}
P({a,b,c,d})={0,{c},{d},{c,d},{a},{a,c},{a,d},{a,c,d},{b},{b,c},{b,d},{b,c,d},{a,b},{a,b,c},{a,b,d},{a,b,c,d}}
P(A+B)=P(A)*P(B)
People do not know the real multiplication of sets.

Title: The cross product of two sets


问:
Standard error and multicollinearity
"The two most correlated features and A and B, and neither of them are significant in the linear regression. This could be due to the inflation of standard errors caused by multicollinearity."

Two questions regarding the statement above:

Why does multicollinearity cause the inflation of standard errors?

Why can the inflation of standard errors cause variables to be insignificant?

Thanks!

答:
A regression is meant to measure the relationship between each independent variable and the outcome, holding all other covariates constant. If you have multiple independent variables that are highly correlated, the estimates of the individual relationships become more difficult to tease apart from each other, since your data doesn't have a lot of variation in one variable while holding another constant. This doesn't affect the point estimate of any of the coefficients, but makes the ones from highly correlated variables more uncertain, hence the higher standard errors.

Statistical significance in linear regression is typically measured by a $t$-statistic of the form $$ t = \frac{\hat{\beta}_i}{\text{se}_i} , $$ where $\text{se}_i$ is the standard error of the estimate of $\beta_i$. If the standard error is higher, the observed $t$ statistic will be lower, potentially to the point that the estimate of $\beta_i$ is no longer statistically significant.

Title: Standard error and multicollinearity


问:
Find rational numbers $c_0$, $c_1$, ..., $c_{d-1}$ such that $2 a^{-1} - 3 a^6 = c_0 + c_1 a + \cdots + c_{d-1} a^{d-1}$.
Let $a = \sqrt{2 - \sqrt{2}}$.

(1) Find the minimal polynomial of $a$ over $\mathbb{Q}$.

(2) Let $d = [\mathbb{Q}(a) ~:~ \mathbb{Q}]$. Compute $d$ and find rational numbers $c_0$, $c_1$, ...,
$c_{d-1}$ such that

$\hspace{10mm} 2 a^{-1} - 3 a^6 = c_0 + c_1 a + \cdots + c_{d-1} a^{d-1}$.

(3) Prove or disprove that all roots of the minimal polynomial of $a$ over $\mathbb{Q}$ belongs to $\mathbb{Q}(a)$.

$\textbf{My Attempt:}$

For question (1):

Since, $a = \sqrt{2 - \sqrt{2}} \implies (2 - a^2)^2 = 2 \implies a^4 - 4a^2 + 2 = 0$.

Notice that $x^4 - 4 x^2 + 2$ is irreducible over $\mathbb{Q}$ and $a^4 - 4a^2 + 2 = 0$ and $x^4 - 4 x^2 + 2$ is monic polynomial.

So, the minimal polynomial of $a$ over $\mathbb{Q}$ is $x^4 - 4x^2 + 2$.

For question (2):

Since, the minimal polynomial of $a$ over $\mathbb{Q}$ is $x^4 - 4x^2 + 2$. Then, the degree of it is $4$.

So, $d = [\mathbb{Q}(a) ~:~ \mathbb{Q}] = $ the degree of the minimal polynomial of $a$ over $\mathbb{Q}$ which equals to $4$.

For question (3):

Since, the minimal polynomial of $a$ over $\mathbb{Q}$ is $x^4 - 4x^2 + 2$.

Which $x^4 - 4x^2 + 2 = (x - \sqrt{2 - \sqrt{2}})(x + \sqrt{2 - \sqrt{2}})(x - \sqrt{2 + \sqrt{2}})(x + \sqrt{2 + \sqrt{2}})$.

Then, $\pm \sqrt{2 - \sqrt{2}}$, $\pm \sqrt{2 + \sqrt{2}}$ are all roots to $x^4 - 4x^2 + 2$.

But $-\sqrt{2 + \sqrt{2}}$ and $\sqrt{2 + \sqrt{2}}$ are not belongs to $\mathbb{Q}(a)$.

So, we disproved that all roots of the minimal polynomial of $a$ over $\mathbb{Q}$ belongs to $\mathbb{Q}(a)$.

$\textbf{My Question:}$

First, for question (2) I am stuck on finding the rational number $c_0$, $c_1$, ...,
$c_{d-1}$ such that

$\hspace{10mm} 2 a^{-1} - 3 a^6 = c_0 + c_1 a + \cdots + c_{d-1} a^{d-1}$.

I can only find out that $2a^{-1} - 3a^6 = (2 + \sqrt{2}) a - (18 - 12\sqrt{2}) a$

which I can not further convert to $c_0 + c_1 a + c_2 a^2 + c_3 a^3$ where $c_0, ..., c_3 \in \mathbb{Q}$.

Also, are the other question correct ?

Body:
Note that
$$
a^4 - 4a^2 + 2 = 0 \implies  2 = -a^4 + 4a^2 \implies 2a^{-1} = -a^3+4a \\
a^4 - 4a^2 + 2 = 0 \implies a^4 = 4a^2-2 \implies a^6 = 4a^4-2a^2 = 4(4a^2-2)-2a^2
$$
and, since $a^2 = 2-\sqrt2$,

$$
\begin{align}
\pm \sqrt{2+\sqrt2} &= \pm \sqrt{\frac{(2+\sqrt2)(2-\sqrt2)}{2-\sqrt2}} \\ &= \pm \sqrt{\frac{2}{2-\sqrt2}}
\\ &= \pm \frac{\sqrt2}{\sqrt{2-\sqrt2}} \\ &= \pm \frac{2-a^2}{a} \in \mathbb Q(a).
\end{align}$$


问:
Difficulty in understanding "Find $E[N]$ , where $N=\min\{n>0: X_n=X_0\}$"
It is Question 30 on page 168 in Ross's book (Introduction to Probability Models-11th edition)

Let $X_i, i\ge 0$ be independent and identically distributed random variables with probability mass function

$p(j)=P(X_i=j), j=1,\ldots,m, \sum_{j=1}^m P(j) = 1$

Find $E[N]$, where $N=\min\{n>0: X_n = X_0\}$

The same question was here, but I don't have enough credit to comment on. Find $E[N]$, where $N = \min\{n>0: X_n = X_0\}$

Here are my questions:

What is the sample space for it? Can anyone give me a simple example? or any references?

What does $X_n = X_0$ mean for 2 random variables with the same distribution?

Is it possible to resolve it with conditional probability?

Are there any applications of such a question? What is the point behind the question?

Any help would be greatly appreciated.

Update 1:
The Solution says:

E[N] = $\sum_{j=1}^m$ E[N|Xo = j] * p(j) = $\sum_{j=1}^m (1/p(j)) * p(j) = m$

I am totally confused, so I want to confirm a few things.

$N=\min\{n>0: X_n = X_0\}$: Is this the least index of indices of Xi that are equal to X0 (having the same j with X0)?

答:
Thank you @Fimpellizieri for your answer. I was thinking to explain Solution in simple words.

Assume the 0th variable $X_0=j$, what means its probability is $p(j)$. Now the question is: at what $n$ is the next occurrence of $j$? If $X_n=j$ that means $X_1 \neq j, X_2 \neq j, ..., X_{n-1} \neq j$. What is clearly a Geometric distribution, so $P(X_n=j)=(1-p(j))^{n-1}p(j)$. In case of Geometric distribution, we know $E[N]=\frac{1}{p(j)}$.  The value of a random variable $X_0$ depends on its probability $P\{X_0=j\}=p(j)$, that is why we condition on $X_0=j$:
$$ E[N]=E\left[E[N|X_0=j]\right]=E\left[\frac{1}{p(j)}\right]=\sum^m_{j=1}\frac{1}{p(j)}p(j)=m$$


答:
You need to make use of the Chain Rule: https://en.wikipedia.org/wiki/Chain_rule.

$\frac{dy}{dx} = \frac{dy}{d\theta} \frac{d\theta}{dx}$.

The first term on the RHS is simple to calculate since you have been given $y$ as a function of $\theta$ but the second term requires a bit more attention. You have been given $x(\theta)$ and you want to get $\frac{d\theta}{dx}$.

$x = x(\theta)$

$\implies \frac{d(x)}{dx} = \frac{d(x(\theta))}{dx}$

$\implies \frac{d(x)}{dx} = \frac{d(x(\theta))}{d\theta} \frac{d\theta}{dx}$

$ \therefore \frac{d\theta}{dx} = \frac{1}{\frac{d(x(\theta))}{d\theta}}$

Apply these steps and also be mindful of you trigonometric identities. Good luck!

Title: Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2\sin\theta}{2\sin\theta-1}$

Body:
Parametric equation of curve is $ x=2\sin\theta +\cos(2\theta) $ , $y=1+\cos(2\theta)$ , for $0\leq\theta\leq\pi/2$

Show that $$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2\sin\theta}{2\sin\theta-1}$$
I drived $x$ and $y$ seperatly for $x$,

$$ x=2\cos\theta+\sin\theta\pm\sin\theta$$ and for the

derivative of $y$ i got $y=-2\sin(2\theta)$

and then I did

$$dy/dx= y=-2\sin(2\theta), \qquad   x=2\cos\theta+\sin\theta\pm \sin\theta$$

but I dont know if I did it wronge or I don't know how to continue. Any help would be much appreciated. Thank you!


问:
What is the standard way of showing that some point does not have a countable basis?
I apologize that I cannot post the exact question and I know this is too vague, but I am asked to show that, around some point $x$ in some quotient space, it does not have a countable basis around that point.
I was guessing that, then around $x$, it does have to have an uncountable basis. But I'm getting nowhere from here, and I cannot really think of any trick.

Maybe I need to use the contradiction, but I'm not sure what "having countably many basis" will imply. Could someone give me any input?

答:
The commonest way I've seen is a kind of diagonalisation argument: given countably many $(B_n)_n$ open neighbourhoods of $x$ we can find an open neighbourhood $O$ of $x$ such that $B_n \nsubseteq O$ for all $n$.

You could also contradict a consequence of first countability: e.g. $x \in \overline{A}$ for some set $A$ while there is no sequence from $A$ converging to $x$.

It all depends on who the topology is defined. For quotients the first way is the most common, maybe because open sets are easy to recognise.


问:
Conditional probability of joint events
Let $X, Y, Z$ be three random variables. Is the equality
$$
P(X=x, Y=y |Z=z) = P(X=x|Y=y,Z=z) P(Y=y|Z=z)
$$
true?
Note that $P(X,Y)$ denotes the joint probability of random variables $X$ and $Y$.

答: The equality holds true given $P(Z=z)>0$ and $P(Y=y,Z=z)>0$. This immediately follows from the definition of the conditional probability: If $A, \ B$ are events, such that $P(B)>0,$ then $P(A|B)=P(A,B)/P(B)$. So we have $$ \frac{P(X=x,Y=y|Z=z)}{P(Y=y|Z=z)} = \frac{P(X=x,Y=y,Z=z)}{P(Z=z)}\frac{P(Z=z)}{P(Y=y,Z=z)} = \frac{P(X=x,Y=y,Z=z)}{P(Y=y,Z=z)} = P(X=x|Y=y,Z=z).$$

Title: Conditional probability of joint events


答:
By pairing adjacent terms, you’ve actually changed the series with which you’re working: you’re working with $\sum_{n\ge 1}\left(\frac1{2^n}+\frac1{3^n}\right)=\sum_{n\ge 1}\frac{2^n+3^n}{6^n}$, the series whose $n$-th term is $\frac{2^n+3^n}{6^n}$. The ratio test works fine on this series:

$$\begin{align*}
\lim_{n\to\infty}\frac{\frac{2^{n+1}+3^{n+1}}{6^{n+1}}}{\frac{2^n+3^n}{6^n}}&=\lim_{n\to\infty}\frac{2^{n+1}+3^{n+1}}{6(2^n+3^n)}\\
&=\lim_{n\to\infty}\frac{2^{n+1}}{6(2^n+3^n)}+\lim_{n\to\infty}\frac{3^{n+1}}{6(2^n+3^n)}\\
&=\lim_{n\to\infty}\frac{2^n}{3(2^n+3^n)}+\lim_{n\to\infty}\frac{3^n}{2(2^n+3^n)}\\
&=\lim_{n\to\infty}\frac1{3\left(1+\left(\frac32\right)^n\right)}+\lim_{n\to\infty}\frac1{2\left(\left(\frac23\right)^n+1\right)}\\
&=0+\frac12\\
&=\frac12\;,
\end{align*}$$

which is essentially the calculation that you made. However, if you apply the ratio test to the original series, you find that if the $n$-th term is $a_n$, then

$$a_n=\begin{cases}
\frac1{2^{(n+1)/2}},&\text{if }n\text{ is odd}\\\\
\frac1{3^{n/2}},&\text{if }n\text{ is even}\;,
\end{cases}$$

so that

$$\begin{align*}
\frac{a_{n+1}}{a_n}&=\begin{cases}
\frac{1/3^{(n+1)/2}}{1/2^{(n+1)/2}},&\text{if }n\text{ is odd}\\\\
\frac{1/2^{(n+2)/2}}{1/3^{n/2}},&\text{if }n\text{ is even}
\end{cases}\\\\
&=\begin{cases}
\left(\frac23\right)^{(n+1)/2},&\text{if }n\text{ is odd}\\\\
\frac12\left(\frac32\right)^{n/2},&\text{if }n\text{ is even}\;.
\end{cases}
\end{align*}$$

Thus, $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$ does not exist: the terms with odd indices are approaching $0$, but those with even indices are increasing without bound. Since the limit does not exist, the ratio test is inconclusive.

The ratio test applied to your modified series gives the correct answer because the original series is absolutely convergent; this permits you to combine the adjacent terms as you did, but since you don’t know ahead of time that the series converges, you can’t make use of the fact to show that it converges.

Title: Infinite series and its upper and lower limit.

Body:
I am learning analysis on my own and I am puzzled with the following question.

Consider the series $$\frac{1}{2}+\frac{1}{3}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^3}+\cdots$$ Indicate whether this series converges or diverges.

The following is what I was able to get.

First I noticed that $$a_n = \frac{1}{2^n}+\frac{1}{3^n}$$ would be a simple and nice form of this sequence, so according to the ratio test, $$\lim_{n \to \infty} \frac{1}{6} \left( \frac{3+2(\frac{2}{3})^n}{1+(\frac{2}{3})^n}\right) = \frac{1}{2}$$ therefore the series must converge.

However, I later understood the concept of upper an lower limits a little, so taking $$ a_n =
\begin{cases}
\frac{1}{2^n},  & \text{if $n$ is odd} \\
\frac{1}{3^{n-1}}, & \text{if $n$ is even}  \\
\end{cases}$$ I figured out the following $$\lim_{n \to \infty} \frac{1}{2}\left(\frac{2}{3}\right)^n = 0$$ when $n$ is odd and $$\lim_{n \to \infty} \frac{1}{2}\left(\frac{3}{2}\right)^n = +\infty$$ when $n$ is even.

Through this experience I realized and really understood how a subsequence of a sequence may have multiple limits, and the supremum and the infimum of those limits give us the upper and lower limits, respectively. But the book does something different than I did, and I would like to have the following questions answered.

1),$\quad$ From my understanding, the ratio test tells us that $\Sigma a_n$ diverges when the upper limit is greater than 1. Clearly one of the limits goes to $+\infty$, so why cannot we conclude that the series diverge ?

2),$\quad$  I learned that by choice of the subsequence, the limit may or mayn't be different. In the case that it is different, how do we know what is the supremum and infimum? It would be clear that if they were $\pm \infty$ but in my case, why is it guaranteed that there is no other subsequence that has a negative limit ? Is there an algebraic way to guarantee them ? Or do we always have to conceptually understand all of the subsequential limits.

3),$\quad$  The book says the ratio test is inconclusive, which turns me back to question 1), but it used the ratio test to determine that the series converges. According to my even-odd wise definition of $a_n$, the "larger limit" (I am purposefully not saying supremum because I don't know how it's guaranteed that it is the supremum) using the ratio test can be found as $$\lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \sqrt[n]{\frac{1}{2^n}} = {1 \over 2}$$ But the book says $$\lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \sqrt[2n]{\frac{1}{2^n}} = {1 \over \sqrt{2}}.$$ Where did the $2n$ come from ?

Can someone help me out ?

Edit:Thanks for helping me out everyone! I really appreciate it.


问:
Inverse limit without category theory
In Hirsch's book Differential Topology, he by and large does not use any category theory, with the exception of one passage on pg. 52 which I am trying to understand.  It is as follows (paraphrased):

Let $X$ be a set and $\mathcal{A}$ a family of subsets, such that $\mathcal{A}$ contains $X$ and is closed under unions.  Suppose we have a contravariant functor from the partially ordered (by inclusion) set $\mathcal{A}$ to the category of sets.  That is, to every $A\in\mathcal{A}$ there is associated a set $\mathcal{F}(A)$, and to every pair of sets $A,B\in\mathcal{A}$ with $A\subset B$ there is assigned a map of sets $\mathcal{F}_{AB}:\mathcal{F}(B)\rightarrow \mathcal{F}(A)$, denoted by $x\mapsto x|A$, where $x\in\mathcal{F}(B)$ such that $\mathcal{F}_{AB}\mathcal{F}_{BC}=\mathcal{F}_{AC}$ whenever $A\subset B\subset C$, and $\mathcal{F}_{AA}=$ identity map of $\mathcal{F}(A)$.

...

A structure functor is continuous if the following holds.  If $\{Y_\alpha\}$ is any simply ordered family of elements of $\mathcal{A}$, and $\cup Y_\alpha=Y$, then the inverse limit of the maps $\mathcal{F}_{Y_{\alpha}Y}:\mathcal{F}(Y)\rightarrow \mathcal{F}(Y_\alpha)$ is a map $\mathcal{F}(Y)\rightarrow \text{inv lim} \mathcal{F}(Y_\alpha)$.

(It seems kind of strange that he explains what a functor is, but assumes the reader knows what an inverse limit is, but whatever.)

So I am trying to understand the last paragraph--in particular, what an inverse limit is, and what a continuous structure functor is--without knowing anything really about category theory beyond the definition of a functor given in the first paragraph.  Is there a way to understand the inverse limit in this specific instance without knowing category theory more generally?

(If the answer is simply that I need to go read up on category theory, that's fine, and I would appreciate if someone would tell me so.)

答: In topological spaces we can realise inverse limits of spaces as subspaces of products. So as a means to construct spaces we don't really need category theory. But I don't know the text you're referring to, and this might use these concepts more extensively than just as a way to construct a space.</p>

Title: Inverse limit without category theory


问:
Leibniz test and $\sum\frac{(-1)^k k^2}{\sqrt{k!}}$
I would like to check if the following series $\sum\frac{(-1)^k k^2}{\sqrt{k!}}$ is convergent.

My original idea was to employ Leibniz's criterion, the one that says: An alternating series $(-1)^k a_k$ such that $a_k$ is a non-increasing sequence that satisfies $\lim a_k=0$ then the series converges, however I'm having difficults when showing that my $a_k$ is non-increasing through induction.

I would appreciate some help.

Body:
[Note] : $a_{n}>0$ is non-increasing if and only if $\displaystyle\frac{a_{n+1}}{a_{n}}\leq1$.
$$
\frac{a_{n+1}}{a_{n}}=\frac{\frac{(n+1)^{2}}{\sqrt{(n+1)!}}}{\frac{n^{2}}{\sqrt{n!}}}=\frac{(n+1)^{2}}{n^{2}}\frac{\sqrt{n!}}{\sqrt{(n+1)!}}=\frac{(n+1)^{2}}{n^{2}}\frac{1}{\sqrt{n+1}}
$$
$$
\lim_{n\to\infty}\frac{(n+1)^{2}}{n^{2}}\frac{1}{\sqrt{n+1}}=1\cdot0=0<1
$$
Clearly non-increasing.


问:
If the union of infinite sequences of countable sets is countable, then how is the set of subsets of the rational numbers uncountable?
I'm reading Rudin's book and this won't quite click for me. On one hand, you can prove quite easily that the set of subsets of rational numbers is uncountable (almost like cantor's diagonal argument.) On the other hand the union of infinite sequences of countable sets is countable. But the set of subsets of rational numbers is just the union of subsets of rational numbers with cardinality n, for all natural numbers n. So it should be countable, but it clearly isn't. I'm confused, can someone explain please?</p>

答:
For every real number $r$ there is a sequence of all distinct rationals $q_n(r)$ that converges to $r$.

So there are at least as many sequences in $\Bbb Q$ (and infinite subsets of $\Bbb Q$) as there are reals. So uncountably many.

A countable union of countable subsets is countable, but that's another unrelated fact.


问:
problems on Lebesgue integral

1) Given a measure space, $f$ a non-negative measurable function and $A$ in the $\sigma$-algebra such that $\mu(A)=0$, prove that $\displaystyle\int_{A} f\;d\mu=0$ .

My try:

$$0\leq \int_{A}f\;d\mu=\int fI_{A}\;d\mu\leq \int \sup{f(A)}I_{A}\;d\mu=\sup{f(A)}\int I_{A}\;d\mu=\sup{f(A)}\mu(A)=0\;,$$
then $\displaystyle\int_{A}f\;d\mu=0\,.$

2) Given a measure space, let $f$ be an integrable function. Show that $\displaystyle \int_{\{|f|>t\}} |f|\;d\mu\rightarrow 0$ as $t\rightarrow\infty$ .

My try: since $f$ is integrable, $|f|\lt \infty$ almost everywhere, $\mu (\{|f|>t\}) = \mu(\Omega)- \mu(\{|f|\leq t\})$, and as $t\rightarrow\infty$,  $\mu (\{|f|>t\})<\epsilon$, then
$$\int_{\{|f|>t\}} |f|\;d\mu=\int |f|I_{\{|f|>t\}}\;d\mu\leq\max\{|f|\}\int I_{\{|f|>t\}}=\max\{|f|\}\;\mu(\{|f|>t\})<\max\{|f|\}\;\epsilon\;.$$
Then it's done; I can take $\epsilon=\frac1t$.

Well, both demonstrations seem right for me. I will be thankful if someone tells me what's wrong on them.

Greetings.

答:
Solution to (2): I replaced t with n cause that's how I like it, haha. Let $E_n = \{ x: |f(x)|>n \}$. Note that $E_n \supset E_{n+1}$, we have a decreasing
sequence of sets. And $\bigcap_n E_n = \{ x: |f(x)| = \infty \}$.
Now since $f$ is integrable, so is $|f|$. We define a measure $\lambda(E) = \int_E |f| d\mu$
for all $E$ in the $\sigma$-algebra.
Note that $\lambda(E_1) = \int_{E_1} 1 d\mu\ \leq \int_{E_1} |f| d\mu\ < \infty$.
And since $\mu(\bigcap_n E_n) = 0, \lambda(\bigcap_n E_n) = 0$. Thus
\begin{equation*}
\lim_{n \to \infty} \int_{E_n} |f| d\mu\ =
\lim_{n \to \infty} \lambda(E_n) = \lambda(\bigcap_n E_n) = 0.
\end{equation*}

Remark: Since $f$ is integrable, so is $|f|$. Therefore if it so happens that $|f| = \infty$ at some set, that set has measure zero! So for this problem, we need not worry about it.

Title: problems on Lebesgue integral


问:
Prove that $\exists a,g,h\in G\colon h=aga^{-1}, g\neq h ,gh=hg$ in a finite non-abelian group $G$.
Let $G$ be a finite and non-abelian group.  How do I prove the following statement? $$\exists a,g,h\in G \colon\quad h=aga^{-1},\ g\neq h ,\ gh=hg.$$

Thanks in advance.

答:
Here is an outline which avoids the Miller-Moreno classification.  It requires a couple (challenging!) exercises.

By induction on group order, we can assume every proper subgroup of $G$ is abelian.  This is enough to conclude $G$ is solvable.

Because $G$ is solvable, there is a maximal subgroup which is also normal - call it $M$.

So $M$ is a maximal subgroup, which is also normal.  Thus $G/M$ is cyclic of prime order.  Choose a $g\in G$ such that $\bar{g}$ generates $G/M$.

Then we cannot have $g$ centralizing $M$, for otherwise $G$ would be abelian. So there is an $x\in M$ such that $x^g\neq x$.  However, both $x^g$ and $x$ are in $M$, so $[x^g,x]=1$.


问:
Let $a = \sqrt{2 - \sqrt{2}}$. Prove that $\mathbb{Q}(\sqrt{2})$ is a subfield of $\mathbb{Q}(a)$
Let $a = \sqrt{2 - \sqrt{2}}$.

Find the minimal polynomial of $a$ over $\mathbb{Q}(\sqrt{2})$ and prove that $\mathbb{Q}(\sqrt{2})$ is a subfield of $\mathbb{Q}(a)$.

Notice that $\mathbb{Q}(\sqrt{2}) = \{ b_0 + b_1 \sqrt{2} ~|~ b_0, b_1 \in \mathbb{Q} \}$.

Also, $\mathbb{Q}(a) = \{ c_0 + c_1 a + c_2 a^2 + c_3 a^3 ~|~ c_0, c_1, c_2, c_3 \in \mathbb{Q} \}$.

$\textbf{My Attempt:}$

Since, $a = \sqrt{2 - \sqrt{2}} \implies a^2 + (\sqrt{2} - 2) = 0$.

Notice that $x^2 + (\sqrt{2} - 2)$ is irreducible over $\mathbb{Q}(\sqrt{2})$ and $a^2 + (\sqrt{2} - 2) = 0$ and $x^2 + (\sqrt{2} - 2)$ is monic polynomial.

So, the minimal polynomial of $a$ over $\mathbb{Q}(\sqrt{2})$ is $x^2 + (\sqrt{2} - 2)$.

Prove that $\mathbb{Q}(\sqrt{2})$ is a subfield of $\mathbb{Q}(a)$:

Notice that $\mathbb{Q}(\sqrt{2}) = \{ b_0 + b_1\sqrt{2} ~|~ b_0,b_1 \in \mathbb{Q} \}$ and $\mathbb{Q}(a) = \{ c_0 + c_1 a + c_2 a^2 + c_3 a^3 ~|~ c_0, c_1, c_2, c_3 \in \mathbb{Q} \}$.

Since, $\mathbb{Q}(\sqrt{2}) \subseteq \mathbb{Q}(a)$ [which I don't know how to prove this is true].

Then, $\mathbb{Q}(\sqrt{2})$ is a subfield of $\mathbb{Q}(a)$ can be prove by using the Subfield Test.

Let $x, y \in \mathbb{Q}(\sqrt{2})$ and $y \neq 0$.

Then, $x = a + b\sqrt{2}$ and $y = c + d\sqrt{2}$ where $a,b,c,d \in \mathbb{Q}$.

Since, $x - y = (a + b\sqrt{2}) - (c + d\sqrt{2}) = (a-c) + (b-d)\sqrt{2} \in \mathbb{Q}(\sqrt{2})$.

Also, $xy^{-1} = (a + b\sqrt{2})(c + d\sqrt{2})^{-1} = \frac{ac-2bd}{c^2 - 2d^2} + \frac{bc-ad}{c^2-2d^2}\sqrt{2}$.

Since, $\frac{ac-2bd}{c^2 - 2d^2}$ and $\frac{bc-ad}{c^2-2d^2}$ are all in $\mathbb{Q}$.

Then, $xy^{-1} \in \mathbb{Q}(\sqrt{2})$. So, by the Subfield test $\mathbb{Q}(\sqrt{2})$ is a subfield of $\mathbb{Q}(a)$.

$\textbf{The question is how to prove that $\mathbb{Q}(\sqrt{2}) \subseteq \mathbb{Q}(a)$ ?}$

答: Observe that $\sqrt{2} = 2-a^2 \in \mathbb{Q}[a]$. Element in $\mathbb{Q}[\sqrt{2}]$ is of the form $r_1 + r_2 \sqrt{2}$ for some $r_1,r_2\in\mathbb{Q}$. Clearly $r_1+r_2\sqrt{2} = r_1 + r_2(2-a^2)\in\mathbb{Q}[a]$.