问:
How do I fill in the FFT matrix for the nth roots of unity, for example n = 8
I can do this for n=4, $w_4^0 = 1$ $w_4^1 = i$, $w_4^2=-1$, $w_4^3=-i$ but then anything in between my best guess is that I have to find the Cartesian coordinates?  Which already does not make sense to me as I'm saying it because that implies two numbers and not one.

But I looked around and saw an equation like $cos(theta) + i sin(theta)$

This is for an algorithms class FFT for polynomial multiplication but in exams, I'd have to do it by hand so I want to use matrices rather than divide every polynomial into A_even and A_odd and recurse.

Sorry but I have no clue how to write a matrix in MathJax.

n = 4 has a matrix like this

[1  1  1  1]
[1 -i -1  i]
[1 -1  1 -1]
[1  i -1 -i]

What does the n=8 matrix look like?

答: The standard primitive eighth root of unity is the complex number $\frac{\sqrt 2}2+i\frac{\sqrt 2}2$. You can find all other eighth roots of unity by rasing this to a power. Note that its square is simply $i$.


问:
Is $\int\limits_0^\infty\int\limits_0^x x^{-3/2}e^{y-x}dy\ dx$ convergent?
Does $$I= \int\limits_0^\infty\int\limits_0^x x^{-3/2}e^{y-x}dy\ dx$$ converges, how can I prove it?

I know from Wolfram|Alpha that this integral actually converges to $2\sqrt{\pi}$. This is precisely $$-\Gamma\left(\frac{-1}{2}\right)=-\int\limits_0^\infty x^{-3/2}e^{-x} =2\sqrt{\pi}$$
However when I do $$I=\int\limits_0^\infty x^{-3/2}e^{-x}(e^x-1) dx=-\int\limits_0^\infty x^{-3/2}e^{-x}dx+\int\limits_0^\infty x^{-3/2}dx$$
Which is equivalent to $-\Gamma\left(\frac{-1}{2}\right)+\infty=\infty$.

答: You made algebraic mistakes in the last-but-one line. Here is what you can do: $I \leq \int_0^{1} x^{-3/2} (1-e^{-x}) dx+\int_1^{\infty} x^{-3/2}dx <\infty$ since $1-e^{-x} \leq x$ for $0 <x<1$ and $x^{-1/2}$ is integrable on $(0,1)$.

ContentLicense: CC BY-SA 4.0


问:
Maximum Likelihood Estimator for Logarithmic Distribution
I am trying to calculate the MLE for the logarithmic distribution.
It holds

$$ P(X=k) = -\frac{p^k}{k \cdot \ln(1-p} $$

Thus, the ML function is

$$L_p(X_1,\dots, X_n) = \prod_{i=1}^{n} -\frac{p^{x_i}}{x_i \cdot \ln(1-p)} = p^{\sum_{i=1}^n x_i} \left(\frac{1}{\ln(1-p)}\right)^n \prod_{i=1}^n -\frac{1}{x_i} $$

and the log likelihood function is

\begin{align}
\log L_p(X_1,\dots,X_n)& = \sum_{i=1}^n x_i \cdot \log(p) + n \cdot \log\left(\frac{1}{\log(1-p)}\right) + \log\left(\prod_{i=1}^n -\frac{1}{x_i}\right)
\\&= \sum_{i=1}^n x_i \cdot \log(p) + n \cdot -\log(\log(1-p)) + \log\left(\prod_{i=1}^n -\frac{1}{x_i}\right)
\end{align}

So:

$$\frac{\partial L_p}{\partial p} = \frac{1}{p} \sum_{i=1}^n x_i + n \frac{1}{\log(1-p)(1-p)} \overset{!}{=} 0$$

This is equivalent to:

$$\frac{1}{n} \sum_{i=1}^n x_i = - \frac{p}{\log(1-p)(1-p)}$$

Now I don't know how to go on. How do I get the estimator for $p$?

Thanks in advance, for helping !

Body:
I precise that I am not familiar with this domains you tagged.

Let me suppose that you look for $p$ such that
$$\frac{1}{n} \sum_{i=1}^n x_i = - \frac{p}{\log(1-p)(1-p)}$$ where the value of the lhs is known.

This is the same as  solving for $p$ the equation
$$-\frac 1p \log(1-p)(1-p) =a \qquad \text{where} \qquad a=\frac n{\sum_{i=1}^n x_i}$$ and the lhs is a pretty nice function (plot it).

A very simplistic approximation of the lhs is
$$-\frac 1p \log(1-p)(1-p) \sim 1-\frac 12(p^{1.3}+p^{2.6})$$ which would give as an estimate
$$p_0=\Bigg[\frac{\sqrt{9-8 a}-1}{2}\Bigg]^{\frac {10}{13}} $$
The solution will be easily obtained using very few iterations of Newton method
$$p_{n+1}=\frac{p_n ((a+1) p_n+(2-p_n) \log (1-p_n))}{p_n+\log (1-p_n)}$$

Just to show a few numbers

$$\left(
\begin{array}{ccccc}
 a    &  p_0     &  p_1  & p_2    & \text{exact} \\
 0.05 & 0.973965 & 0.990822 & 0.989085 &  0.989044 \\
 0.10 & 0.947100 & 0.975995 & 0.973133 &  0.973082 \\
 0.15 & 0.919331 & 0.956826 & 0.953390 &  0.953343 \\
 0.20 & 0.890577 & 0.933684 & 0.930116 &  0.930080 \\
 0.25 & 0.860741 & 0.906737 & 0.903375 &  0.903350 \\
 0.30 & 0.829712 & 0.876061 & 0.873142 &  0.873127 \\
 0.35 & 0.797359 & 0.841674 & 0.839341 &  0.839333 \\
 0.40 & 0.763526 & 0.803555 & 0.801865 &  0.801861 \\
 0.45 & 0.728024 & 0.761651 & 0.760580 &  0.760578 \\
 0.50 & 0.690622 & 0.715880 & 0.715332 &  0.715332 \\
 0.55 & 0.651034 & 0.666131 & 0.665952 &  0.665952 \\
 0.60 & 0.608897 & 0.612262 & 0.612254 &  0.612254 \\
 0.65 & 0.563738 & 0.554098 & 0.554036 &  0.554036 \\
 0.70 & 0.514924 & 0.491419 & 0.491084 &  0.491084 \\
 0.75 & 0.461572 & 0.423957 & 0.423166 &  0.423166 \\
 0.80 & 0.402390 & 0.351379 & 0.350039 &  0.350039 \\
 0.85 & 0.335341 & 0.273265 & 0.271443 &  0.271441 \\
 0.90 & 0.256830 & 0.189082 & 0.187101 &  0.187099 \\
 0.95 & 0.159033 & 0.098164 & 0.096722 &  0.096722
\end{array}
\right)$$

Edit

A better $p_0$ can be generated approximating
$$f(p)=-(1-p)\log(1-p)\qquad \text{by} \qquad g(p)=p(1-p)(\alpha+\beta p+\gamma p^2)$$
Minimizing, with respect to $(\alpha,\beta,\gamma)$ the norm
$$\Phi(\alpha,\beta,\gamma)=\int_0^1 \big[f(p)-g(p)\big]^2\,dp$$ leads to
$$\alpha=-\frac{49}{20}\qquad \qquad \beta=\frac{203}{60}\qquad \qquad \gamma=-\frac{259}{120}$$ which makes that we need now to solve for $p$ the equation
$$p(1-p)(\alpha+\beta p+\gamma p^2)= ap$$ which is a cubic with one real root given by
$$p_0=\frac{1}{63} \left(29-\sqrt{1298} \sinh \left(\frac{1}{3} \sinh
   ^{-1}\left(\frac{1}{649} \sqrt{\frac{2}{649}} (102060
   a-72403)\right)\right)\right)$$ The iterates will be
$$p_{n+1}=\frac{p_n+\log (1-p_n)}{\log (1-p_n)+1-a}$$
Repeating the same table as before
$$\left(
\begin{array}{ccccc}
 a    &  p_0     &  p_1  & p_2    & \text{exact} \\
 0.05 & 0.981261 & 0.989673 & 0.989050 &  0.989044 \\
 0.10 & 0.961424 & 0.973919 & 0.973087 &  0.973082 \\
 0.15 & 0.940327 & 0.954123 & 0.953346 &  0.953343 \\
 0.20 & 0.917764 & 0.930652 & 0.930081 &  0.930080\\
 0.25 & 0.893478 & 0.903668 & 0.903351 &  0.903350 \\
 0.30 & 0.867132 & 0.873231 & 0.873127 &  0.873127 \\
 0.35 & 0.838282 & 0.839336 & 0.839333 &  0.839333 \\
 0.40 & 0.806326 & 0.801910 & 0.801861 &  0.801861 \\
 0.45 & 0.770425 & 0.760804 & 0.760578 &  0.760578 \\
 0.50 & 0.729386 & 0.715776 & 0.715332 &  0.715332 \\
 0.55 & 0.681491 & 0.666490 & 0.665953 &  0.665952 \\
 0.60 & 0.624389 & 0.612589 & 0.612254 &  0.612254 \\
 0.65 & 0.555658 & 0.554043 & 0.554036 &  0.554036 \\
 0.70 & 0.475971 & 0.491719 & 0.491085 &  0.491084 \\
 0.75 & 0.393972 & 0.426015 & 0.423189 &  0.423166 \\
 0.80 & 0.320678 & 0.353488 & 0.350077 &  0.350039 \\
 0.85 & 0.259426 & 0.272093 & 0.271443 &  0.271441 \\
 0.90 & 0.208371 & 0.189219 & 0.187125 &  0.187099 \\
 0.95 & 0.164984 & 0.117574 & 0.099978 &  0.096722
\end{array}
\right)$$


问:
What kind of distributions are the following
Let $Y_1, . . . , Y_5$ be a random sample of size 5 from a normal population with mean 0 and variance 1 and let $\bar{Y} = (1/5) \Sigma_{i=1}^5 Y_i$

Let $Y_6$ be another independent observation from
the same population. Finally, let $U = \Sigma_{i=1}^5 (Y_i - \bar{Y})^2$ and let $W = \Sigma_{i=1}^5 Y_i^2 $.

What is the distribution of

a. $\frac{\sqrt{5} * Y_6}{\sqrt{W}} $

b. $\frac{2 * Y_6}{\sqrt{U}} $

c. $\frac{2*(5\bar{Y}^2 + Y_6^2)}{U} $

Why?

I don't really get how to examine the problems. Maybe if I understood the answer to a. I could understand the other 2 but I don't have an answer to any of them right now.

Edit:
The distributions are probably either
T distribution
Chi distribution
normal
but it could be something else I guess

答:
Let's see our components:

$$U=\frac{(n-1)S^2}{\sigma^2}\sim\chi_{(4)}^2$$

$$Y_6\sim N(0;1)$$

Obviously $Y_6^2\sim \chi_{(1)}^2$

$$\overline{Y}\sim N\left(0;\frac{1}{5}\right)$$

That is

$$\sqrt{5}\cdot\overline{Y}\sim N(0;1)$$

and obviuosly

$$5\overline{Y}^2\sim\chi_{(1)}^2$$

Thus

$$V=5\overline{Y}^2+Y_6^2\sim\chi_{(2)}^2$$

Being in a gaussian model $V\perp\!\!\!\!\!\!\perp U$ your distribution is

$$\frac{\frac{V}{2}}{\frac{U}{4}}\sim F(2;4)$$


Title: Is $\int\limits_0^\infty\int\limits_0^x x^{-3/2}e^{y-x}dy\ dx$ convergent?

Body:
Does $$I= \int\limits_0^\infty\int\limits_0^x x^{-3/2}e^{y-x}dy\ dx$$ converges, how can I prove it?

I know from Wolfram|Alpha that this integral actually converges to $2\sqrt{\pi}$. This is precisely $$-\Gamma\left(\frac{-1}{2}\right)=-\int\limits_0^\infty x^{-3/2}e^{-x} =2\sqrt{\pi}$$
However when I do $$I=\int\limits_0^\infty x^{-3/2}e^{-x}(e^x-1) dx=-\int\limits_0^\infty x^{-3/2}e^{-x}dx+\int\limits_0^\infty x^{-3/2}dx$$
Which is equivalent to $-\Gamma\left(\frac{-1}{2}\right)+\infty=\infty$.

Body: $\int_0^\infty x^{-3/2}e^{-x}\,dx\neq\Gamma(-1/2)$ because the integral diverges. However, $$I=\int_0^\infty x^{-3/2}(1-e^{-x})\,dx=-2x^{-1/2}(1-e^{-x})\Bigg|_0^\infty+2\int_0^\infty x^{-1/2}e^{-x}\,dx=2\Gamma(1/2)=2\sqrt\pi.$$


问:
Best Maths Books for Non-Mathematicians
I'm not a real Mathematician, just an enthusiast. I'm often in the situation where I want to learn some interesting Maths through a good book, but not through an actual Maths textbook. I'm also often trying to give people good Maths books to get them "hooked".

So the question: What are good books, for laymen, which teach interesting Mathematics, but actually does it in a "real" way. For example, "Fermat's Last Enigma" doesn't count, since it doesn't actually feature any Maths, just a story, and most textbook don't count, since they don't feature a story.

My favorite example of this is "Journey Through Genius", which is a brilliant combination of interesting storytelling and large amounts of actual Mathematics. It took my love of Maths to a whole other level.

Edit:

A few more details on what I'm looking for.

The audience of "laymen" should be anyone who has the ability (and desire) to understand actual mathematics, but does not want to learn from a textbook. Obviously I'm thinking about myself here, as a programmer who loves mathematics, I love being exposed to real maths, but I'm not going to get into it seriously. That's why books that show actual maths, but give a lot more exposition (and much clearer explanations, especially of what the intuition should be) are great.

When I say "real maths", I'm talking about actual proofs, formulas, or other mathematical theories. Specifically, I'm not talking about philosophy, nor am I talking about books which only talk about the history of maths (Simon Singh style), since they only talk about maths, they don't actually show anything. William Dunham's books and Paul J. Nahin's books are good examples.

答: As an undergrad, I read a fair number of pop math books. The best by far was Ash and Gross' "Fearless Symmetry". This book is very beautiful. It sustains a nice level of rigor while being approachable by those who aren't professionals. Additionally, it weaves the tale of one of the most beautiful recent stories in mathematics. Everyone I know of who have read the book have found it wonderful.</p>

ContentLicense: CC BY-SA 3.0

Title: Best Maths Books for Non-Mathematicians

ContentLicense: CC BY-SA 2.5


问:
Inequality for embedding in Sobolev space
For $\Omega=(0,1). $Prove that there exists $M>0$ such that
$$||u||_{C^0(\overline{\Omega})}\le M||u||_{H^1(\Omega)}$$
for all $u\in H^1(\Omega).$</p>

答:
Assume that $u(0)=0$. Then
$$
u(x)=\int_0^x u'(s)ds=\int_0^1 \textbf{1}_{(0,x)}u'(s)ds.
$$
Now you use Cauchy-Schwarz inequality.
$$
u(x)\leq \|u'\textbf{1}_{(0,x)}\|_{L^2}=\int_0^1 (u'(s))^2\textbf{1}_{(0,x)}ds\leq \int_0^1 (u'(s))^2 ds.
$$
Right?</p>


问:
Poisson Process - Joint Distribution and estimation of $p_i$
Two copy editors read a $300$-page manuscript. The first found $100$ typos, the second found $120$, and their lists contain $80$ errors in common. Suppose that the author's typos follow a Poisson process with some unknown rate $\lambda$ per page, while the two copy editors catch errors with unknown probabilities of success $p_1$ and $p_2$. Let $X_0$ be the number of typos that neither found. Let $X_1$ and $X_2$ be the number of typos found only by $1$ or only by $2$, and let $X_3$ be the number of typos found by both.

($a$) Find the joint distribution of ($X_0;X_1;X_2;X_3$), expressed in $\lambda$, $p_1, p_2$.

($b$) Use the answer to ($a$) to find an estimates of $p_1, p_2$ and then the number of undiscovered typos. (Hint: Let $N(s)$ be the Poisson process, with Ti being the "time" when the i-th typo occurs, and the "time" variable is the page number.)

Here's the thing...

Well, I've got that ($a$) is, given that each error is found by neither editor, $1$ only, $2$ only, and both editors with probabilities $(1-p_1)(1-p_2),p_1(1-p_2),(1-p_1)p_2$ and $p_1p_2$, respectively, and assuming that $x_0+x_1+x_2+x_3=x$,

$\Pr[(X_0,X_1,X_2,X_3)=(x_0,x_1,x_2,x_3)]=\Pr[N=x]f(x_0,x_1,x_2,x_3)$

where $\Pr[N=x]=(300\lambda)^x\exp(-300\lambda)/x!$

and

$f(x_0,x_1,x_2,x_3) ={x\choose x_0,x_1,x_2,x_3}p_1^{x_1+x_3}p_2^{x_2+x_3}(1-p_1)^{x_0+x_2}(1-p_2)^{x_0+x_1}$

is the mass function of the multinomial distribution.

But, i'm not sure how to obtain the values for (b). Can someone help me?
Searching, i've got the following answer:

For (b), we are given that $X_1=20,X_2=40,X_3=80$. It is reasonable to guess that $p_1p_2=2(1-p_1)p_2=4p_1(1-p_2)$, which gives $p_2=4/5$, $p_1=2/3$. Then, $(1−p_1)(1−p_2)=1/15$, so a reasonable estimate for $X_0$ is $80/8=10$.

But why?

答: Another way to obtain your joint density is to recognize that we're dealing with a split Poisson process with independent $X_0,X_1,X_2,X_3$ having distributions $$X_0 \sim \text{Poisson}\Big(300\lambda(1-p_1)(1-p_2)\Big)$$ $$X_1 \sim \text{Poisson}\Big(300\lambda p_1(1-p_2)\Big)$$ $$X_2 \sim \text{Poisson}\Big(300\lambda(1-p_1)p_2\Big)$$ $$X_3 \sim \text{Poisson}\Big(300\lambda p_1 p_2\Big)$$ So the joint pdf of $(X_1,X_2,X_3,X_4)$ which we'll denote by $p_{X_1X_2X_3X_4}$ can be established using independence and without needing to condition on $X_1+X_2+X_3+X_4$: $$p_{X_1X_2X_3X_4}(x_1,x_2,x_3,x_4)=P(X_0=x_0)P(X_1=x_1)P(X_2=x_2)P(X_3=x_3)$$ Next, observe the joint pdf of $(X_1,X_2,X_3)$ simply equals $$p_{X_1X_2X_3}(x_1,x_2,x_3)=P(X_1=x_1)P(X_2=x_2)P(X=x_3)$$ and so $$p_{X_1X_2X_3}(20,40,80)\propto \lambda^{140}p_1^{100}p_2^{120}(1-p_1)^{40}(1-p_2)^{20}e^{-300\lambda(p_1-p_1p_2+p_2)}$$ A natural way to estimate $p_1,p_2,$ and $\lambda$ is to find such parameters which maximizes $p_{X_1X_2X_3}(20,40,80)$ on the domain $$(\lambda,p_1,p_2)\in [0,\infty) \times [0,1]\times [0,1]$$ We find that $p_{X_1X_2X_3}(20,40,80)$ is maximized whenever $$(\lambda,p_1,p_2)=\Big(\frac{1}{2},\frac{2}{3},\frac{4}{5}\Big)$$ I think this is how the problem is intended to be solved as it makes explicit use of the joint density. The technique you and @BruceET utilize to solve this problem is the method of moments which requires that you solve the system of equations $$E(X_1)=20$$ $$E(X_2)=40$$ $$E(X_3)=80$$ This becomes $$300\lambda p_1 (1-p_2)=20$$ $$300 \lambda (1-p_1)p_2=40$$ $$300 \lambda p_1 p_2=80$$ The solution to this system is $(\lambda,p_1,p_2)=\Big(\frac{1}{2},\frac{2}{3},\frac{4}{5}\Big)$ and provides us with the same estimate we obtained by using the joint density explicitly.


问:
(Wrong site) Can't find mistake in resolution of second order non-homogenous linear differential equation
I'm trying to solve: $y'' + 10y' + 25y = (3x+2)e^{-5x}$

I start by rewriting as $(D^2 + 10D + 25)y = (3x + 2)e^{-5x}$

We have a non-distinct real root of multiplicity 2 which is -5.

The homogeneous solution is then $y_{h} = (C_{1} + C_{2})e^{-5x}$

We now find the right candidate for the undermined coefficients method.

$F(x) = (3x+2)e^{-5x}$

$(3x+2)$ is in the form of a first-degree polynomial and $e^{-5x}$ is an exponential function.

So our particular solution is $y_{p} = (Ax + B)e^{-5x}$

My problem is that when I replace $y_{p}$ into the differential equation of line 1, I get 0, so I can't continue the problem...

Thank you.

答:
Misprint:

The characteristic equation is given by $\left(\lambda^{2}+5\right)$, whose solution is $\lambda=-5$ with multiplicity two.

Sorry it's my post (Eduardo), but I didn't read that this question was in another community.


问:
Computing conditional expected value of a dice
Let $X$ be a random variable denoting the number of dots that show up when you roll a fair, six-sided dice. Compute $E(X \mid (X - 3)^2$). Hint: It might help to present it as a table for different values of $(X - 3)^2$.

I thought about doing casework on the value of $(X - 3)^2$ since it equals $0$ each with probability $1/6$ it equals $1$ with probability $1/3$, it equals $4$ with probability $1/3$, and it equals $9$ with probability $1/6$.

If $(X - 3)^2 = 0$, then $E(X) = 3$

If $(X - 3)^2 = 1$, then $E(X) = \frac{2 + 4}{2} = 3$

If $(X - 3)^2 = 4$, then $E(X) = \frac{1 + 5}{2} = 3$

If $(X - 3)^2 = 9$, then $E(X) = 9$.

Is this right? I'm confused because this isn't really a number. I didn't account for the probability of $\{(X - 3)^2 = k\}$ occurring at all.

答: It is right, except for the $E(X|(X-3)^2=9)\ne 9$. It should be $E(X|(X-3)^2=9)=6$.


答:
You are almost there...

If $(X-3)^2=0$ then $E[X|(X-3)^2=0]=3$, not $E(X)$. What you wrote is wrong because $E(X)$ is a marginal expectation

and so on...

The latest is wrong, it's 6 not 9

in fact if $(X-3)^2=9$ then $X=6$ with probability 1

Title: Computing conditional expected value of a dice

Body:
Let $X$ be a random variable denoting the number of dots that show up when you roll a fair, six-sided dice. Compute $E(X \mid (X - 3)^2$). Hint: It might help to present it as a table for different values of $(X - 3)^2$.

I thought about doing casework on the value of $(X - 3)^2$ since it equals $0$ each with probability $1/6$ it equals $1$ with probability $1/3$, it equals $4$ with probability $1/3$, and it equals $9$ with probability $1/6$.

If $(X - 3)^2 = 0$, then $E(X) = 3$

If $(X - 3)^2 = 1$, then $E(X) = \frac{2 + 4}{2} = 3$

If $(X - 3)^2 = 4$, then $E(X) = \frac{1 + 5}{2} = 3$

If $(X - 3)^2 = 9$, then $E(X) = 9$.

Is this right? I'm confused because this isn't really a number. I didn't account for the probability of $\{(X - 3)^2 = k\}$ occurring at all.


问:
Recurrence Relation Confusion
Consider the recurrence relation $$a_{n + 1} = \frac{n - 2}{n^2 - 3n - 4} a_n$$  It seems that all $a$'s with index greater than or equal to $3$ would end up as $0$, regardless of the numeric value of $a_0$, because $a_3 = 0$, and each $a$ indexed after that is ultimately a product involving $a_3$.  However, as has been pointed out in the comments, this would be overridden for $a_4$ and higher indices because the denominator of the recurrence relation is undefined at $n = 4$.

However, the solution given by Wolfram (after lumping constants and assuming $n$ an integer) is $$a_n = A\frac{(n - 4)(n - 3)}{n!}$$  This is defined and nonzero for $n \geq 5$ so long as the constant $A$ is nonzero.  What is going on here?

I noticed that by making the substitution $n = m + 10$, yielding $$a_{m + 11} = \frac{m + 8}{m^2 + 17m + 66} a_{m + 10},$$ Wolfram's answer remains the same, but the zeros and undefined points of the recurrence relation are numerically different.  Is Wolfram doing this or something else to infer what the recurrence relation solution should be for larger values of $n$?

答:
We do not know what you did provide to Wolfram Alpha. If you write (have a looh here)

RSolve[{a[n + 1] == (n - 2)/(n^2 - 3 n - 4) a[n], a[0] == A}, a[n], n]

the given result is (using Pochhammer symbols)

$$a_n=\frac{ (n-4) (n-3)}{1440\,\, (6)_{n-5}} A=\frac{ (n-4) (n-3)}{12 \, n!}A$$ and the sequence is
$$\left\{A,\frac{A}{2},\frac{A}{12},0,0,\frac{A}{720},\frac{A}{1440},\frac{A}{5040},
   \frac{A}{24192},\frac{A}{145152},\frac{A}{1036800},\cdots\right\}$$

Title: Recurrence Relation Confusion


问:
Factoring $a^4+b^4+(a-b)^4$
I'm trying to factor $$a^4+b^4+(a-b)^4$$ so the result would be $2(a^2-ab+b^2)^2$ but I can't get that.

I rewrite it as:

$$a^4+b^4+(a-b)^4=(a^2+b^2)^2-2a^2b^2+(a-b)^4=(a^2-\sqrt2 ab+b^2)(a^2+\sqrt2 ab+b^2)+(a-b)^4$$
But I can't use difference of squares anymore because $(a-b)^4$ is not negative.

Title: Factoring $a^4+b^4+(a-b)^4$

Body:
It's a variation of Candido's identities, its proof without words are as follows:

Now

\begin{align}
  [x^2+y^2+(x+y)^2]^2 &= 2[x^4+y^4+(x+y)^4] \\
  [a^2+(-b)^2+(a-b)^2]^2 &= 2[a^4+(-b)^4+(a-b)^4] \\
  [2a^2-2ab+2b^2]^2 &= 2[a^4+b^4+(a-b)^4] \\
  4(a^2-2ab+b^2)^2 &= 2[a^4+b^4+(a-b)^4] \\
  a^4+b^4+(a-b)^4 &= 2(a^2-ab+b^2)^2
\end{align}

Refer to link 1 and link 2 for your further interests.


问:
Equivalence of metric
Definition 1.  Two metrics $d_{1}$, $d_{2}$ on a set $X$ are equivalent, if there exist two positive numbers $c$, $c'\in\mathbb{R}_{> 0}$, such that if $x$, $y\in X$ then$$
c d_{2}(x, y)\leq d_{1}(x, y)\leq c'd_{2}(x, y).
$$

Question. What does the equivalence of metrics mean? I think there are some types of equivalence:

$d_{1} = d_{2}$ i.e. $d_{1}(x, y) = d_{2}(x, y)$ for all $x$, $y\in X$.

The induced topological structure with respect to $d_{1}$ is equal to that of $d_{2}$.

The induced uniform structure with respect to $d_{1}$ is equal to that of $d_{2}$.

The induced "structure of metric space" with respect to $d_{1}$ is equal to that of $d_{2}$, although I don't know what it means.

I tried to show 2, but it seems to be false. As for 3, I don't know enough about uniform spaces. If 4 is true, does the definition 1 defines the equivalence of "structure of metric space", or is there a clearer definition ?

答:
2 is certainly true, as I explain in this answer, but more can be said: having this "equivalence via constants", it's quite clear that the uniformities induced by $d_1$ and $d_2$ are the same: for every entourage of the form $\{(x,y) \mid d_1(x,y) < r\}$ we can find another entourage of the form $\{(x,y)\mid d_2(x,y) < r'\}$ that is a subset of it and vice versa, and as these kinds form a (filter) base for the respective uniformities from $d_1$ resp. $d_2$, we get that the uniformities are the same.

So $d_1$ and $d_2$ induce the same uniform properties: uniform continuity, total boundedness and Cauchy-ness and completeness means exactly the same under either.

3 alwyas implies 2 as well, but 3 is stronger and so probably the intended answer, as 4 is false:

If a map is an isometry wrt $d_1$ it need not be an isometry w.r.t. $d_2$ or vice versa. So really metric stucture (where the actual values of the metric matter!) can be different. We can alraedy see this by the "shapes" of the balls under the different equivalent metrics on $\Bbb R^2$. Ths shows that point 4 is false.

So the most informative is 3: the uniform structure is the same (and hence the topologies too).

Title: Equivalence of metric


问:
Prove that $P(A\cup B) = P (B) + P (A)\cdot P (B^c)$
Suppose A and B are independent events. Show that $P(A\cup B) = P (B) + P (A)\cdot P (B^c)$

Could someone tell me how i can prove this?

Remembering that two events are independent if $P (A \cap B) = P (A)\cdot  P (B)$

答:
$$\begin{split}P(A\cup B)&=P(A)+P(B)-P(A\cap B)\\
&=P(A)+P(B)-P(A)P(B)\text{ independence}\\
&=P(B)+P(A)(1-P(B)) \text{ factorization}\\
&=P(B)+P(A)P(B^c) \text{ definition of complement}\end{split}$$


问:
True/false : If $K$ be compact neighbourhood of $x$ and $U \subset K$ be a open neighbourhood of $x$ then $ K \setminus U$ is compact
Given $X $ is locally compact Hausdorff space

Is the following statement is true/false

If $K$ be compact neighbourhood of  $x$ and $U \subset K$ be a open neighbourhood of $x$ then $ K \setminus U$ is compact ?

My attempt : I think this statement is false

take $X=[0,1]$, $K= [1/3,1/2]$ and  $U= (1/3,1/2)$

$K \setminus U= [1/3,/1/2]-(1/3,1/2)=\{1/3,1/2\}$

$\{1/3,1/2\} $is not compact because $f :[0,1] \to \{1/3,1/2\}$   will not be continious

答:
$K \setminus U= K \cap (X\setminus U)$ is a closed subset of $K$ so compact.
And a finite set is certainly compact; nothing to do with continuity, just definitions.


问:
If a subset of the square grid can be tiled by $1\times n$ rectangles for every $n$, can it be tiled by infinite rays?
Suppose that we have some set $S$ of grid-aligned squares in the plane; equivalently, we can think of our set as $S\subset \mathbb{Z}^2$. Suppose that for every positive integer $n$, $S$ can be tiled by disjoint copies of a $1\times n$ rectangle (possibly rotated). Is it necessarily the case that $S$ can be tiled by congruent copies of an infinite "ray" of squares, i.e., a region congruent to the set $\{(k,0)\ |\ k\ge0\}$?

For example, the region given by removing from $\mathbb{Z}^2$ the points $(0,0), (1,1)$, and $(-1,2)$ can be tiled by every $1\times n$ rectangle, and also by infinite rays, as shown:

I'd like to say that this follows from some kind of compactness argument, but my attempts to apply such reasoning ran into trouble (I can't rule out that the $1\times n$ tilings have their endpoints move away from the origin without a well-defined limiting behavior). Any thoughts on how to prove this (or a counterexample, if my intuition is badly wrong)?

I believe the same statement should hold in $\mathbb{Z}^n$ with $1\times1\times\ldots\times1\times n$ tiles and the same kind of rays; if the original statement proves easy to show, I would be interested in this generalization.

As a remark, the converse is obviously true, since the infinite ray can be tiled by any $1\times n$ rectangle.

答:
Apologies in advance for shooting sparrows with cannons, but I can't give away such a funny opportunity to see "pure" math being applied to "simple" geometry problems.

I'll be using Gödel's compactness theorem, which is well-explained here, so check it out. In essence, the theorem is as follows: In first-order logic, if a property is witnessed by arbitrarily large finite numbers, then there is a way to witness it by infinities. It certainly feels relevant in the post's context.

Consider the theory of the abelian group $\mathbb{Z}^2$ with generators $a,b$. Thus the following sentences hold:

$ab=ba$

$\forall y,\,ey=ye=y$

$a\ne b\ne e$

$a^{-1}\ne a^{1729}$

etc.

We collect all valid sentences into the set $\operatorname{Th}(\mathbb{Z}^2,a,b)$.

We introduce the relation $R(x,y)$, which says that "the group elements $x,y$ of $\mathbb{Z}^2$ belong to the same tile". Then the following things can be expressed as formulas/sentences:

$R$ is an equivalence relation (left as exercise)

Every tile is a single strip of rectangle, or equivalently, any two elements not on a same row/column cannot be in the same tile. We express this with infinitely many sentences. For example, $\forall x,\lnot R(x,a^{-3} b^4 x)$ would be one of them. We also need the rectangles to be connected, and they can also be expressed by infinitely many sentences, for example, $\forall x,R(x,a^6 x)\rightarrow \bigwedge_{i=1}^5 R(x,a^i x)$.

For a particular group element $x$, we can ask for it not to be tiled. We use the sentence $\phi(x)$, which says that $\forall y\ne x,\lnot R(x,y)$.

For a particular group element $x$, we can also ask for it to be tiled, and we get to choose how big the tile is! For an integer $n>0$, we use the sentence $\psi_n(x)$ to mean that $x$ is in a tile of size $\ge n$, which is expressed as:
$$\left(\bigvee_{i=1-n}^0\bigwedge_{j=i}^{i+n-1} R(x,a^j x)\right)\lor\left(\bigvee_{i=1-n}^0\bigwedge_{j=i}^{i+n-1} R(x,b^j x)\right)$$

Now consider making a theory $T$ in the language of groups with the marked elements $a,b$ and the relation $R$. The consistency of this theory will claim that $S\subseteq \mathbb{Z}^2$ can be tiled by infinite-sized tiles. It will contain all appropirate axioms listed above, but we take special care when putting in any $\phi(x)$ or $\psi_n(x)$ into $T$. In particular, if $x\in S$ is an element of $\mathbb{Z}^2$ expressed as a word in $a,b$, then we put $\psi_n(x)\in T$ for all $n$ (but of course $\phi(x)\notin T$). Otherwise if $x\notin S$, then we only put $\phi(x)\in T$.

By the compactness theorem, the consistency of $T$ will follow from the consistency of every finite segment $T_0\subseteq T$. However as $T_0$ is finite, there is some bound $N\in\mathbb{N}$ such that every $\psi_n(x)\in T_0$ will be of the form $n\le N$. In particular, $T_0$ is now a segment of a theory whose consistency claims that $S$ can be tiled by rectangular tiles, and every tile has size at least $N$. By our assumption, we can simply tile $S$ by rectangular tiles of size $N$, and see that $T_0$ is indeed consistent. By the compactness theorem, $T$ is consistent.

Thus we know that $S$ can indeed be tiled by rectangles of infinite size. Any such rectangle is either a ray or can be tiled by two rays. Hence $S$ can also be tiled by rays.


问:
Prove that $x^{2} \equiv -1 \pmod p$ has no solutions if prime $p \equiv 3 \pmod 4$.
Assume: $p$ is a prime that satisfies $p \equiv 3 \pmod 4$

Show: $x^{2} \equiv -1 \pmod p$ has no solutions $\forall x \in \mathbb{Z}$.

I know this problem has something to do with Fermat's Little Theorem, that $a^{p-1} \equiv 1\pmod p$. I tried to do a proof by contradiction, assuming the conclusion and showing some contradiction but just ran into a wall. Any help would be greatly appreciated.

答: It is equivalent to $\nexists n: \frac{n^2+1}{p}\in \mathbb{N}$, which is a case of the sum of squares theorem.


问:
My paradox in determining the probability of distributing similar balls into different boxes
Inspired by the discussion between @Matthew Pilling and me in this post What is the probability that there more rabbits than chickens in each of these three cages?, I am trying to find and understand the answer to a simple problem I am thinking for a while. (Feel free to read the comments under my answer in that post if you are interested and want to know more.)

Let's say I have two identical balls and I want to distribute them into two different boxes (box1 and box2). What is the probability of having one ball in each box?

Approach 1:
We can talk about the total number of solutions for $n_{1} + n_{2} = 2$, where $n_{1},n_{2} \in W $. And, then, the number of solutions to $n_{1} + n_{2} = 2$, where $n_{1},n_{2} \ge 1$. The probability will be calculated as: $\frac{1}{3}$.

Approach 2:
Or, we can think of this problem as follows: we have two boxes. The probability of choosing the first box is $\frac{1}{2}$. So, there is $\frac{1}{4}$ probability of choosing the first box for both balls. Same reasoning is true for choosing the second box for both balls. So, in total, there is $\frac{1}{2}$ probability to have both balls in only one box. And, thus, the probability of having only one ball in each box is $\frac{1}{2}$.

Could someone please help me understand what's happening here? Are we dealing with two different problems (in these two approaches) or one of the approaches is wrong because ... ?

Thanks for any input you may provide.

Nima

答: The probability that you get balls in unique boxes is $1\cdot \frac 1 2=\frac 1 2$. It doesn't matter what box the first ball goes into, and then the second ball must go into the box the first ball did not go into.


问:
Show that the set $\{x \in \mathbb{R}| f(x)=x\}$ is closed.
Show that the set $\{x \in \mathbb{R}| f(x)=x\}$ is closed, assuming that $f:\mathbb{R} \to \mathbb{R}$ is continuous. So here is my attempt:

Let $\{x_k\} \subseteq \mathbb{R}$ such that $x_k \in \{x \in \mathbb{R}| f(x)=x\}$ and $x_k \to x$ as $k \to \infty$.Then, as $f:\mathbb{R} \to \mathbb{R}$ is continuous by a theorem in my text, the sequence $\{y_k\}$ where $\{y_k\} = f(x_k)$ converges to $f(x)$. So we have that $f(x) = \lim_{x \to \infty} f(x_k) =x $.  So by definition $x$ is in the set, and we can conclude now that since the set contains all of its limit points, that the set is a closed subset of $\mathbb{R}$.

I have a feeling I made some flaws in logic somewhere around here, any help is appreciated. I only want some hints on what I did wrong so I can think about the problem longer instead of just having the solution.

答:
Your proof is fine, but I would write it differently, plus I'll add missing details:

Let $C=\{x\in \Bbb R\mid f(x) =x\}$. To show $C$ is closed we take a sequence $(x_n)_n \to x$ with $x_n \in C$ for all $n$ and we show $x$ is in $C$ as well (so $C$ is closed under sequential limits, which is enough to show closedness in all metric spaces).

As $f$ is continuous, $f(x_n) \to f(x)$ as well. But, as $x_n \in C$, $f(x_n) = x_n$ and so $f(x)=x$ because limits of sequences are unique (here we are using a property of metric spaces again (Hausdorffness): we have the same sequence $(x_n)$ both converging to $x$ and $f(x)$; in a non-metric space we could have $x \neq f(x)$ ) and as $f(x)=x$, $x \in C$ by definition and we're done.

FWIW (and for your later study): this proof would straightforwardly generalise to all Hausdorff topological spaces (a much wider class than just $\Bbb R$), using nets instead of sequences.


问:
Riemann-Stieltjes integral of a continuous function w.r.t. a step function
$
\newcommand{\para}[1]{\left( #1 \right)}
\newcommand{\abs}[1]{\left| \, #1 \, \right|}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\ds}{\displaystyle}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\norm}[1]{\left \Vert #1 \right \Vert}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\vp}{\varphi}
\newcommand{\ve}{\varepsilon}
\newcommand{\a}{\alpha}
\newcommand{\g}{\Gamma}
\newcommand{\og}{\overline{\Gamma}}
$Definition: I define the notion of an "increasing partition" $P$ of $[a,b]$ to be a finite partition $\set{z_k}_{k=0}^n \subseteq [a,b]$ where $a = z_0 < z_1 < \cdots < z_n = b$. It's a minor thing that helps reduce the repetition in the wording herein.

Problem Statement:  Let $\vp : [a,b] \to \Bbb R$ be a step function with $\set{\a_i}_{i=1}^m$ its discontinuities forming an increasing partition of $[a,b]$. Thus $\vp$ is constant on each interval $(\alpha_{i-1},\alpha_i)$. Let $f : [a,b] \to \Bbb R$ be continuous, and let
\begin{alignat*}{99}
\vp(\alpha_i^+) &:=\;&& \lim_{x \to  \a_i^+} \vp(x) &&\quad \forall i \in \set{0,1,\cdots,m-1} \\
\vp(\a_i^-) &:=\;&& \lim_{x \to \alpha_i^-} \vp(x) &&\quad \forall i \in \set{0,1,\cdots,m-1} \\
d_i &:=\;&& \vp(\a_i^+) - \vp(\a_i^-) &&\quad \forall i \in \set{1,2,\cdots,m-1} \\
d_0 &:=\;&& \vp(\a_0^+) - \vp(\a_0) \\
d_m &:=\;&& \vp(\a_m) - \vp(\a_m^-)
\end{alignat*}
Show
$$
\int_a^b f \, \dd \vp = \sum_{i=1}^m f(\a_i) d_i
$$

Context: This ultimately comes up as a homework assignment for me, and I just wanted to check my solution, because I have some doubts on it. I'll express those at the end, though, since they tie more intimately into my proof.

Attempted Proof:

Let the following be:

Let $\g_\a := \set{\a_i}_{i=0}^m$ be an increasing partition of $[a,b]$. (This is the set of discontinuities of $\vp$, i.e. $\vp$ is constant on $(\a_{i-1},\a_i)$ $\forall i$.)

Let the definitions in the problem statement hold.

Let $\ve > 0$.

Let $\ds \delta := \min_{1 \le i \le m} \left| \a_i - \a_{i-1} \right|$.

Let $\g := \set{x_i}_{i=0}^n$ be an increasing partition of $[a,b]$ with $\norm \g < \delta$.

Let $\ol \g := \g \cup \g_\a$, and let it be written $\ol \g := \set{\ol{x_i}}_{i=0}^p$.

Let $\ol{\xi_i} \in [\ol{x_i}, \ol{x_{i-1}}]$ for each $i$. (These are our representative valuation points in the definition of the integral.)

By the merit of $\ol \g$ being a refinement of $\g$, and by merit of definition, we have
$$
\norm{ \ol \g } \le \norm \g < \delta \le \norm{\g_\a}
$$
By the Cauchy criterion for Riemann-Stieljes integration, then, where $R_\g$ denotes the usual Riemann-Stieltjes sum,
$$
\abs{ R_\g - R_{\ol \g} } < \ve
$$
and so $R_{\ol \g}$ equals the desired integral. It remains to see if this equals the desired sum.

Write
$$
R_{\og} = \sum_{j=1}^p f \para{ \ol{ \xi_j } } \Big( \vp \para{ \ol{x_j} } - \vp \para{ \ol{x_{j-1}} } \Big)
$$
From definition, we may have either $\ol{x_i} = \alpha_j \in \g_\a$ for some $i$ (i.e. is a discontinuity of $\vp$), or $\ol{x_i} = x_i \in \g$ for some $i$. Note that for any consecutive pairing $\ol{x_{i-1}},\ol{x_{i}}$ that not both are in $\g_\a$. (This is because each are in $\og$ for which $\norm \og < \delta$, and thus the smallest distance between any two discontinuities of $\vp$ is bigger than the distance between any two members of $\og$.) Thus, we only have three cases to consider, based on how many of a given pair are, indeed, discontinuities:

Case 1 (neither): In the case neither $\ol{x_{i-1}},\ol{x_{i}}$ are discontinuities of $\vp$, then trivially we will have
$$
\vp \para{ \ol{x_j} } - \vp \para{ \ol{x_{j-1}} } = 0
$$

Case 2 (right discontinuity): Suppose $\ol{x_{i}} = \a_j$ for some $j$ (but, again, that means $\ol{x_{i-1}}$ is not a discontinuity). Then in such a case, using the piecewise continuity and constantness of $\vp$,
$$
\vp \para{ \ol{x_j} } - \vp \para{ \ol{x_{j-1}} } = \vp\para{ \a_j } - \vp \para{ \a_{j}^- }
$$

Case 3 (left discontinuity): Similar to the previous, we have $\ol{x_{i-1}} = \a_j$ for some $j$, and take
$$
\vp \para{ \ol{x_j} } - \vp \para{ \ol{x_{j-1}} } = \vp\para{ \a_j^+ } - \vp \para{ \a_{j} }
$$

Let $\ol{\xi_{i,j}}$ be the $\ol{\xi_i}$ whose terms are not eliminated by merit of Case 1 but lie in Case 2; similarly, $\ol{\xi'_{i,j}}$ for those whose terms lie in Case 3. Then we have, once the terms have negated,
$$
R_{\og} = \underbrace{\sum_{j=1}^m f \para{ \ol{\xi_{i,j}} } \Big( \vp\para{ \a_j } - \vp \para{ \a_{j}^- } \Big) }_{\text{case 2 terms}} + \underbrace{\sum_{j=1}^m f \para{ \ol{\xi'_{i,j}} } \Big( \vp\para{ \a_j^+ } - \vp \para{ \a_{j} } \Big)}_{\text{case 3 terms}}
$$
Note that, for the Case 2 terms, each $\ol{\xi_{i,j}}$ is within $\delta$ of $\a_j$; specifically, $\ol{\xi_{i,j}} \in (\a_j - \delta,\a_j)$. Likewise, $\ol{\xi'_{i,j}} \in (\a_j,\a_j+\delta)$. Since $f$ is continuous, we may take each to be its corresponding $\a_j$, and write
$$
R_{\og} = \sum_{j=1}^m f \para{ \a_j } \Big( \vp\para{ \a_j } - \vp \para{ \a_{j}^- } \Big)  +\sum_{j=1}^m f \para{ \a_j } \Big( \vp\para{ \a_j^+ } - \vp \para{ \a_{j} } \Big)
$$
Some clear cancellation results, giving
$$
R_{\og} = \sum_{j=1}^m f \para{ \a_j } \Big( \vp\para{ \a_j^+ } - \vp \para{ \a_{j}^- } \Big)
$$
Taking the usual convention for what occurs for the $\a_j$ which are $a,b$, we see the $\vp$ difference here is indeed $d_i$. So we see
$$
\int_a^b f \, \dd \vp  = R_{\og} = \sum_{j=1}^m f \para{ \a_j } d_i
$$
as desired.

My Questions / Issues:

While I feel like this proof is on the right path, I have some anxieties about some steps. Some in particular I feel aren't quite "there" yet. I was wondering if anyone would be able to help me flesh these ideas uot.

The use of the Cauchy criterion for Riemann-Stieltjes integrals (stated here) feels far too un-fleshed out, and I doubt it actually lets me conclude that $R_\og$ equals the integral in question. I imagine it only lets me conclude the integral exists at best. I do see that this proof seems to use something similar towards the end of their proof, but their motivation, choice of bound, and the weird $\delta$'s all escape me.

The second chunk of the proof (verifying the alternate form for $R_\og$) feels right, but is there any issue there? Something feels sketchy about how I waited until then to define the $\ol{\xi_{i,j}}$ terms to be $\a_j$ in particular, but I don't see how it would be problematic either...

I feel like I barely used the continuity of $f$ at all, which means its purpose is lurking elsewhere. Where would that be?

Thanks for the insights and thoughts you guys can give!

答:
To suggest a better strategy and see the continuity of $f$ enter as it should,  consider the proof for a single step discontinuity at $\alpha_1 \in (a,b)$.  The objective is to show that for any $\epsilon > 0$ there exists $\delta > 0$ such that for any partition with $\|\Gamma\| < \delta$, we have $|R_\Gamma - f(\alpha_1)d_1| < \epsilon$.

Since $f$ is uniformly continuous on the compact interval $[a,b]$ , there exists $\delta > 0$ such that $|f(x) - f(y)| < \epsilon/|d_1|$ when $|x-y| < \delta$.

Take any partition $\Gamma$ with points $a = x_0 < x_1 < \ldots < x_n = b$ such that $\|\Gamma\|< \delta$. There exists and index j such that one of the following three cases must hold:

$$\tag{1} \quad x_{j-1} < \alpha_1 < x_j, \quad \varphi(x_{j-1}) = \varphi(\alpha_1^-),\quad\varphi(x_{j}) = \varphi(\alpha_1^+)$$
$$\tag{2} \quad x_{j-1} < \alpha_1 = x_j,\quad  \varphi(x_{j-1}) = \varphi(\alpha_1^-),\quad\varphi(x_{j}) = \varphi(\alpha_1^+)$$
$$\tag{2} \quad x_{j-1} = \alpha_1 < x_j,\quad  \varphi(x_{j-1}) = \varphi(\alpha_1^-),\quad\varphi(x_{j}) = \varphi(\alpha_1^+)$$

In each case we have for some intermediate point $\xi \in [x_{j-1},x_j]$,

$$\left| R_\Gamma - f(\alpha_1)d_1\right|= \left|f(\xi)(\varphi(x_j) - \varphi(x_{j-1})-f(\alpha_1) (\varphi(\alpha_1^+) - \varphi(\alpha_1^-) \right|\\ = |f(\xi) - f(\alpha_1)|\, |d_1|<  \epsilon$$

To generalize the proof for multiple points where the step function is discontinuous, choose $\delta$ such that in addition to controlling the oscillation of $f$ on subintervals of partitions with $\|\Gamma\| < \delta$, we also have no more than one point of discontinuity $\alpha_k$ in a subinterval.


问:
Axiom of Regularity - Transitive set
I just managed to confuse myself completely while studying for Set Theory.
We have the Axiom of regularity:
$$\forall S (S\not= \emptyset \rightarrow (\exists x\in S)(S\cap x=\emptyset))$$
Now a set is transitive, if $x\in T$ implies $x\subset T$.

I don't understand anymore how that can be while we have the Axiom of regularity.

Doesn't follow from the definition of a transitive set, that $x\cap T = x$?

Maybe if someone knows a good source for a more detailed explanation of the Axiom of regularity or has the time to explain me some more?

Thanks and best,

Luca

答:
You are right. If $x$ is transitive, and $y\in x$ then $x\cap y=y$.

Hint: Show that a transitive set $x$ must either be empty, or $\varnothing\in x$.

Title: Axiom of Regularity - Transitive set


问:
Prove the diagonals of a parallelogram bisect each other
When the two diagonals intersect together, aren't they cutting it into 4 equal parts and not 2?

And to prove that they bisect, you would just need to find the midpoint and see if half of the length of the diagonal is exactly half so that it reaches the midpoint. It just doesn't seem like a proof at all to me?

Can someone help me with these 2 seperate but related questions?

答:
Suppose you have a parallelogram with diagonals drawn through E (not shown due to difficulty).

```<code>    /A----------------B
   /                 /
  /        E        /
 /                 /
D----------------C/
</code>```

Since AB and DC are parallel, angle ACD and CAB are congruent.

Likewise, angles BDC, ABD are congruent

We know that CD and AB are congruent because they are the sides of a  parallelogram

By ASA (Angle-Side-Angle), triangles ABE and CDE are congruent. Therefore

EC is congruent to AE

BE is congruent to DE

They are not equal unless the parallelogram is a rectangle.

Body:
Suppose you have a parallelogram with diagonals drawn through E (not shown due to difficulty).

    /A----------------B
   /                 /
  /        E        /
 /                 /
D----------------C/

Since AB and DC are parallel, angle ACD and CAB are congruent.

Likewise, angles BDC, ABD are congruent

We know that CD and AB are congruent because they are the sides of a  parallelogram

By ASA (Angle-Side-Angle), triangles ABE and CDE are congruent. Therefore

EC is congruent to AE

BE is congruent to DE

They are not equal unless the parallelogram is a rectangle.


问:
Question concerning isomorphism between two direct product of a pair of sets.
For the following question: prove that any two direct products of a pair of sets A and B are isomorphic.

The text where this question comes from gives the following definitions and a theorem which the proof was presented in text:

Given two sets $A$ and $B$, we defined their cartesian products as: $A \times B := \{(x,y)\mid x\in A,  y\in B\}$.

We define their corresponding projection maps on sets $A$ and $B$ respectively as:

$\pi_{A}: A\times B \rightarrow A; (x,y) \mapsto x)$ and

$\pi_{B}: A\times B \rightarrow B; (x,y) \mapsto y)$

Theorem 1: Given any set $C$ and maps $f:C\rightarrow A$, $g:C\rightarrow B$, there is a unique map $(f,g):C\rightarrow A\times B$ such that $(f,g)\pi_{A}=f$ and $(f,g)\pi_{B}=g$.

Definition:  The $\textit{direct product } A\times B$ of two sets $A$ and $B$ is a set $A\times B$ equipped with maps $\pi_{A}: A\times B \rightarrow A$ and $\pi_{B}: A\times B \rightarrow B$ such that, for any set $C$ and set maps $f:C\rightarrow A$, $g:C\rightarrow B$, there is a unique map $(f,g):C\rightarrow A\times B$ such that $(f,g)\pi_{A}=f$ and $(f,g)\pi_{B}=g$.

I am confused about the wording of the question.  Is the question asking the following:  Given sets $M$, $N$, $E$ and $F$, the two cartesian products respectively $A= M \times N$ and $B=E\times F$ We define the function $h:A\rightarrow B$
 and the corresponding pairs of projection maps associated with the product set $A$ and the product set $B$ such that
$\pi_{M}: M\times N \rightarrow M$

$\pi_{N}: M\times N \rightarrow N$

$\pi_{E}: E\times F \rightarrow E$

$\pi_{F}: E\times F \rightarrow F$

Show that the function $h$ is an isomorphism.  I am suppose to define direct product for sets $A$ and sets $B$.  I am not sure how to do that properly and also, do I have to show that the function $h$ is homomorphic,  The text at this point have not defined what homomorphism suppose to look like for direct product of sets.  I am really having trouble rephrasing the question in more formal math notations.  Thank you in advance

答:
I believe what you're being asked to prove is: if you have two sets $S$ and $T$, and both of them satisfy the definition of the direct product (the maps $\pi_A$ and $\pi_B$ and the uniqueness property with respect to other sets $C$), then $S$ and $T$ are isomorphic.

(Hint for the proof: use the uniqueness property for $S$ with $C$ chosen to be $T$, and vice versa.)

This fact, that any two such sets are isomorphic, is what justifies naming such a set "the" direct product of $A$ and $B$ and giving it a name $A\times B$.


问:
Using fermat's little theorem and the chinese remainder theorem?
So I'm trying to find $9^{65}\pmod{77}$.

What I've done:

I used Fermat's little theorem and separated them into $9^{65}\pmod7$ and $9^{65}\pmod{11}$, where I got the remainders to be $4$ and $1$ respectively. Now, I have no idea what to do from this point. What is the best way to combine both to get the final answer?

答:
I will show how to solve the congruences in a simple way that can be used for smaller values:

$$\begin{align}
x&\equiv4 \pmod 7\tag1
\\x &\equiv 1 \pmod {11}\tag2
\end{align}$$

Now note that from $(1)$, $x=7a+4$, where $a$ is an integer. From $(2)$, $x \equiv 1 \pmod {11}$ so:

$$\begin{align}
7a+4 &\equiv 1 \pmod {11}
\\7a &\equiv 8 \pmod {11}
\\7a &\equiv 63 \pmod {11}
\\a &\equiv 9 \pmod {11}
\end{align}$$
So from the above congruence, $a=11b+9$ where $b$ is an integer. Therefore, $$x=7a+4 = 7(11b+9)+4 = 77b+67$$
Because $b$ is an integer, we can deduce that $$x \equiv 67 \pmod {77}$$


问:
Subrings and homomorphisms of unitary rings
Let $(R,+,\cdot)$ be a ring (in the definition i use multiplication is associative operation and it's not assumed that there is unity in the ring).

I've seen two definitons of subring.

1) non-empty subset $S \subset R$ is called subring of ring $(R,+,\cdot)$ iff $(S,+,\cdot)$ is a ring

2) Let $(R,+,\cdot)$ be unitary ring with unity $e$. Non-empty subset $S \subset R$ is called subring of ring $(R,+,\cdot)$ iff $(S,+,\cdot)$ is a ring and $e \in S$.

I'm looking for an example of such ring R and its subset S that $(S,+,\cdot)$ is a ring but not a unitary ring.

I will be also very grateful for an example of such UNITARY rings $(R_1,+_1,\cdot_1)$, $(R_2,+_2,\cdot_2)$ and function $f: ~~ R_1 \longrightarrow R_2$ that

$(1) \forall a,b \in R_1 ~~~ f(a+_1 b) = f(a) +_2 f(b) $

$(2) \forall a,b \in R_1 ~~~ f(a\cdot_1 b) = f(a) \cdot_2 f(b)$

$(3) f(e_1) \neq e_2$

where $e_1$ is unity in $R_1$; $e_2$ - in $R_2$.

Thanks in advance.

答:
Take any $S$ without $1$ and join to it extra unity $1$, so you get the unitary ring $R=\{a+n\cdot 1|a\in S, n\in \mathbb{Z}\}$ in which $S$ is a subring.

Let $M$ be a monoid with an idempotent $e\ne 1$, $\mathbb{Z}M $ its semigroup ring. Then the embedding $\mathbb{Z}e\to \mathbb{Z}M$ is what you want.


答:
Use the extended Euclidean algorithm:
$11x+7y=1$ , find $x = 2 , y = -3 $.

$9^{65}$ = $ (7y)+4(11x$) mod 77

Basic plug in
$9^{65}$ = 67

Title: Using fermat's little theorem and the chinese remainder theorem?

Body:
So I'm trying to find $9^{65}\pmod{77}$.

What I've done:

I used Fermat's little theorem and separated them into $9^{65}\pmod7$ and $9^{65}\pmod{11}$, where I got the remainders to be $4$ and $1$ respectively. Now, I have no idea what to do from this point. What is the best way to combine both to get the final answer?


问:
Why are roots of unity always distinct?
As per the fundamental theorem of algebra, $z^n-1=0$ has $n$ complex solutions--but, for a general polynomial, some of those solutions may have multiplicity $> 1$, they may repeat.

Why is it that roots of unity never repeat, i.e., that the equation above always has $n$ distinct solutions for all $n$?

答:
If a polynomial $\phi$ has a repeated root $a$ then $\phi(X)=(X-a)^k \psi(X)$ for some polynomial $\psi$ and some $k\geqslant 2$. Then $\phi'(X)=k(X-a)^{k-1}\psi(X)+(X-a)^k \psi'(X)$, and $\phi'(a)=0$.

Apply this to $X^n-1$ whose derivative is $nX^{n-1}$: the only candidate for a repeated root is $0$ and it is not even a root.

[Note that over some fields, where the characteristic divides $n$, we do get repeated roots.]

Title: Why are roots of unity always distinct?


答:
Because we know explicitely its $n$ roots $$\exp(2i\pi k/n), \ \ \ k=0,1,...(n−1)$$

and that all of them are distinct.

Title: Why are roots of unity always distinct?

Body:
As per the fundamental theorem of algebra, $z^n-1=0$ has $n$ complex solutions--but, for a general polynomial, some of those solutions may have multiplicity $> 1$, they may repeat.

Why is it that roots of unity never repeat, i.e., that the equation above always has $n$ distinct solutions for all $n$?


问:
Ascending or descending property of a sequence according to the value of its first element
Suppose that $(a_n)_{n≥1}$ is a sequence of real numbers satisfying $a_{n+1}=\frac{3a_n}{2+a_n}$.

Suppose $0<a_1<1$, then prove that the sequence $a_n$ is increasing and hence show that $lim_{n \rightarrow \infty}a_n=1$.

Suppose $a_1>1$, then prove that the sequence $a_n$ is decreasing and hence show that $lim_{n \rightarrow \infty}a_n=1$.

I attempted to write $a_{n+1}-a_n=\frac{a_n(1-a_n)}{2+a_n}$ but I am kind of stuck here.

答:
Note that$$(\forall n\in\Bbb N):\frac{a_{n+1}}{a_n}=\frac3{2+a_n}.$$Therefore, if $a_n\in(0,1)$, then $\frac{a_{n+1}}{a_n}>1$, which means that $a_{n+1}>a_n$. So, $(a_n)_{n\in\Bbb N}$ is indeed increasing. But$$a_n<1\implies1-a_{n+1}=2\frac{1-a_n}{2+a_n}>0,$$and therefore $a_{n+1}<1$ too. So, $\lim_{n\to\infty}a_n=1$.

Can you deal with the case in which $a_1>1$ now?


问:
Showing that $\sum_{n=1}^{\infty}e^{-n^2 \pi y}$ $\leq $ $2e^{-\pi y}$ When $y \geq 1$.
How can I show that $\sum_{n=1}^{\infty}e^{-n^2 \pi y}$ $\leq $ $2e^{-\pi y}$ When $y \geq 1$. On using the inequality with the integral I get: $\sum_{n=1}^{\infty}e^{-n^2 \pi y} \leq \int_{1}^{\infty}e^{-\pi y t}dt = \frac{e^{-\pi y}}{\pi y}$.  From here I can conclude that the summation have the desired upper bound?

答:
You can try the following approach. For reasons of personal preference I shall replace the $y$ in your notation with $\alpha \geqslant 1$ and introduce $a \colon=\mathrm{e}^{-\pi \alpha}$. It is clear that $0\leqslant \mathrm{e}^{-\pi}<\mathrm{e}^{-1}=\frac{1}{\mathrm{e}}<\frac{1}{2}$.

We can produce the following estimate:
$$\sum_{n=1}^{\infty}a^{n^2}=\sum_{n \in \mathbb{N}^{\times}}a^{n^2}=\sum_{m \in \left(\mathbb{N}^{\times}\right)^{\underline{2}}}a^m \leqslant \sum_{m \in \mathbb{N}^{\times}}a^m=\frac{a}{1-a},$$
which relies on the fact that for a sequence of positive reals, the sum of the associated series coincides with the sum of the sequence understood in the general sense of summation of arbitrary families in topological groups. The notation $\left(\mathbb{N}^{\times}\right)^{\underline{2}}=\left\{m^2\right\}_{m \in \mathbb{N}^{\times}}$ expresses the set of all nonzero square natural numbers and in order to obtain the sum of the right-most (geometric) series we relied on the condition that $a<1$.

This upper bound can be used to reach the desired conclusion as long as we manage to establish that in the given situation $\frac{a}{1-a} \leqslant 2a$, which amounts to claiming that $a \leqslant \frac{1}{2}$ and is indeed a condition we have already ascertained $a$ to satisfy.

This concludes the solution.

P.S. An equally valid manner of justifying the inequality $\displaystyle\sum_{n=1}^{\infty}a^{n^2} \leqslant \displaystyle\sum_{n=1}^{\infty}a^n$ is to simply the remark the term-wise comparison $a^{n^2} \leqslant a^n$ occurring between the two series in question, as $n^2 \geqslant n$ for any $n \in \mathbb{N}$ and $a<1$.

Body:
You can try the following approach. For reasons of personal preference I shall replace the $y$ in your notation with $\alpha \geqslant 1$ and introduce $a \colon=\mathrm{e}^{-\pi \alpha}$. It is clear that $0<a \leqslant \mathrm{e}^{-\pi}<\mathrm{e}^{-1}=\frac{1}{\mathrm{e}}<\frac{1}{2}$.

We can produce the following estimate:
$$\sum_{n=1}^{\infty}a^{n^2}=\sum_{n \in \mathbb{N}^{\times}}a^{n^2}=\sum_{m \in \left(\mathbb{N}^{\times}\right)^{\underline{2}}}a^m \leqslant \sum_{m \in \mathbb{N}^{\times}}a^m=\frac{a}{1-a},$$
which relies on the fact that for a sequence of positive reals, the sum of the associated series coincides with the sum of the sequence understood in the general sense of summation of arbitrary families in topological groups. The notation $\left(\mathbb{N}^{\times}\right)^{\underline{2}}=\left\{m^2\right\}_{m \in \mathbb{N}^{\times}}$ expresses the set of all nonzero square natural numbers and in order to obtain the sum of the right-most (geometric) series we relied on the condition that $a<1$.

This upper bound can be used to reach the desired conclusion as long as we manage to establish that in the given situation $\frac{a}{1-a} \leqslant 2a$, which amounts to claiming that $a \leqslant \frac{1}{2}$ and is indeed a condition we have already ascertained $a$ to satisfy.

This concludes the solution.

P.S. An equally valid manner of justifying the inequality $\displaystyle\sum_{n=1}^{\infty}a^{n^2} \leqslant \displaystyle\sum_{n=1}^{\infty}a^n$ is to simply the remark the term-wise comparison $a^{n^2} \leqslant a^n$ occurring between the two series in question, as $n^2 \geqslant n$ for any $n \in \mathbb{N}$ and $a<1$.


问:
Existence theorem for Laplace's equation.
Regarding Laplace's equations, the following two are my questions:
1) If Dirichlet boundary conditions are specified on the surfaces, is it guaranteed that a solution will exist? Or is there some condition that the boundary conditions themselves must satisfy?
2)  If Neumann boundary conditions are specified on the surfaces, is it guaranteed that a solution will exist? Or is there some condition that the boundary conditions themselves must satisfy?</p>

答:
You have
$$
\Delta u= 0, \nabla u\cdot n=g,
$$
then you need the following compatibility condition
$$
\int_{\partial \Omega} g(s)ds=0.
$$
For the Dirichlet problem I refer you the book of Evans 'Partial Differential Equations'.</p>


问:
Solutions to sum of exponential
I am reading a paper where the authors state that there exist infinite solutions to

$$
\sum_{n=1}^N\frac{\sin(m\pi a_n)}{m\pi}e^{-im\pi(2b_n+a_n)}=0, m\in\mathbb{Z}, m\neq0
$$

where $a_n$ and $b_n$ are between $0$ and $1$. I have spent lots of time figuring out how the solutions, i.e., $a_n$'s and $b_n$'s, look like for this problem but I have failed to proceed. I was hoping I could find some advice on this here.

答:
Not an answer but too long for a comment.

Suppose $a_1=\ldots = a_N=\frac{1}{2}$, then we see that the sum is
\begin{align}
\frac{\sin\frac{\pi m}{2}}{m\pi}e^{-im\pi/2}\sum^N_{n=1}e^{-i2\pi mb_n}.
\end{align}
If $m$ is even then any thing will work, i.e. the sum will be zero (not in a very interesting way). If $m$ is odd then, we have that
\begin{align}
\frac{\sin\frac{\pi (2k+1)}{2}}{(2k+1)\pi}e^{-i(2k+1)\pi/2} = \frac{1}{(2k+1)\pi i}\ne 0
\end{align}
which means that the sum is zero if
\begin{align}
\sum^N_{n=1}e^{-i2\pi mb_n}=0
\end{align}
which means that $e^{-i2\pi mb_n}$ are the $N$th roots of unity (and any rotation of them).

Title: Solutions to sum of exponential


问:
how to find the second derivative of the $\ln \frac{1-m^{x}}{x}$ at $x=0, m>1$?
I tried to write them as Taylor series, but then I obtain the series which is divergent.</p>

Body:
$$y=\log \left(\frac{1-m^x}{x}\right)=\log(1-m^x)-\log(x)$$

$$\log(1-m^x)=\log (-\log (m))+\log (x)+\frac{1}{2} x \log (m)+\frac{1}{24} x^2 \log
   ^2(m)+O\left(x^4\right)$$
$$y=\log (-\log (m))+\frac{1}{2} x \log (m)+\frac{1}{24} x^2 \log
   ^2(m)+O\left(x^4\right)$$ Differentiate twice.

Edit

Around $x=0$
$$m^x=e^{x \log(m)}=1+x \log (m)+\frac{1}{2} x^2 \log ^2(m)+O\left(x^3\right)$$
$$1-m^x=-x \log (m)-\frac{1}{2} x^2 \log ^2(m)+O\left(x^3\right)$$
$$\log(1-m ^x)=\log (-\log (m))+\log (x)+\frac{1}{2} x \log (m)+O\left(x^2\right)$$


问:
If we regard a field $F$ as a vector space then it can only have two subspaces.
I would like to show that given a field $F$ and a vector space $V=F$ (so we are viewing this field as a vector space), then $V$ has only two subspaces. I know that these two subspaces are $\{0\}$ and $V$ and I know how to show that each one of them is a subspace. However, my issue is to show that these two are the only subspaces. I have tried to assume that there is a third subspace other than these two and somehow arrive at a contradiction, but so far no luck. Would someone please be able to help?

答:
Assuming that $F$ is being viewed as a vector space over itself, note that it has dimension $1$. Why? Take the multiplicative identity $e \in F$. Note that $\{e\}$ forms a basis for $F$. Then, any subspace of $F$ either has dimension $1$ or $0$. If it's $0$, the only subspace it can be is $\{0\}$. If it's $1$, then it's all of $F$.

Edit: An Alternative Approach

Let $V'$ be a subspace of $F$. Suppose $V'$ is not $\{0\}$. Therefore, there exists non-zero $v \in V' \subset F$. If $m \in F$, then note that $m = (mv^{-1})v$ where $v^{-1}$ exists because $v \neq 0$, and $V'$ is a subspace (so we have a notion of scalar multiplication in that subspace). Hence, $m$ can be viewed as a scalar multiple of $v \implies F \subset V'$ (remember your subspace axioms! $v \in V'$ implies $cv \in V'$ for all $c \in F$). It follows $F = V'$


问:
How many 6 digit positive integers have their digits in weakly decreasing order?
How do I solve this combinatorically? I'm not quite sure how to approach this since each digit is dependent on the digit before it.

For example, if the first digit is a 9, then there are 10 possible choices for the second digit (0-9).
9 _ _ _ _ _.
If we choose 5 to be the second digit, then the third digit has 6 possible choices (0-5).
9 5 _ _ _ _.
If we choose 0 to be the third digit, then the fourth digit has 1 possible choices (0).
9 5 0 _ _ _.
And for the rest of the digits, only 0 is possible.
9 5 0 0 0 0

What formula do I use to solve this? Could I use Stirling numbers or Permutations and Combinations in this situation?

答:
HINT: This is the same as counting the $8$-digit numbers that start with $9$, end with $0$, and have their digits in weakly decreasing order, and the latter are easier to count.

Let $d_0d_1d_2d_3d_4d_5d_6d_7$ be such a number, so that

$$9=d_0\ge d_1\ge\ldots\ge d_6\ge d_7=0\,.$$

For $k=0,\ldots,6$ let $x_k=d_k-d_{k+1}$; each of these numbers is non-negative.

Show that $x_0+x_1+\ldots+x_6=9$.

Explain why there is a bijection between the numbers that we’re counting and solutions to this equation in non-negative integers.

Use the stars and bars method to compute the number of such solutions.


问:
determine whether a point of a set is in the interior point
Let $A \subseteq \Bbb R^2$ with $A = \{(x,y): 1<x<4, 1<y<3 \}$.
Is the point $(2,2)$ be an interior point of $A$?
Is the point $(4,2)$ be an interior point of $A$?
Justify!

Attempt: Edit:
I know that the answer is yes and no, respectively.
But, I'm not sure how to show it.

To show: $(2,2)$ is an interior point of $A$.
Let $w \in B(u,r)$ with $u=(2,2)$. Let $r=\frac{1}{2}>0$ and $k=\frac{1}{2} - ||w-u||$. Then,
$k > 0$ since $||w-u|| < \frac{1}{2}$. Let $x \in B(w,k)$. Then, by the triangle inequality,
\begin{equation*}
||x-u|| \le ||x-w|| + ||w-u|| \le k + ||w-u|| = \frac{1}{2}.
\end{equation*}
Thus, $x \in B(u,\frac{1}{2})$ so that $B(w,k) \subseteq B(u,\frac{1}{2}) \subseteq A$.
Hence, $w$ is an interior point of $A$ such that $B(u,\frac{1}{2}) \subseteq int(A)$ and so, $u$ is an interior point of $A$.

How to show it?

On the other hand, a point $x \in A$ is an interior point of $A$ if there exists $r>0$ such that $B(x,r) \subseteq A$ i.e.
there exists $r>0$ such that for all $y \in \Bbb R^2, ||y-x||<r \implies y \in A$.

答: $A = (1,4) \times (1,3)$ is open in $\Bbb R^2$ so all of its points are interior points.


问:
Existence and uniqueness of $SL(n,\mathbb R)$ left-invariant Borel probability measure on $SL(n,\mathbb R)/SL(n,\mathbb Z)$.
Consider the action of $SL(n,\mathbb R)$ on the homogeneous space $SL(n,\mathbb R)/SL(n,\mathbb Z)$ by left translation. Are there any good refences where I can find the proof of the existence and uniqueness of the left $SL(n,\mathbb R)$-invariant Borel probability measure on $SL(n,\mathbb R)/SL(n,\mathbb Z)$.

If there is some powerful theorem addressing this for general $G$ and $G/\Gamma$, please let me know how $SL(n,\mathbb R)$ and $SL(n,\mathbb R)/SL(n,\mathbb Z)$ satisfy the hypothesis of that theorem.

答:
Generally, if $G$ is a unimodular locally compact topological group, and $\Gamma$ is a unimodular closed subgroup of $G$, there is a unique up to scalar left $G$-invariant Borel measure on the quotient space $\Gamma \backslash G$.  This measure $\mu_{G/\Gamma}$ has the property that
$$\int\limits_G f(g)dg = \int\limits_{\Gamma \backslash G} \int\limits_{\Gamma} f(\gamma g)d \gamma d\mu_{G/\Gamma}(g) \tag{$f \in L^1(G)$}$$
In your case, $G = \operatorname{SL}_2(\mathbb R)$ is unimodular, and $\Gamma = \operatorname{SL}_2(\mathbb Z)$ is a discrete and hence unimodular topological group, so the existence and uniqueness of the measure on $\Gamma \backslash G$ is no problem.  The more interesting question is why this measure space $\Gamma \backslash G$ has finite measure.

An indirect, but enlightening way of seeing this is to let $K = \operatorname{SO}_2(\mathbb R)$, and identify $G/K$ with the upper half plane $\mathbb H = \{ a+bi : b > 0\}$.  This is an identification of topological spaces and $\Gamma$-actions.  Since $K$ is compact and hence unimodular, $G/K$ has a right $K$-invariant measure, and one can check (painfully) that this measure coincides with the hyperbolic measure $\frac{dx dy}{y^2}$ on $\mathbb H$.

Let $D$ be the closure of the usual fundamental domain for $\Gamma$ acting on $\mathbb H$.  It has finite hyperbolic measure on $\mathbb H$, and for every $z \in \mathbb H$, there exists a $\gamma \in \Gamma$ such that $\gamma.z \in D$.  Identifying $D$ with a subset of $G/K$, we therefore also have $\mu_{G/K}(D) < \infty$.  Moreover,   Since $K$ is compact, the preimage $E = \{ g \in G : gK \in D\}$ has finite measure, since if $f$ is the characteristic function of $E$, we have

$$\operatorname{meas}(E) = \int\limits_G f(g)dg = \int\limits_{G/K} \int\limits_K f(gk)dkd\mu_{G/K}(k) = \operatorname{meas}(K) \mu_{G/K}(D) < \infty.$$

On the other hand, we can look at this finite number in a different way:

$$\operatorname{meas}(E) = \int\limits_G f(g)dg = \int\limits_{\Gamma \backslash G} \bigg(\sum\limits_{\gamma \in \Gamma} f(\gamma g) \bigg))d\mu_{G/\Gamma}(g).$$

Since for every $g \in G$, there is at least one $\gamma \in \Gamma$ such that $\gamma g \in E$, we have $\sum\limits_{\gamma \in \Gamma} f(\gamma g) \geq 1$ for every $g \in G$, and therefore

$$\operatorname{meas}(E) \geq \int\limits_{\Gamma \backslash G} d\mu_{G/\Gamma}(g).$$

This forces $\Gamma \backslash G$ to have finite measure, and it also implies an upper bound on each sum $\sum\limits_{\gamma \in \Gamma} f(\gamma g)$, except possibly for $g$ in a set of measure zero.


问:
If $\operatorname{Im}A + \operatorname{Im}B = \Bbb R^n$ then $\ker A \cap\ker B$ is not necessarily $\{0\}$?
Let $A,B$ be square matrices of order $n\geq 2$. If $\operatorname{Im}A + \operatorname{Im}B = \Bbb R^n$, why $\textbf{can't}$ we say that $\ker A \cap \ker B = \{0\}$?

答:
Because there is no relation between the geometric positions of the kernel and the image of a linear map $V\to V$, just their dimensions are linked (as they are complementary) through the rank-nullity theorem. For instance, one can make a linear map $\def\R{\Bbb R}\R^2\to\R^2$ of rank $1$ (so one-dimensional kernel and image) with kernel the line given by the equation $ax+by=0$ and image given by the span of the nonzero vector $\binom cd$, by having its matrix be $$\pmatrix{c\\d}\pmatrix{a&b}=\pmatrix{ca&cb\\da&db}.$$
Now for $A$ and $B$ simply keep $a,b$ unchanged (so their kernels will be the same) and respectively take $\binom cd=\binom10$ for $A$ and $\binom cd=\binom01$ for $B$. That gives you the counterexamples, for any nonzero pair $(a,b)$
$$
  A=\pmatrix{a&b\\0&0},\qquad B=\pmatrix{0&0\\a&b}.
$$


问:
Showing that $\{(x, y) \in \mathbb{R}^2 \mid x< 0, y<0 \}$ is open

Let $A = \{(x, y) \in \mathbb{R}^2 \mid xy > 0 \}$. Show that $A$ is a open subset of the plane.

I have that $A = \{(x, y) \in \mathbb{R}^2 \mid x > 0,y>0 \} \cup \{(x, y) \in \mathbb{R}^2 \mid x< 0, y<0 \}$

Letting $(x,y) \in \{(x, y) \in \mathbb{R}^2 \mid x > 0,y>0 \}$ and $r=\min\{x,y\},$$B((x,y), r)$ I have that $$|x-x'| \leqslant \sqrt{(x-x')^2 + (y-y')^2} = d((x,y), (x',y')) \leqslant r < \frac{r}{2}$$ so $x' >\frac{r}2 >0.$ Similar reasoning can be done for $y'$. Which would show that $B((x,y),r) \subset \{(x, y) \in \mathbb{R}^2 \mid x > 0,y>0 \}$.

How should I deal with the set $\{(x, y) \in \mathbb{R}^2 \mid x< 0, y<0 \}$? I cannot seem to show this... If I let $(x,y) \in \{(x, y) \in \mathbb{R}^2 \mid x< 0, y<0 \}$ and again $r = \min\{x,y\}$ and construct a ball $B((x,y), r)$ I can pick $(x',y') \in B((x,y), r)$, but I'm not sure how to proceed from here?

答:
$f(x,y)= xy$ is continuous as a map from $\Bbb R^2 \to R$.

$A = f^{-1}[(0, \rightarrow\rangle]$ is thus open, as the inverse image of an open set under a continuous function.


问:
Comparison between the limits of two real functions
I know

If $f:D(\subset\mathbb R)\to\mathbb R,c$ is a limit point of $D,$ and $f(x)\ge(\text{resp.}\le)~a~\forall~x\in D-\{c\},$ then $\displaystyle\lim_{x\to c}f(x)\ge(\text{resp.}\le)~a.$ (Provided the limit exists)

If $f,g,h:D(\subset\mathbb R)\to \mathbb R,c$ is a limit point of $D,$ and $f(x)\le g(x)\le h(x)~\forall~x\in D-\{c\},$ then $\displaystyle\lim_{x\to c}f(x)=\displaystyle\lim_{x\to c}h(x)\implies \displaystyle\lim_{x\to c}f(x)=\displaystyle\lim_{x\to c}g(x)=\displaystyle\lim_{x\to c}h(x).$

I wounder whether the following result holds:

$\diamond$ Let $f,g:D(\subset\mathbb R)\to\mathbb R,c$ is a limit point of $D,f(x)\ge g(x)~\forall~x\in D-\{c\}.$ Do all these imply $\displaystyle\lim_{x\to c}f(x)\ge\displaystyle\lim_{x\to c}g(x)?$ (Provided the limits exist)

For then it will be more powerful result from which both 1 and 2 follow.

My Attempt:

$F-\epsilon<f(x)<F+\epsilon~\forall~x\in D\cap N'(\delta_1)$

$G-\epsilon<g(x)<G+\epsilon~\forall~x\in D\cap N'(\delta_1)$

Set $\delta=\min\{\delta_1,\delta_2\}.$ Then $G-\epsilon<g(x)\le f(x)<F+\epsilon~\forall~x\in D\cap N'(\delta)$ i.e. $G<F+\epsilon.$

Since $\epsilon>0$ is arbitrary so $G\le\displaystyle\inf_{\epsilon>0}(F+\epsilon)=F.$

Is my attempt correct?

答: Basically, you can just use 1. for the function $h = f - g$ and $a=0$.


问:
Solve the equation $x=1-5(1-5x^2)^2$
Solve the equation
$$x=1-5(1-5x^2)^2$$

###My work

Let $f(x)=1-5x^2$. Then we have tha equation $f(f(x))=x$. But in this case we don't use the equation $f(x)=x$ because $f(x)$ is not monotonic function

答:
You are on the right track; a modification of your approach will get you the rest of the way.

Let $y = f(x) = 1 - 5x^2$.  Then $f(f(x)) = x$ implies $f(y) = x$, hence the solution to the simultaneous system $$\begin{align} y &= 1 - 5x^2, \\ x &= 1 - 5y^2, \end{align}$$ will yield the desired $x$-values.  To this end, we take the difference to obtain $$y - x = 5(y^2 - x^2) = 5(y-x)(x+y),$$ hence $$(y-x)(5(x+y) - 1) = 0.$$  Then $x = y$ amounts to solving $x = 1 - 5x^2$, which are the "obvious" solutions.  The other set corresponds to $x + y = \frac{1}{5}$, which are the "nontrivial" solutions, and require solving the quadratic $$5x^2 - x - \frac{4}{5} = 0.$$

Title: Solve the equation $x=1-5(1-5x^2)^2$


问:
Riemann-Stieltjes integral of a continuous function w.r.t. a step function
$
\newcommand{\para}[1]{\left( #1 \right)}
\newcommand{\abs}[1]{\left| \, #1 \, \right|}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\ds}{\displaystyle}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\norm}[1]{\left \Vert #1 \right \Vert}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\vp}{\varphi}
\newcommand{\ve}{\varepsilon}
\newcommand{\a}{\alpha}
\newcommand{\g}{\Gamma}
\newcommand{\og}{\overline{\Gamma}}
$Definition: I define the notion of an "increasing partition" $P$ of $[a,b]$ to be a finite partition $\set{z_k}_{k=0}^n \subseteq [a,b]$ where $a = z_0 < z_1 < \cdots < z_n = b$. It's a minor thing that helps reduce the repetition in the wording herein.

Problem Statement:  Let $\vp : [a,b] \to \Bbb R$ be a step function with $\set{\a_i}_{i=1}^m$ its discontinuities forming an increasing partition of $[a,b]$. Thus $\vp$ is constant on each interval $(\alpha_{i-1},\alpha_i)$. Let $f : [a,b] \to \Bbb R$ be continuous, and let
\begin{alignat*}{99}
\vp(\alpha_i^+) &:=\;&& \lim_{x \to  \a_i^+} \vp(x) &&\quad \forall i \in \set{0,1,\cdots,m-1} \\
\vp(\a_i^-) &:=\;&& \lim_{x \to \alpha_i^-} \vp(x) &&\quad \forall i \in \set{0,1,\cdots,m-1} \\
d_i &:=\;&& \vp(\a_i^+) - \vp(\a_i^-) &&\quad \forall i \in \set{1,2,\cdots,m-1} \\
d_0 &:=\;&& \vp(\a_0^+) - \vp(\a_0) \\
d_m &:=\;&& \vp(\a_m) - \vp(\a_m^-)
\end{alignat*}
Show
$$
\int_a^b f \, \dd \vp = \sum_{i=1}^m f(\a_i) d_i
$$

Context: This ultimately comes up as a homework assignment for me, and I just wanted to check my solution, because I have some doubts on it. I'll express those at the end, though, since they tie more intimately into my proof.

Attempted Proof:

Let the following be:

Let $\g_\a := \set{\a_i}_{i=0}^m$ be an increasing partition of $[a,b]$. (This is the set of discontinuities of $\vp$, i.e. $\vp$ is constant on $(\a_{i-1},\a_i)$ $\forall i$.)

Let the definitions in the problem statement hold.

Let $\ve > 0$.

Let $\ds \delta := \min_{1 \le i \le m} \left| \a_i - \a_{i-1} \right|$.

Let $\g := \set{x_i}_{i=0}^n$ be an increasing partition of $[a,b]$ with $\norm \g < \delta$.

Let $\ol \g := \g \cup \g_\a$, and let it be written $\ol \g := \set{\ol{x_i}}_{i=0}^p$.

Let $\ol{\xi_i} \in [\ol{x_i}, \ol{x_{i-1}}]$ for each $i$. (These are our representative valuation points in the definition of the integral.)

By the merit of $\ol \g$ being a refinement of $\g$, and by merit of definition, we have
$$
\norm{ \ol \g } \le \norm \g < \delta \le \norm{\g_\a}
$$
By the Cauchy criterion for Riemann-Stieljes integration, then, where $R_\g$ denotes the usual Riemann-Stieltjes sum,
$$
\abs{ R_\g - R_{\ol \g} } < \ve
$$
and so $R_{\ol \g}$ equals the desired integral. It remains to see if this equals the desired sum.

Write
$$
R_{\og} = \sum_{j=1}^p f \para{ \ol{ \xi_j } } \Big( \vp \para{ \ol{x_j} } - \vp \para{ \ol{x_{j-1}} } \Big)
$$
From definition, we may have either $\ol{x_i} = \alpha_j \in \g_\a$ for some $i$ (i.e. is a discontinuity of $\vp$), or $\ol{x_i} = x_i \in \g$ for some $i$. Note that for any consecutive pairing $\ol{x_{i-1}},\ol{x_{i}}$ that not both are in $\g_\a$. (This is because each are in $\og$ for which $\norm \og < \delta$, and thus the smallest distance between any two discontinuities of $\vp$ is bigger than the distance between any two members of $\og$.) Thus, we only have three cases to consider, based on how many of a given pair are, indeed, discontinuities:

Case 1 (neither): In the case neither $\ol{x_{i-1}},\ol{x_{i}}$ are discontinuities of $\vp$, then trivially we will have
$$
\vp \para{ \ol{x_j} } - \vp \para{ \ol{x_{j-1}} } = 0
$$

Case 2 (right discontinuity): Suppose $\ol{x_{i}} = \a_j$ for some $j$ (but, again, that means $\ol{x_{i-1}}$ is not a discontinuity). Then in such a case, using the piecewise continuity and constantness of $\vp$,
$$
\vp \para{ \ol{x_j} } - \vp \para{ \ol{x_{j-1}} } = \vp\para{ \a_j } - \vp \para{ \a_{j}^- }
$$

Case 3 (left discontinuity): Similar to the previous, we have $\ol{x_{i-1}} = \a_j$ for some $j$, and take
$$
\vp \para{ \ol{x_j} } - \vp \para{ \ol{x_{j-1}} } = \vp\para{ \a_j^+ } - \vp \para{ \a_{j} }
$$

Let $\ol{\xi_{i,j}}$ be the $\ol{\xi_i}$ whose terms are not eliminated by merit of Case 1 but lie in Case 2; similarly, $\ol{\xi'_{i,j}}$ for those whose terms lie in Case 3. Then we have, once the terms have negated,
$$
R_{\og} = \underbrace{\sum_{j=1}^m f \para{ \ol{\xi_{i,j}} } \Big( \vp\para{ \a_j } - \vp \para{ \a_{j}^- } \Big) }_{\text{case 2 terms}} + \underbrace{\sum_{j=1}^m f \para{ \ol{\xi'_{i,j}} } \Big( \vp\para{ \a_j^+ } - \vp \para{ \a_{j} } \Big)}_{\text{case 3 terms}}
$$
Note that, for the Case 2 terms, each $\ol{\xi_{i,j}}$ is within $\delta$ of $\a_j$; specifically, $\ol{\xi_{i,j}} \in (\a_j - \delta,\a_j)$. Likewise, $\ol{\xi'_{i,j}} \in (\a_j,\a_j+\delta)$. Since $f$ is continuous, we may take each to be its corresponding $\a_j$, and write
$$
R_{\og} = \sum_{j=1}^m f \para{ \a_j } \Big( \vp\para{ \a_j } - \vp \para{ \a_{j}^- } \Big)  +\sum_{j=1}^m f \para{ \a_j } \Big( \vp\para{ \a_j^+ } - \vp \para{ \a_{j} } \Big)
$$
Some clear cancellation results, giving
$$
R_{\og} = \sum_{j=1}^m f \para{ \a_j } \Big( \vp\para{ \a_j^+ } - \vp \para{ \a_{j}^- } \Big)
$$
Taking the usual convention for what occurs for the $\a_j$ which are $a,b$, we see the $\vp$ difference here is indeed $d_i$. So we see
$$
\int_a^b f \, \dd \vp  = R_{\og} = \sum_{j=1}^m f \para{ \a_j } d_i
$$
as desired.

My Questions / Issues:

While I feel like this proof is on the right path, I have some anxieties about some steps. Some in particular I feel aren't quite "there" yet. I was wondering if anyone would be able to help me flesh these ideas uot.

The use of the Cauchy criterion for Riemann-Stieltjes integrals (stated here) feels far too un-fleshed out, and I doubt it actually lets me conclude that $R_\og$ equals the integral in question. I imagine it only lets me conclude the integral exists at best. I do see that this proof seems to use something similar towards the end of their proof, but their motivation, choice of bound, and the weird $\delta$'s all escape me.

The second chunk of the proof (verifying the alternate form for $R_\og$) feels right, but is there any issue there? Something feels sketchy about how I waited until then to define the $\ol{\xi_{i,j}}$ terms to be $\a_j$ in particular, but I don't see how it would be problematic either...

I feel like I barely used the continuity of $f$ at all, which means its purpose is lurking elsewhere. Where would that be?

Thanks for the insights and thoughts you guys can give!

答:
Actually this problem admits a direct solution. Notice that $$\int_a^b fd\phi=\sum_{i=1}^m\int_{\alpha_{i-1}}^{\alpha_i}fd\phi $$ because $\{\alpha_i\}$ it's a partition of $[a,b]$. Since $\phi$ is constant on every $[\alpha_{i-1},\alpha_i)$ and possible different only on $\alpha_i$, any Riemann Stieltjes sum is zero except for one single term (the right extreme), i.e., $$\int_{\alpha_{i-1}}^{\alpha_i}fd\phi=f(\alpha_{i})d_i $$ from where your result follows, here you need to use the uniform continuity of $f$. I realized that this idea is very similar to the other answer, with the advantage that if you prove the result for one single integral, then by additivity the general result follows. I will add the proof since it's a little bit different:

Let $\varepsilon>0$, and let $\delta>0$ such that the uniform continuity of $f$ holds on $[\alpha_{i-1}, \alpha_i]$ with $\varepsilon/|d_i|$  take $\Gamma=\{n_j\}_{j=0}^k$ a partition of $[\alpha_{i-1}, \alpha_i]$ with $\max_{1\leq j\leq k}\{|n_{j}-n_{j-1}|\}<\delta$ and $\{p_j\}$ a sequence of points such that $n_{j-1}\leq p_j\leq n_j$, then $$R_{\Gamma} =\sum_{j=1}^mf(p_j)(\phi(n_{j-1})-\phi(n_{j}))=f(p_k)(\phi(n_{k-1})-\phi(n_k))=f(p_k)(\phi(n_k^+)-\phi(n_k^-))$$ so we have $R_\Gamma=f(p_k)(\phi(\alpha_i^+)-\phi(\alpha_i^-))=f(p_k)d_i$ now since $n_{k-1}\leq p_k\leq \alpha_i$ we have $$|R_\Gamma-f(\alpha_i)d_i|\leq |d_i||f(p_k)-f(\alpha_i)|<|d_i|\varepsilon/|d_i|=\varepsilon $$ by the uniform continuity of $f$, this completes the proof.


问:
$P(x)=x^n+a_1x^{n−1}+\dots+a_n$ is a polynomial with real coefficients. If $a_1^2<a_2$ prove that $P(x)$ has no real roots.

$P(x)=x^n+a_1x^{n−1}+\dots+a_n$ is a polynomial with real coefficients. If $a_1^2<a_2$ prove that $P(x)$ has no real roots.

I tried solving it with contradiction assuming real roots and working out coefficients $a_1$ and $a_2$ but nothing seems to work. I tried using calculus too differentiating it n-2 times but I couldn't find any substantial result.

答:
We may suppose that $a_n\not =0$. Suppose that $P$ has all its roots real. Then it is the same for $Q(x)=x^nP(1/x)=1+a_1 x+a_2 x^2+\cdots$. Put $Q(x)=\prod (x-b_k)$ with $b_k\in \mathbb{R}$. Then $$\frac{Q^{\prime}(x)}{Q(x)}=\sum_{k=1}^{n}\frac{1}{x-b_k},$$ and taking the derivative we get  $$\frac{Q(x)Q^{\prime\prime}(x)-Q^{\prime}(x)^2}{Q(x)^2}=-\sum_{k=1}^n\frac{1}{(x-b_k)^2}$$
Now, putting $x=0$:
$$2a_2-a_1^2=-\sum_{k=1}^n\frac{1}{b_k^2} $$
and it is easy to finish.


答:
Clearly $\{0\}$ is a subspace. Now assume that $W \neq \{0\}$ is a subspace of $V$. We show that $W = V$. since $W \neq \{0\}$, $\exists c \in W, \text{ such that } c \neq 0$. but then from axioms of vector space it follows that $\dfrac{1}{c} \cdot c = 1 \in W$ (as $\dfrac{1}{c} \in F$). But $W$ itself is a vector space over field $F$. Thus, again from axioms  of vector spaces, it follows that $k \cdot 1 \in W, \forall k \in F$. but $k \cdot 1 = k$. Hence the result follows.

(This proof does not invoke any result related to dimension. Also, I have used same symbols  for vectors in $V (=F)$ and scalars in $F$. Hope it does not lead to any confusion.)

Title: If we regard a field $F$ as a vector space then it can only have two subspaces.

Body: I would like to show that given a field $F$ and a vector space $V=F$ (so we are viewing this field as a vector space), then $V$ has only two subspaces. I know that these two subspaces are $\{0\}$ and $V$ and I know how to show that each one of them is a subspace. However, my issue is to show that these two are the only subspaces. I have tried to assume that there is a third subspace other than these two and somehow arrive at a contradiction, but so far no luck. Would someone please be able to help?


问:
Real polynomials forming a basis
How do you show that a sequence of polynomials form a basis? I know that the rref criterion of a basis (no $0$ rows and has a leading 1 in every column) only works for $K^n$. Particularly,

Consider the set Consider the set $V$ of real polynomials $f$ of degree at most
3 which satisfy $f(1) = f(2)$, and $V$ is a vector space over $\mathbb{R}$.

Show that the following polynomials $f_1(x) = 1$, $f_2(x) = x^2-3x$
and $f_3(x) = x^3-7x$ form a basis of V .

Note: I can't use the determinant method to determine independence.

PROVING V is a vector space:

Let $f(1)=f(2)$, $g(1)=g(2)$ and $\alpha, \beta \in \mathbb{R}$

$(\alpha f+ \beta g)(1)$

= $(\alpha f)(1) + (\beta g)(1)$

= $\alpha (f(1)) + \beta (g(1))$

= $\alpha (f(2)) + \beta (g(2))$

= $(\alpha f)(2) + (\beta g)(2)$

= $(\alpha f + \beta g)(2)$

答:
You need to show:

The dimension of $V$ is $3$

$f_1, f_2$ and $f_3$ are in $V$

$f_1, f_2$ and $f_3$ are linearly independent.

i.e. The only linear combination $a_1f_1 + a_2f_2 + a_3f_3=0$ is the trivial combination.

Title: Real polynomials forming a basis


问:
If $V$ is $k$-dimensional then show that $\dim \wedge^n V = \binom {k} {n}.$

If $V$ is $k$-dimensional then show that $\dim \wedge^n V = \binom {k} {n}.$

I don't understand this result. If $k \lt n$ then how does the quantity $\binom {k} {n}$ make sense? Can anybody please help me in this regard?

Thanks for your time.

EDIT $:$ Let $V$ be a vector space. Then for any $n \in \Bbb N$ we can identify $V^{\times n}$ by $V^{\{1,2, \cdots, n \}}.$ Since $\{1,2, \cdots, n \}$ is an $S_n$-set it follows that $V^{\{1,2, \cdots, n \}}$ is an $S_n$-vector space. For any $\sigma \in S_n$ and any function $f : \{1,2, \cdots, n \} \longrightarrow V$ the action of $\sigma$ on $f$ is given by $(\sigma \cdot f) (i) = f (\sigma^{-1} (i)).$ This shows that $V^{\times n}$ is a $S_n$-vector space. Let $\mu$ be the action of $S_n$ on $V^{\times n}$ then for each $\sigma \in S_n$ the map $\otimes_V \circ \mu (\sigma, \cdot) : V^{\times n} \longrightarrow V^{\otimes n}$ will be a bilinear map. Then by universal property of tensor products there exists a unique bijective linear map $\rho (\sigma) : V^{\otimes n} \longrightarrow V^{\otimes n}.$ This gives a group homomorphism $\rho : S_n \longrightarrow GL(V^{\otimes n}).$ Hence $V^{\otimes n}$ is also a $S_n$-vector space where for any element $\sigma \in S_n$ and for any $v \in V^{\otimes n}$ the action of $\sigma$ on $v$ is given by $\sigma \cdot v : = \rho (\sigma) (v).$ If $\rho (\sigma) (v) = \left (\text {sign} (\sigma) \right ) v,$ for all $\sigma \in S_n$ and for some $v \in V^{\otimes n}$ then $v$ is said to be an alternating/anti-symmetric tensor. The collection of all alternating/anti-symmetric tensors forms a subspace of $V^{\otimes n}$ which is denoted by $\wedge^n V$ or $V^{\wedge n}.$

答: When $k<n$, the binomial coefficient ${{k}\choose{n}}$ is $0$. Also, when $k<n$ and $k = \dim V$, note that it is impossible to choose $n$ different basis vectors of $V$ to wedge together. As the space $\bigwedge^n V$ is spanned by all such exterior products with $n$ factors, and an exterior product is $0$ if the same basis vector appears in it twice, $\bigwedge^n V = \lbrace 0 \rbrace$, which has dimension $0 = {{k}\choose{n}}$.