答:
Suppose that I have a pool of $n$ people and want to form two disjoint committees, $A$ and $B$, using a total of $k$ of the $n$ people. $\binom{n}i\binom{n-i}{k-i}$ is the number of ways to choose $i$ people for committee $A$ and then choose $k-i$ of the remaining $n-i$ members for committee $B$. Summing over $i=0,\ldots,k$ gives the total number of ways to form these committees, including those in which $A$ or $B$ is empty.

This is the same as the number of ways to choose $k$ people from the pool and then select a subset of them to make up committee $A$; committee $B$ will then comprise the people who were among the $k$ initially chosen but are not on committee $A$. This can be done in $2^k\binom{n}k$ ways. Thus,

$$\sum_{i=0}^k\binom{n}i\binom{n-i}{k-i}=2^k\binom{n}k\,.$$

Title: looking for combinatorial identity

Body: Does anybody know a nicer expression for the sum $\sum_{i=0}^{k} \binom{n}{i} \binom{n-i}{k-i}$?


问:
What is the derivative of: $f(x)=x^{2x^{3x^{4x^{5x^{6x^{7x^{.{^{.^{.}}}}}}}}}}$?
I happened to ponder about the differentiation of the following function:
$$f(x)=x^{2x^{3x^{4x^{5x^{6x^{7x^{.{^{.^{.}}}}}}}}}}$$
Now, while I do know how to manipulate power towers to a certain extent, and know the general formula to differentiate $g(x)$ wrt $x$, where $$g(x)=f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{.{^{.^{.}}}}}}}}}}$$
I'm still unable to figure out as to how I can adequately manipulate the function to differentiate it within its domain of convergence.

General formula: $$g'(x)=\frac{g^2(x)f'(x)}{f(x)\left[1-g(x)\ln(f(x))\right]}$$

答:
First of all, we need to see how to define this function and if it exists.
The truth is that there are two functions behind it, not one.

Define the family of functions:

$$f_n(x)=x^{(n+1)f_{n+1}(x)}$$

But, there is an instant trouble. First of all we have

$$f_1(1)=1$$ and
$$f_1(1^{+})=\infty$$

The trouble is how to define $f(0)$. Let us define a partial function

$$
f_{m,n}(x)=\begin{cases}
 &  x^{f_{m,n+1}(x)},\text{ if } n \leq m  \\
 &  1,\text{ if } n>m
\end{cases}
$$

and then try to find the limiting function. And here is the problem.

If $m$ is even then the function starts at $1$ and has a jump at $\frac{1}{m}$ towards $0$.
If $m$ is odd then the the function starts at $0$ and has a jump at $\frac{1}{m}$ towards $1$.

Going towards infinity, we will obtain two versions of the functions: even that starts at $0$ and is monotonically increasing and odd that has a minimum between $0$ and $1$ and starts at $1$.

(We can agree that neither of two is continuous at $0$ or we need to extend the function at $0$ ourselves, because the limiting process is never going to reach $0$ or $1$ value at $f_1(0)$ by itself.)

odd and even case

odd and even case

Now regarding first derivative, there is no closed form of it. We can simply state it the same way the function is defined and then use an infinite chain to define it in full.

$$f_n'(x)=(n+1)x^{(n+1)f_{n+1}(x)-1} \left (x\ln(x)f'_{n+1}(x)+f_{n+1}(x) \right )$$

This makes it obvious that $f_1'(1)=2 \cdot 3 \cdot 4 \cdot 5 ... = \infty$ while $f_1'(0) = 0$

Now one thing is certain and that is that the difference between $f_{2n}$ and $f_{2n+2}$ is all smaller and smaller as $n$ tends infinity. The same for $f_{2n+1}$ and $f_{2n+3}$.

so we can try to use that and define that at $n \gg 1$

$$f_n(x)=y \sim x^{(n+1)x^{(n+2)y}}$$

Now, here comes the story. First, since $x$ is between $0$ and $1$ $x^n$ for very large $n$ tends to $0$. So we arrive at the expression

$$y = 0^{0^y}$$

which can lead only to $y=0$ or $y=1$

So at infinity we have $f_n(x)$ reaching flat $1$. (We could take $0$ as well but that is the same as saying that $f_{n-1}(x)$ reaches $1$. So $1$ is more universal.)

For this reason, our analysis of starting with the above definition for $f_{m,n}(x)$ is just right.

Other than this it is not likely that it is possible to write this function in any known and simple closed form even with all other special functions available. It is simply a beast on its own starting from its dual nature.

Another curious question is how to represent these exponential functions. It is difficult to find a fine cut and the difference between

$$x^{2^{x^{3}}}$$

and

$$x^{2x^{3}}$$

and

$$x^{(2x)^{3}}$$

This notation should reduce ambiguity and do notice that extended exponential is no longer commutative.

$$x^{2 \cdot x^{3 \cdot x^{4 \cdot x^{5 \cdot x...}}}}=\operatorname*{E}\limits_{n=1}^{\infty}x^{n+1}$$

$$x^{(2x)^{(3x)^{(4x)^{(5x)...}}}}=\operatorname*{E}\limits_{n=1}^{\infty}nx$$

$x^{2^{x^{3}}}$ is obviously a special case with alteration
$$
(x,n) = \begin{cases}
 &  x, n \mod 2=1\\
 &  \frac{n}{2}+1, n \mod 2=0
\end{cases}
$$

$$\operatorname*{E}\limits_{n=1}^{\infty}(x,n)$$

A general problem with this extended exponentiation is that you technically need to read it right to left in order to clear the confusion.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
Confusion about linearity of expectation
I am following the course CS 285 Deep Reinforcement Learning from UC Berkeley. In lecture 4, part 1 (around 17:00), prof. introduces an expectation of rewards of a trajectory over a policy. He first introduces the probability of a trajectory over theta as follows.
\begin{aligned}
\underbrace{p_{\theta}\left(\mathbf{s}_{1}, \mathbf{a}_{1}, \ldots, \mathbf{s}_{T}, \mathbf{a}_{T}\right)}_{p_{\theta}(\tau)}=p\left(\mathbf{s}_{1}\right) \prod_{t=1}^{T} \underbrace{\pi_{\theta}\left(\mathbf{a}_{t} \mid \mathbf{s}_{t}\right) p\left(\mathbf{s}_{t+1} \mid \mathbf{s}_{t}, \mathbf{a}_{t}\right)}_{\text {Markov chain on }(\mathbf{s}, \mathbf{a})}
\end{aligned}

Then he says that we can use linearity of expectation at this point and writes the following equality.

\begin{aligned}
\theta^{\star} &=\arg \max _{\theta} E_{\tau \sim p_{\theta}(\tau)}\left[\sum_{t} r\left(\mathbf{s}_{t}, \mathbf{a}_{t}\right)\right] \\
&=\arg \max _{\theta} \sum_{t=1}^{T} E_{\left(\mathbf{s}_{t}, \mathbf{a}_{t}\right) \sim p_{\theta}\left(\mathbf{s}_{t}, \mathbf{a}_{t}\right)}\left[r\left(\mathbf{s}_{t}, \mathbf{a}_{t}\right)\right]
\end{aligned}

And tells that $p_{\theta}(\mathbf{s}_t, \mathbf{a}_t)$ is state-action marginal. I am trying to understand this. Even though we have markov assumption, $p_{\theta}(\mathbf{s}_t, \mathbf{a}_t)$ still depends on $(\mathbf{s}_{t-1}, \mathbf{a}_{t-1})$ so the elements of the joint distribution $p_\theta(\tau)$ are not independent.
Don't we need to use conditional probabilities for such expansion of the joint probability in the second equation? Isn't this kind of equal to saying $E_{x,y \sim p(x,y)}[f(x,y)] = E_x[f(x,y)] + E_y[f(x,y)]$. Maybe this is doable when $f(x,y)$ is something like $x+y$ which is I believe similar to the example.

答: Linearity of expectation does not require independence: it holds whenever the marginal expectations are finite, even if there is dependence among the random variables.</p>


问:
Repeatedly assigning people to subgroups so everyone knows each other
Say a teacher divides his students into subgroups once every class. The profile of subgroup sizes is the same everyday (e.g. with 28 students it might be always 8 groups of 3 and 1 group of 4). How can the teacher specify the subgroup assignments for a classes so that in the shortest number of classes, everyone has been in a subgroup with everyone else? Exhaustive search seems intractable, so what would be good optimization approach?

Edit: To clarify, I'm assuming the subgroup structure is given and fixed. The optimization is just the assignment of students to groups across the days.

答:
The shortest number of classes is 1, with a single subgroup that contains all the students.

The second shortest number of classes is 3, which is achieved with one small subgroup and a large one.  For example, ABCDEFG/HI, ABCDEHI/FG, ABCFGHI/DE.

The proof that 2 classes is impossible: let X, Y be the two largest subgroups from the first day.  Each member of X must meet each member of Y on the second day, but to do this would create a subgroup larger than either X or Y.


问:
how do I justify this $\int_{0}^{\infty}\frac {e^{-2x}-e^{-x}}{x}dx=\lim_ {s\to1^{-}} \int_{0}^{\infty}\frac{e^{-2x}-e^{-x}}{x^s}dx$
I was trying to compute
$\displaystyle\int_{0}^{\infty}
\frac{{\rm e}^{-2x} - {\rm e}^{-x}}{x}\,{\rm d}x$ when I had an idea, the gamma function is $\Gamma(1+z)=\int_{0}^{\infty}x^ze^{-x}dx$ and for $s<1$ the integral $\int_{0}^{\infty}\frac{e^{-2x}-e^{-x}}{x^s}dx$
converges, so call
$$I(s)=\int_{0}^{\infty}\frac{e^{-2x}-e^{-x}}{x^s}dx$$
and hope that it is continuous from the left at $s=1$.

Assuming that it is, for $s<1$ we have
$$I(s)=(2^{s-1}-1)\Gamma(1-s)$$
and so:
$$I(1)=\int_{0}^{\infty}\frac {e^{-2x}-e^{-x}}{x}dx = \lim_{s\to1^-}I(s)= \lim_{s\to1^-}(2^{s-1}-1)\Gamma(1-s)$$

Now, write $$2^{s-1}-1=1+\ln(2)(s-1)+o((s-1)^2)-1=\ln(2)(s-1)+o((s-1)^2)$$
and so:
$$\lim_{s\to1^-}(2^{s-1}-1)\Gamma(1-s) = -\ln(2)\lim_{s\to 1^-}(1-s)\Gamma(1-s)+\lim_{s\to 1^-}o((1-s)^2)\Gamma(1-s)$$

Here comes the cool part, use the reflection formula ($\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$) and obtain:
$$(1-s)\Gamma(1-s) = \frac {\pi (1-s)}{\sin(\pi s)\Gamma(s)}$$
and:
$$o((1-s)^2)\Gamma(1-s) = o(1-s)\frac {\pi (1-s)}{\sin(\pi s)\Gamma(s)}$$

Now, $\Gamma$ is continuous at 1 and $\Gamma(1)=1$ so:
$$\lim_{s\to 1^-}\frac {\pi (1-s)}{\sin(\pi s)\Gamma(s)} = \lim_{s\to 1^-}\frac {\pi (1-s)}{\sin(\pi s)}$$

this limit is finite and is equal to 1 (with l'hopital) and so
$$\lim_{s\to 1^-}o((1-s)^2)\Gamma(1-s) = 0$$

And we can finally finish up and say $I(1)=-\ln(2)$.
Now, I don't know how to justify the fact that $I(s)$ is continuous at $s=1$ from the left and so I don't know if my solution is correct.

BUT! the answer is correct which gives me hope that this solution is correct with some additional justifications about the continuity of $I$ at $s=1^-$ which is why I am looking for help.

答: $\int_0^{\infty} \frac {e^{-x}-e^{-2x}} {x^{s}}dx=\int_0^{1} \frac {e^{-x}-e^{-2x}} {x^{s}}dx+\int_1^{\infty} \frac {e^{-x}-e^{-2x}} {x^{s}}dx$. In the second term use the DCT to justify interchange of limit and integral. (Note that $x^{s} >1$ for $x >1$ and $s>0$). In the first term use Monotone Convergence Theorem.


问:
population of ants is given $P(t) = 100 e^{0.3t}$ where $t$ is days. Find rate of change in population when $t = 3$ and $t = 9$.
The population, $P(t)$, of rare Brazilian ants is given by $P(t) = 100e^{0.3t}$ where $t$ is in days. Find the rate of change in population when $t = 3$ and $t = 9$.
What would be answer for this?

答: The word " The rate of change " refers to the derivative. So the question asks you to calculate the derivative of $P(t)$ when $t = 3,9$ respectively. Hence $P'(t) = 0.3\times 100e^{0.3t}\implies P'(3)= 30\times e^{0.3\cdot 3}= 30\times e^{0.9}\approx 73.79$. In other words, the population increases at a rate of about $74$ person per day on the third day of the experiment. The other one is done in the same way.


问:
Find this limit without using L'Hopital's rule or Taylor Series
I want to solve $$\lim_{x\to 2}{{\log_2(\sqrt{4x-x^2})-\sqrt{\cos{\pi x}}}\over(\frac{x}{2})^{(\sin{\pi x})}-1}$$
without using differentiation in any context. Basically without using L'Hopital's rule and Taylor's theorem.

I have solved the given limit by using Taylor's theorem in order to find equivalent functions and I have gotten the correct answer. However I am required to do so, without actually differentiating. I am guessing that already established equivalent function formulae and substitutions are the only two tools that I am allowed to use.

Any hints or help will be very much appreciated.

答:
The method is to decompose this function into several whose limits are easier.

First, substituting $x = 2+y$ (to better see what's going on) we have
$$ \lim_{x\to 2} \frac{\log_2(\sqrt{4x-x^2}) - \sqrt{\cos\pi x}}{(\frac{x}{2})^{\sin \pi x} - 1} = \lim_{y\to 0} \frac{\log_2\sqrt{4-y^2}-\sqrt{\cos \pi y}}{(1+\frac{y}{2})^{\sin\pi y} - 1} $$
We have then
$$ \frac{\log_2\sqrt{4-y^2}-\sqrt{\cos \pi y}}{(1+\frac{y}{2})^{\sin\pi y} - 1} = \frac{\log_2\sqrt{4-y^2}-\sqrt{\cos \pi y}}{y^2} \frac{y^2}{(1+\frac{y}{2})^{\sin\pi y} - 1}$$
$$ \frac{\log_2\sqrt{4-y^2}-\sqrt{\cos \pi y}}{y^2} = \frac{\log_2\sqrt{4-y^2}-1}{y^2} + \frac{1-\sqrt{\cos \pi y}}{y^2}$$
\begin{align} \frac{\log_2\sqrt{4-y^2}-1}{y^2} &=\frac{\log_2\sqrt{4-y^2}-\log_2 2}{y^2} =\frac{\log_2(\frac12\sqrt{4-y^2})}{y^2}= \frac{\log_2\sqrt{1-\frac{y^2}{4}}}{y^2} = \\
&= \frac{\log_2\sqrt{1-\frac{y^2}{4}}}{\sqrt{1-\frac{y^2}{4}} - 1} \cdot \frac{\sqrt{1-\frac{y^2}{4}} -1}{y^2} =\\
&= \frac{\log_2\sqrt{1-\frac{y^2}{4}}}{\sqrt{1-\frac{y^2}{4}} - 1} \cdot \frac{(1-\frac{y^2}{4})-1}{y^2(\sqrt{1-\frac{y^2}{4}} +1)} = \\
&= \frac{1}{\ln 2}\frac{\ln\sqrt{1-\frac{y^2}{4}}}{\sqrt{1-\frac{y^2}{4}} - 1} \cdot \frac{-\frac14}{\sqrt{1-\frac{y^2}{4}} +1} \end{align}
We have
$$ \lim_{y\to 0} \frac{-\frac14}{\sqrt{1-\frac{y^2}{4}} +1} = \frac{-\frac14}{1+1} = -\frac18$$
and using $t = \sqrt{1-\frac{y^2}{4}} - 1$ we have
$$ \lim_{y\to 0} \frac{\ln\sqrt{1-\frac{y^2}{4}}}{\sqrt{1-\frac{y^2}{4}} - 1} = \lim_{t\to 0} \frac{\ln(1+t)}{t} = 1 $$
(assuming we already know this limit), so
$$ \lim_{y\to 0} \frac{\log_2\sqrt{4-y^2}-1}{y^2} = -\frac{1}{8\ln 2}$$
Similarily, we have
\begin{align} \frac{1-\sqrt{\cos \pi y}}{y^2} &= \frac{1-\cos \pi y}{y^2(1+\sqrt{\cos \pi y})} = \frac{2\sin^2 \frac{\pi y}{2}}{y^2(1+\sqrt{\cos \pi y})} = \\
&= \left(\frac{\sin\frac{\pi y}{2}}{\frac{\pi y}{2}}\right)^2 \frac{\pi^2}{2(1+\sqrt{\cos \pi y})}\end{align}
Assuming we know that $\lim_{t\to 0}\frac{\sin t}{t} =1$, we get
$$ \lim_{y\to 0} \frac{1-\sqrt{\cos \pi y}}{y^2} = 1^2 \frac{\pi^2}{2(1+\sqrt{1})} = \frac{\pi^2}{4} $$
Finally we have
\begin{align} \frac{y^2}{(1+\frac{y}{2})^{\sin\pi y} - 1} &= \frac{y^2}{e^{(\ln(1+\frac{y}{2}))(\sin\pi y)} - 1} = \frac{y^2}{(\ln(1+\frac{y}{2}))(\sin\pi y)}\frac{(\ln(1+\frac{y}{2}))(\sin\pi y)}{e^{(\ln(1+\frac{y}{2}))(\sin\pi y)} - 1} = \\
&= \frac{2}{\pi}\frac{y/2}{\ln(1+\frac{y}{2})}\frac{\pi y}{\sin\pi y}\frac{(\ln(1+\frac{y}{2}))(\sin\pi y)}{e^{(\ln(1+\frac{y}{2}))(\sin\pi y)} - 1}\end{align}
Assuming we know that $$\lim_{t\to 0}\frac{t}{\ln(1+t)} = \lim_{t\to 0} \frac{t}{\sin t} = \lim_{t\to 0}\frac{t}{e^t-1} = 1 $$ we get
$$ \lim_{y\to 0} \frac{y^2}{(1+\frac{y}{2})^{\sin\pi y} - 1} = \frac{2}{\pi}\cdot 1\cdot 1\cdot 1 = \frac{2}{\pi}$$
In total, we get
$$ \lim_{y\to 0} \frac{\log_2\sqrt{4-y^2}-\sqrt{\cos \pi y}}{(1+\frac{y}{2})^{\sin\pi y} - 1} = (-\frac{1}{8\ln 2}+\frac{\pi^2}{4}) \frac{2}{\pi}$$


问:
Can there exist a function that has a weak derivative, but is not differentiable almost everywhere?
I am currently learning about Sobolev spaces, and I am trying to build some intuition of weak derivatives. My current intuition is imagining the weak derivative of f as a function equal to f's derivative almost everywhere. However, this assumption assumes that f is differentiable almost everywhere. Is this always the case for weakly differentiable functions?</p>

答:
Fellow time travelers, allow me to highlight a good example which I found in the notes that Hass Saidane refers to:

The Heaviside step function is differentiable almost everywhere. However, it is not weakly differentiable.

[Edit: thus, we see that even for functions which are classically differentiable almost everywhere, OP's proposed intuition still isn't entirely air-tight].

Indeed, it can be said that $H'=h$ weakly if two conditions are met: First, (integrating against) $h$ is the distributional derivative of (integrating against) $H$. Second, $h$ lives in some appropriate $L^p$ space.

However, the distributional derivative of (integrating against) the Heaviside functional is the dirac-delta, in a sense which is very different from the technically legitimate $L^p$ function $\infty \cdot \chi_{\{0\}}$ (which is precisely the zero element of $L^p$).
Therefore, the distributional derivative of $H$ fails to be of the form "integrate against an $L^p$ function."

Good luck, fellow students.


问:
On finding determinantal representations of polynomials
Consider the ring $\mathbb{Q}[x_1,\ldots,x_n]$ of multivariate polynomials with rational coefficients. Given a positive integer $d$, let $\mathcal{C}_d \subset \mathbb{Q}[x_1,\ldots,x_n]$ be the class of polynomials $f$ which can be written as

$$f = \operatorname{det}(A)$$

where $A$ is a $d$-by-$d$ matrix with entries which are affine in the $x_1, \ldots, x_n$ with rational coefficients. It is of course possible to consider entries which are simply in $\mathbb{Q}[x_1,\ldots,x_n]$, but in the problem I am interested in, I think it is enough to consider entries that are affine in the $x_1, \ldots, x_n$.

If $f \in \mathcal{C}_d$, let us say that $f$ has a degree $d$ determinantal representation (with affine entries).

Question 1: concretely, how to devise an algorithm to tell whether or not a polynomial $f \in \mathbb{Q}[x_1,\ldots,x_n]$ of degree $d$ is actually in $\mathcal{C}_d$ or not?

Question 2: assuming there exists at least one such determinantal representation, corresponding to some $d$-by-$d$ matrix $A$, then there are infinitely many other representations, since for instance we could replace $A$ with $PAQ$, where $P,Q \in \mathrm{SL}(d,\mathbb{Q})$. Is there a way to concretely devise an algorithm that would give the "nicest" such representation? For example, if there is a representation $A$ with entries that are affine with integer coefficients, then this is "nicer" than a representation with rational coefficients which are not integers. It is ambiguous, I agree, but if someone could make this more precise, either in the comments or as an answer, then I would appreciate it.

答: One result from an encyclopedic book by Schinzel. He has his own Hypothesis. You seem to be talking about complete factorization, which is more restrictive, while the tests get simpler to describe.


问:
Prove that sum is finite
Let $j \in \mathbb{N}$. Set
$$
a_j^{(1)}=a_j:=\sum_{i=0}^j\frac{(-1)^{j-i}}{i!6^i(2(j-i)+1)!}
$$
and $a_j^{(l+1)}=\sum_{i=0}^ja_ia_{j-i}^{(l)}$.

Please help me to prove that the following sum is finite
$$
\sum_{j=1}^{\infty}j!\, a_j^{(l)}
$$

Thank you.

答:
Define the generating functions for $a_n^{(l)}$ like so
$$ g_l(z) = \sum_{n\geq0} a_n^{(l)}z^n$$
Then from the recurrence relations that you wrote down, it is easy to deduce that
$$g_l(z) = \alpha(z)^l \beta(z)^l, \qquad \alpha(z) = e^{z/6}, \qquad \beta(z) = \frac{\sin\sqrt{z}}{\sqrt{z}}. $$

Write the sum that you have in the form
$$ \sum_{n\geq0} a_n^{(l)}n! z^n = \sum_{n\geq0} \int_0^\infty dt\, e^{-t}(tz)^n a_n^{(l)} = \int_0^\infty e^{-t} g(tz). $$
If your sum converges absolutely, then this integral must converge also and have the same value, and if the integral converges (if it does it will converge absolutely), then the sum will converge. However, asymptotically
$$ g(tz) = \Theta(e^{t z l /6} (tz)^{-l/2}), $$
so the integral will have an exponential term of the form $e^{(-1+zl/6)t}$. For the integral to converge the real part of $-1+zl/6$ must be negative, so the sum will converge if
$$ \Re(z) \leq 6/l. $$

Setting $z=1$ to get your desired sum, it will converge absolutely when $l\leq 6$.

Title: Prove that sum is finite


问:
Gelfand transform relation between the spectrum of $ab$ and the spectra of $a$ and $b$
I want to have a better understanding of the Gelfand transform. This is a follow up regarding Julien's reply to the question on the spectrum of the sum of two commuting elements in a Banach algebra:

Assume that $A$ is commutative. Then by Gelfand, for every $x\in A$, we have
$$
\sigma(x)=\{\phi(x)\;;\;\phi\in \hat{A}\}
$$
where $\hat{A}$ denotes the set of characters (nonzero algebra homomorphisms from $A$ to $\mathbb{C}$).

It follows readily that for all $x,y\in A$:
$$
\sigma(x+y)\subseteq\sigma(x)+\sigma(y)\qquad \mbox{and}\qquad\sigma(xy)\subseteq\sigma(x)\sigma(y).
$$

My question is, why is clear from the above that we have

$$\sigma(x+y)\subseteq\sigma(x)+\sigma(y)\qquad  \text{and} \qquad\sigma(xy)\subseteq\sigma(x)\sigma(y)?$$

I understand that $\phi$ are homomorphisms, but where do the spectrums take place in the definition and how did we get "$\subseteq$"?

Thank you.

答:
If $\lambda\in \sigma (x+y)$,  then there is some $\varphi $ in $\hat A$, such that $\lambda =\varphi (x+y)$.  Therefore
$$
  \lambda =\varphi (x+y) = \varphi (x)+\varphi (y)\in  \sigma (x)+\sigma (y).
  $$
Similarly for $xy$.


问:
Sufficient criterion for convergence of series
Let $(a_k)$ be a sequence decreasing to $0$, for which
$$ b_k:=\left(\sum_{l=1}^ka_l\right)-ka_{k+1}=\sum_{l=1}^k(a_l-a_{k+1})$$
is bounded. Does $\sum_{k\geq 0}a_k$ necessarily converge?

Some immediate observations:

$(b_k)$ is increasing, hence converges.

By considering $b_k-b_{k-1}$, one obtains that $k(a_k-a_{k+1})$ tends towards $0$.

For $\sum a_k$ to converge, it suffices to prove that $ka_k$ converges.

答:
We know that $(b_k)_{k\in\mathbb N}$ is convergent. Let $b$ be its limit.

Let $p$ be a positive integer.

$\forall k \geqslant p \quad  , \quad  b_k = \displaystyle \sum_{\ell = 1}^k (a_{\ell}-a_{k+1})\geqslant \sum_{\ell =1}^p (a_{\ell}-a_{k+1})  $

Then $\quad  \lim b_k \displaystyle \geqslant \lim_{k\rightarrow +\infty} \sum_{\ell=1}^p (a_{\ell}-a_{k+1})$

So $\quad b \geqslant \displaystyle\sum_{\ell = 1}^p a_{\ell}   $

And we can conclude that $\sum a_k$ is convergent.


问:
Is $K_0(Q(B))=0$? $Q(B)=M(B\otimes\mathcal K)/B\otimes\mathcal K(\mathscr H)$,$M$ denotes the multiplier algebra.
$$M(A)=\{x\in A'':xA,Ax\subseteq A\}$$
$\mathscr H$ is an infinite dimensional separable Hilbert space and $\mathcal K$ is the compact operator algebra.

$$ Q(B)=M(B\otimes\mathcal K)/B\otimes\mathcal K
$$

I think I only need to show $K_0(M(B\otimes \mathcal K))=0$, since $Q(B)$ is a unital quotient algebra of $M(B\otimes \mathcal K$).

First suppose $p$ is a projection in $M(B\otimes \mathcal K)$. Let $f_1,f_2,...$ be orthogonal projections such that each of them is equivalent to $1_M=\text{diag}(1,1,...)$.

Then there is a projection $q\leq f_1$ which is equivalent with $p$. It sufficient to show $q\oplus 1_M\sim 1_M\oplus 0$ in $M_2(M(B\otimes\mathcal K))$.

Write $$u=\left(\begin{array}{}\left(\begin{array}{}q&&&\\ &0\\&&0\\ &&&...\end{array}\right)\left(\begin{array}{}1-q&&&\\q&1-q\\&q&1-q\\ &&&...\end{array}\right)\\\left(\begin{array}{}0&&&\\ &0\\&&0\\ &&&...\end{array}\right)\left(\begin{array}{}0&&&\\ &0\\&&0\\ &&&...\end{array}\right)\end{array}\right)
$$

whose real meaning is

$$u=\left(\begin{array}{}q&(f_1-q)+f_{21}q+f_{21}(f_1-q)f_{12}+f_{31}qf_{12}+f_{31}(f_1-q)f_{13}+...\\0&0\end{array}\right)$$

where $f_{ij}^*f_{ij}=f_i$ and $f_{ij}f_{ij}^*=f_j$.

Then by calculation, $u^*u=q\oplus 1_M$ and $uu^*=1_M\oplus 0$. This shows both $u\in M_2(M(B\otimes\mathcal K))$ (since it is bounded) and $q\oplus 1_M\sim 1_M\oplus 0$. Therefore $[q]=0$.

Generally for projection $p\in M_n(M(B\otimes\mathcal K))$, since $1_M\oplus 0\sim 1_{M}\oplus 1_M$, there is a projection $q\leq 1_M$ which is equivalent with $p$. Therefore $[p]=0$.

Thus $K_0(M(B\otimes\mathcal K))=0$ and so $K_0(M(B\otimes\mathcal K)/B\otimes\mathcal K)$ should be.

Am I right?

I am asking this question because the book I am reading is discussing properties of $K_0(Q(B))$ which seems $0$ to me.

答:
The fact that a C*-algebra has trivial $K_0$ does not mean that every quotient of it has the same property!  You are correct in saying that $K_0(M(K\otimes B))$ is zero, so you can plug this informtation in the six-term exact sequence for $$0  \to K\otimes B\to M(K\otimes B) \to M(K\otimes B)/K\otimes B\to0
$$
and conclude that
$$K_0(Q(B))=K_1(K\otimes B) = K_1(B).$$


问:
how can I find a orthogonal matrix Q such that Q$x_i=y_i$
How can I find a orthogonal matrix Q such that Q$x_i=y_i$ given that $\|x_i\|=\|y_i\|$ and $\|x_i-x_j\|= \|y_i-y_j\|$.
I know that Householder reflectors would do the trick here Q=$I-2pp^T$ where p is ($x_i-y_i$)/$\|x_i-y_i\|$.  but I am unable to understand how can someone please explain this to me.

Here $x_i , y_i \in R^n$

I also read this Rotation by Householder matrices solution but couldn't understand the how part.

答:
Assuming the vectors $x_i$ form a basis, let $Q$ be the unique linear transformation that maps $x_i$ to $y_i$.

We need to show that $Q$ is orthogonal, that is, respects the inner product.

But, by hypothesis the inner product of basis elements are preserved because if $i\ne j$
$$\langle x_i,x_j\rangle=-\frac12\big(\|x_i-x_j\|^2-\|x_i\|^2-\|x_j\|^2\big)\,.$$
if $i\ne j$.

If the given vectors $x_i$ are only linearly independent, then we can extend them, say by an orthonormal basis of $\big({\rm span}(x_i)_i\big)^\perp$ and similarly extend $y_i$ by an orthonormal basis of $\big({\rm span}(y_i)_i\big)^\perp$.


问:
Derivative Integral Problem
Preface: I am sorry if there is a blatently obvious error or problem with my logic, I am only just learning about the fundamental theorem of Calculus and have been stumped by this example problem and why it has the solution that it does.

I will first show my understanding and what I originally thought was the solution, the problem is:
$$\frac{d}{dx}\int _{a}^{f(x)} g(t)dt = ?$$
My solution:
$$ \text{let } h(x) = \int_{a}^{f(x)}  g(t)dt$$
By the chain rule:
$$ \frac{d}{dx}h(f(x) = h'(f(x)\cdot{}f'(x)$$
$$ h'(x) = d/df \int_a^{f(x)}g(t)dt $$
By the fundamental theorem of Calculus, the derivative and the integral should cancel correct?
$$\frac{d}{df} \int_a^{f(x)}g(t)dt = g(t) $$
Therefore:
$$\frac{d}{dx}\int _{a}^{f(x)} g(t)dt = g(t)\cdot{}f'(x)$$

This however is not the correct answer, I have been told it is really this:

$$\frac{d}{dx}\int _{a}^{f(x)} g(t)dt = g(f(x))\cdot{}f'(x)$$

Again, I apologize if there is some obvious error, Any and all help is appreciated thanks.

答:
okay lets define:
$$G(x)=\int_0^xg(t)dt$$
so the integral you have can be defined as:
$$\frac d{dx}\left[G(f(x))-G(a)\right]$$
Obviously if we take the derivative of a constant this disappears, so we just care about the first part:
$$\frac{d}{dx}G(f(x))=\frac{dG(f)}{df}\frac{df(x)}{dx}=G'(f(x))f'(x)$$
Now notice that by FTC:
$$G'(x)=g(x)$$
and so:
$$G'(f(x))f'(x)=g(f(x))f'(x)$$
so you got the correct result in the end there were just some steps to justify which were wrong

Title: Derivative Integral Problem


答:
Not an answer, but another way to calculate this integral:
\begin{align} \int_0^\infty \frac{e^{-2x}-e^{-x}}{x}dx &= \lim_{M\to\infty}  \int_0^M \frac{e^{-2x}-e^{-x}}{x}dx = \\
&= \lim_{M\to\infty} \Big( \int_0^M \frac{e^{-2x}-1}{x}dx - \int_0^M \frac{e^{-x}-1}{x} dx \Big) = \\
&= \lim_{M\to\infty} \Big( \int_0^{2M} \frac{e^{-y}-1}{y}dy - \int_0^M \frac{e^{-x}-1}{x} dx \Big) = \\
&= \lim_{M\to\infty} \int_M^{2M} \frac{e^{-x}-1}{x}dx = \\
&= \lim_{M\to\infty} \Big(\int_M^{2M} \frac{e^{-x}}{x}dx -\int_M^{2M}\frac{1}{x}dx \Big) = \\
&= \lim_{M\to\infty} \Big(\int_M^{\infty} \frac{e^{-x}}{x}dx - \int_{2M}^{\infty} \frac{e^{-x}}{x}dx - \ln 2 \Big) = \\
&= -\ln 2\end{align}
where we used the fact that $\int_M^{\infty} \frac{e^{-x}}{x}dx$ is convergent, so $ \lim_{M\to\infty} \int_M^{\infty} \frac{e^{-x}}{x}dx = 0$.

Title: how do I justify this $\int_{0}^{\infty}\frac {e^{-2x}-e^{-x}}{x}dx=\lim_ {s\to1^{-}} \int_{0}^{\infty}\frac{e^{-2x}-e^{-x}}{x^s}dx$

Body:
I was trying to compute
$\displaystyle\int_{0}^{\infty}
\frac{{\rm e}^{-2x} - {\rm e}^{-x}}{x}\,{\rm d}x$ when I had an idea, the gamma function is $\Gamma(1+z)=\int_{0}^{\infty}x^ze^{-x}dx$ and for $s<1$ the integral $\int_{0}^{\infty}\frac{e^{-2x}-e^{-x}}{x^s}dx$
converges, so call
$$I(s)=\int_{0}^{\infty}\frac{e^{-2x}-e^{-x}}{x^s}dx$$
and hope that it is continuous from the left at $s=1$.

Assuming that it is, for $s<1$ we have
$$I(s)=(2^{s-1}-1)\Gamma(1-s)$$
and so:
$$I(1)=\int_{0}^{\infty}\frac {e^{-2x}-e^{-x}}{x}dx = \lim_{s\to1^-}I(s)= \lim_{s\to1^-}(2^{s-1}-1)\Gamma(1-s)$$

Now, write $$2^{s-1}-1=1+\ln(2)(s-1)+o((s-1)^2)-1=\ln(2)(s-1)+o((s-1)^2)$$
and so:
$$\lim_{s\to1^-}(2^{s-1}-1)\Gamma(1-s) = -\ln(2)\lim_{s\to 1^-}(1-s)\Gamma(1-s)+\lim_{s\to 1^-}o((1-s)^2)\Gamma(1-s)$$

Here comes the cool part, use the reflection formula ($\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$) and obtain:
$$(1-s)\Gamma(1-s) = \frac {\pi (1-s)}{\sin(\pi s)\Gamma(s)}$$
and:
$$o((1-s)^2)\Gamma(1-s) = o(1-s)\frac {\pi (1-s)}{\sin(\pi s)\Gamma(s)}$$

Now, $\Gamma$ is continuous at 1 and $\Gamma(1)=1$ so:
$$\lim_{s\to 1^-}\frac {\pi (1-s)}{\sin(\pi s)\Gamma(s)} = \lim_{s\to 1^-}\frac {\pi (1-s)}{\sin(\pi s)}$$

this limit is finite and is equal to 1 (with l'hopital) and so
$$\lim_{s\to 1^-}o((1-s)^2)\Gamma(1-s) = 0$$

And we can finally finish up and say $I(1)=-\ln(2)$.
Now, I don't know how to justify the fact that $I(s)$ is continuous at $s=1$ from the left and so I don't know if my solution is correct.

BUT! the answer is correct which gives me hope that this solution is correct with some additional justifications about the continuity of $I$ at $s=1^-$ which is why I am looking for help.


问:
continuation of integer-valued solution
I have a recurrence relation that is given by

$$f(n) = 2af(n-1)-a,\quad f(1) = a.$$

The solution given in Wolfram alpha is

$$f(n) = \frac{(a-1)2^na^n+a}{2a-1}.$$

(This is not a homework problem).  My question is, supposing that I want to determine the continuous solution $f(x)$, can I find it by knowing $f(n)$ - it's values at the positive integers?  If so, any suggestions on how to do this?

答:
Your question is about a function $\,f(x)\,$ that satisfies
$$f(n) = 2af(n-1)-a\quad \forall n\in\mathbb{Z},\quad f(1) = a. \tag{1} $$
You asked about continuous solutions to equation $(1)$. Let
$\,F(x)\,$ be any continuous function on the interval $\,[0,1]\,$
with $\,F(0)=1,\; F(1)=a.\,$ Use the recurrence relation
$$ F(x) = 2a\,F(x-1)-a\quad \forall x\in\mathbb{R} \tag{2} $$ to extend the function
uniquely and continuously to all reals. Verify using induction that
$$ f(n) = F(n) \quad \forall n\in\mathbb{Z}. \tag{3} $$

Note that this method does not give an explicit formula for any of the
continuous solutions, because allowing $\,F(x)\,$ to be any
continuous function on the interval $\,[0,1]\,$ with
$\,F(0)=1,\,$ $\,F(1)=a.\,$makes this impossible. Given
a particular solution to equation $(1)$ such as
$$ f(n) := \frac{(a-1)(2a)^n+a}{2a-1} \tag{4} $$
it is easy to extend this to all real numbers if $\,a>0\,$ because
$\,(2a)^n\,$ simply becomes $\,(2a)^x.$ In general, it is not possible
to get a unique solution. For a simple example, $\,f(n)=f(n-1)\,$ has
period $1$ functions as solutions and there are very many such functions.

Title: continuation of integer-valued solution


答:
We can prove that $f(s)=\int_0^{\infty} (e^{-x}-e^{-2x})x^{-s} dx$ is analytic on $\Re s \in [\epsilon,2)$ for any $0<\epsilon<1$. Let $g(x)=-e^{-2x}/2+e^{-x}-1/2$ and $g'(x)=e^{-2x}-e^{-x}$. Since $|g(x)|\leq c\min(1,x^2)$ for an absolute constant $c>0$, the integration by parts gives
$$\begin{align}
f(s)&=g(x) x^{-s} \bigg\vert_0^{\infty} + s\int_0^{\infty} g(x) x^{-s-1} dx\\
&=s\int_0^{\infty}g(x) x^{-s-1} dx.\end{align}
$$
That is, $f(s)$ is analytic on $\Re s \in (0,2)$. Then, by your method, we have
$$
f(s)=(2^{s-1}-1)\Gamma(1-s)  \mathrm{ \ if \ }  s \in (0,1).
$$

As $(2^{s-1}-1)\Gamma(1-s)$ is meromorphic with only a removable singularity at $s=1$, we must have $f(s)=(2^{s-1}-1)\Gamma(1-s)$ if $\Re s\in (0,2)$ by analytic continuation. In particular, we have
$$
f(1)=\lim_{s\rightarrow 1} (2^{s-1}-1)\Gamma(1-s)=-\log 2.
$$
Here, $\log=\ln$.

Moreover, $f(s)=\int_0^{\infty} (e^{-x}-e^{-2x})x^{-s} dx$ is analytic on $\Re s \in (-\infty, 1)$. This is because $|e^{-x}-e^{-2x}|\leq c \min(x, e^{-x})$ for some absolute constant $c>0$.

Combining these, we can conclude that $f(s)=\int_0^{\infty} (e^{-x}-e^{-2x})x^{-s} dx$ is analytic on $\Re s \in (-\infty, 2)$, and
$$
f(s)=\int_0^{\infty} (e^{-x}-e^{-2x})x^{-s} dx= (2^{s-1}-1)\Gamma(1-s)
$$
for  $\Re s \in (-\infty, 2)$.


问:
Proof of the Farey sequence
How would I go about proving that the Farey sequence contains all possible numbers? I'm new to number theory, so from observation the sequence looks to have all fractions. But I cannot figure out a proof that any fraction can be created with the fact that numbers in the sequence. I am using the same Farey method where you take fractions $m/n$ and $p/q$ and add them to have $\frac{m+p}{n+q}$

答:
Suppose we construct a list of sequences of rational numbers (and a fake rational number number $\frac10$ which we think of as $+\infty$) by

starting with the sequence $\frac01, \frac10$;

taking the previous sequence, and inserting $\frac{a+c}{b+d}$ between every pair of consecutive terms $\frac ab$, $\frac cd$.

Then you want to show that every positive rational number $\frac pq > 0$ will appear eventually.

To do this, we'll need two facts:

Fact 1. The fractions in each sequence appear in increasing order.

To show this, check that if $\frac ab < \frac cd$, then $\frac ab < \frac{a+c}{b+d} < \frac cd$, so the order property is maintained as we go from each sequence to the next.

Fact 2. Consecutive fractions $\frac ab, \frac cd$ in any sequence satisfy $\frac cd - \frac ab = \frac1{bd}$.

To show this, check that if it's true, then $\frac{a+c}{b+d} - \frac ab = \frac1{b(b+d)}$ and $\frac cd - \frac{a+c}{b+d} = \frac1{(b+d)d}$, so this property is also preserved as we go from each sequence to the next.

Now, let's prove that every fraction $\frac pq$ will always be found.

Initially, $\frac pq$ is somewhere between $\frac01$ and $\frac10$. At each step, we'll keep track of the two consecutive terms $\frac ab , \frac cd$ such that $\frac ab < \frac pq < \frac cd$.

As long as this is true, we have
$$
   \frac1{bd} = \frac cd - \frac ab = \left(\frac cd - \frac pq\right) + \left(\frac pq - \frac ab\right) \ge \frac1{dq} + \frac1{qb},
$$
and by multiplying through by $bdq$, we get $q \ge b+d$. So the sum of the denominators of the fractions bracketing $\frac pq$ can never exceed $q$.

Every time we go from a sequence to the next sequence and insert $\frac{a+b}{c+d}$ between $\frac ab$ and $\frac cd$, either $\frac{a+b}{c+d} = \frac pq$ (and we are done), or $\frac ab < \frac pq < \frac{a+b}{c+d}$ (and we have our next two consecutive fractions between which $\frac pq$ must lie), or $\frac {a+b}{c+d} < \frac pq < \frac cd$ (and we have our next two consecutive fractions between which $\frac pq$ must lie).

The sum of the denominators of the fractions bracketing $\frac pq$ always keeps increasing, but it can't exceed $q$, so there can be only finitely many iterations before we stop.

Title: Proof of the Farey sequence


问:
Show that the bijection $\text{Cat}(A \times B, C) \cong \text{Cat}(A, C^B)$ is natural
This is exercise 2.5.1 in Categories for the Working Mathematician, and it's been asked here a few times but I've never seen the part that confuses me clarified.

So we wish to create a bijection $\phi : \text{Cat}(A \times B, C) \rightarrow \text{Cat}(A, C^B)$. These are both functor categories, so it has elements of the form $(F:A \times B \rightarrow C) \rightarrow (T:A \rightarrow C^B)$.

We can define a bijection $\phi(F)(a)(b) = F(a, b)$, with $\phi^{-1}(T)(a, b) = T(a)(b)$, but my issue is when showing that it is natural.

Since we are dealing with functor categories, the morphisms are natural transformations $\eta : F \rightarrow F'$ and $\mu : T \rightarrow T'$, what is the explicit mapping here?

答:
For simplicity, let's just fix $B$ and $C$ and focus on naturality in $A.$ If we want to look at this in terms of functors, the functors in question are $\operatorname{Cat}(-\times B,C)$ and $\operatorname{Cat}(-, C^B)$ and we want to show that $\phi$ is an arrow between these two functors in the appropriate category (in this case $\mathrm{Set}^{\mathrm{Cat}^{\mathrm{op}}}$), which means we want to show that $\phi_A:\operatorname{Cat}(A\times B,C) \to \operatorname{Cat}(A, C^B)$ can be regarded as the $A$-th component of a natural transformation between the functors $\operatorname{Cat}(-\times B,C)$ and $\operatorname{Cat}(-, C^B).$

Given an arrow $g:A'\to A,$ the first functor takes an arrow $f:A\times B\to C$ to $f\circ (g\times id_{B})$ and given an arrow $h: A\to C^B,$ the second functor takes $h$ to $h\circ g.$ So showing naturality means showing that for any $f:A\times B \to C,$ $\phi_{A'}(f\circ (g\times id_B)) = \phi_A(f)\circ g.$ We can verify this by looking at how each side acts on some $a\in A'$ (either an object or arrow.. no matter). The left-hand side is $$ \phi_{A'}(f\circ (g\times id_B)) (a) = f\circ (g\times id_B)(a,-) = f(g(a), -)$$ and the right-hand side is $$ (\phi_A(f)\circ g) (a) = \phi_A(f)(g(a)) = f(g(a),-),$$ which are the same, so we're done.

Note that for most of the formality here, it was completely irrelevant that the arrows in question were functors. For everything up till the verification of the equation at the end, we may as well have replaced Cat with Set or any other cartesian closed category. Only in verifying the equation did we actually think about what the category actually was, and even then not really... we could have done that step just in terms of the eval arrow, etc, it just might have looked a little more obscure.

Naturality in $B$ and $C$ can be shown similarly. It can be tediously shown that in general, verifying each component's naturality separately is all that is required to show the whole thing is natural, i.e. that $\phi$ is an arrow in the functor category $\operatorname{Set}^{\operatorname{Cat}^{\mathrm{op}}\times \operatorname{Cat}^{\mathrm{op}}\times \operatorname{Cat}}.$ (And it can also be shown that all it takes to be an isomorphism in said category is for each component of the natural transformation to be an isomorphism.)


问:
Prove inequality $\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right) \tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le\frac13 $
How to determine the range of the function

$$f(x)=\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)
$$

It it is straightforward to verify that $f(x)$ is even and

$$f(0)= \frac13, \>\>\>\>\> \lim_{x\to\pm \infty} f(x) \to -\infty$$

which implies $f(x) \in (-\infty,\frac13]$, i.e.

$$\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le \frac13
$$

and is visually confirmed below

However, it is not obvious algebraically that $f(x)$ monotonically decreases away from $x=0$. The standard derivative tests are not viable due to their rather complicated functional forms.

So, the question is how to prove the inequality  $f(x) \le \frac13$ with rigor.

Note that it is equivalent to proving

$$\cot \left(\frac{\pi(1+x)}{3+x^2}\right)
\cot \left(\frac{\pi(1-x)}{3+x^2}\right)\le \frac13
$$

答:
Note: $f(1) \triangleq
\lim_{x\to 1} \tan ( \frac\pi2 \frac{(1+x)^2}{3+x^2})
\tan ( \frac\pi2 \frac{(1-x)^2}{3+x^2}) = 0$.

Since $f(x)$ is even, we only need to prove the case $x \in [0, \infty)$.
Clearly $f(x) < 0$ for all $x > 1$. Also, $f(1) = 0$.
Thus, we only need to prove the case $0\le x < 1$.

(Inspired by @pisoir's 1st equation) Let $A =  \frac\pi2 \frac{(1+x)^2}{3+x^2}, B = \frac\pi2 \frac{(1-x)^2}{3+x^2}$.
Clearly $A, B \in (0, \frac{\pi}{2})$.
It suffices to prove that $\sin A \sin B \le \frac{1}{3}\cos A \cos B$ or
$$\frac{1}{2}[\cos (A - B) - \cos ( A + B)] \le \frac{1}{3}\cdot \frac{1}{2}[\cos (A - B) + \cos (A + B)]$$
or
$$\cos \frac{2\pi x}{x^2 + 3} \le 2 \cos \frac{\pi (x^2 + 1)}{x^2 + 3}. $$

Let $C = \frac{2\pi x}{x^2 + 3}$ and $D = \frac{\pi (x^2 + 1)}{x^2 + 3}$.

We split into two cases:

$\frac{1}{3} \le x < 1$:

Clearly, $\frac{\pi}{3} \le D \le \frac{\pi}{2}$ and $0 \le \frac{\pi}{2} - C \le 2(\frac{\pi}{2} - D) \le \frac{\pi}{2}$. We have
\begin{align}
\cos C &= \sin (\tfrac{\pi}{2} - C) \\
&\le \sin [2(\tfrac{\pi}{2} - D)]\\
&= 2\sin (\tfrac{\pi}{2} - D) \cos (\tfrac{\pi}{2} - D) \\
&\le 2\sin (\tfrac{\pi}{2} - D) \\
&= 2\cos D.
\end{align}

$0 \le x < \frac{1}{3}$:

We give the following auxiliary results (Facts 1-2). The proofs are easy and thus omitted.

Fact 1: $\cos u \le 1 - \frac{1}{3}u^2$ for all $u$ in $[0, \pi/4]$.

Fact 2: $\cos v \ge \frac{1}{2} - \frac{5}{9}\sqrt{3}\, (v - \tfrac{\pi}{3})$ for all $v \in [\pi/3, \pi/2]$.

Let us proceed. Clearly $C \in [0, \pi/4]$ and $D \in [\pi/3, \pi/2]$.
By Facts 1-2, it suffices to prove that
$$1 - \frac{1}{3}C^2 \le
2\left[\frac{1}{2} - \frac{5}{9}\sqrt{3}\, \left(D - \frac{\pi}{3}\right)
\right]$$
that is
$$\frac{4\pi x^2 [9\pi - 5\sqrt{3}(x^2 + 3)]}{27(x^2 + 3)^2}\ge 0$$
which is true.

We are done.


问:
Noncentral Wishart Expected Value - Solving a Matrix Integral
Let $\mathbf{V} \sim ncWish\left(\nu_1, \, \mathbf{\Sigma}, \, \frac{\nu_1}{\nu_2-p-1} \mathbf{\Theta}\right)$ follow a noncentral Wishart distribution according to Theorem 3.5.1. in Gupta, Nagar - Matrix Variate Distributions.

What is
$$
\mathbf{E}\left[ \mathbf{V} |\mathbf{V}|^{\frac{\nu_2}{2}} \right] = \;?
$$

To answer this question we need to find a closed form solution to the following integral, where the part in blue is the kernel density of the noncentral Wishart distribution
$$
\int_{\mathbf{V}>\mathbf{O}} \mathbf{V} \, |\mathbf{V}|^\frac{\nu_2}{2} \color{blue}{ \exp\left(-\frac{1}{2}\text{tr}(\mathbf{\Sigma}^{-1}\mathbf{V})\right) |\mathbf{V}|^{\frac{\nu_1-p-1}{2}} {}_0F_1\left(\frac{\nu_1}{2}; \frac{1}{4}\frac{\nu_1}{\nu_2-p-1} \mathbf{\Theta} \mathbf{\Sigma}^{-1} \mathbf{V} \right)} \mbox{d} \mathbf{V},
$$
where $\nu_1$ and $\nu_2$ are scalars, $\mathbf{\Theta}$ is a real matrix, $\mathbf{V}$ and $\mathbf{\Sigma}$ are real, symmetric, positive definite matrices and ${}_0F_1$ is the hypergeometric function of a matrix argument.

答:
I suppose that all matrices in question are $p \times p$.  It is clear that no closed-form result for the integral can be obtained if $p=1$, so it is doubtful that such an expression exists for general $p$, especially since there is no such expression for the generalized hypergeometric function ${}_0F_1$ of matrix argument.

Next, I note that in multivariate statistical analysis, where this integral was derived, the notation $\color{blue}{\mathbf{\Theta\Sigma^{-1}}}$ is shorthand for a symmetric matrix $\mathbf{M}$ that has the same eigenvalues as the matrix $\color{blue}{\mathbf{\Theta\Sigma^{-1}}}$.

I provide an approach to the integral that is derived from the paper, ``Integral transform methods in goodness-of-fit testing, II: The Wishart distributions,'' Ann. Inst. Statist. Math. 72 (2020), 1317-1370; see p. 1328, Proposition 2 in that paper.

Let us rewrite the integral in the form
$$
f(\mathbf{M}_1,\mathbf{M}_2) := \int_{\mathbf{V}>\mathbf{O}} \mathbf{V} \, |\mathbf{V}|^\alpha  \exp(-{\rm{tr}} \, \mathbf{M}_1^{-1}\mathbf{V}) \, {}_0F_1(\beta; \mathbf{M}_2 \mathbf{V}) \frac{\mbox{d}\mathbf{V}}{|\mathbf{V}|^{(p+1)/2}},
$$
where $\alpha$, $\beta$, $\mathbf{M}_1$, and $\mathbf{M}_2$ are trivially expressible in terms of your original notation, e.g., $\mathbf{M}_1 = 2 \mathbf{\Sigma}$, etc.  With this notation, $\mathbf{M}_1$ is a positive definite (and symmetric) matrix, and $\mathbf{M}_2$ is a symmetric matrix.

Make the substitution $\mathbf{V} \to \mathbf{M}_1^{1/2} \mathbf{V} \mathbf{M}_1^{1/2}$.  It is well-known that the measure $\mbox{d}\mathbf{V}/|\mathbf{V}|^{(p+1)/2}$ is invariant under this substitution; so, after simplification, we obtain
\begin{align*}
f(\mathbf{M}_1,\mathbf{M}_2) = |\mathbf{M}_1|^\alpha \mathbf{M}_1^{1/2} \int_{\mathbf{V}>\mathbf{O}} \mathbf{V} |\mathbf{V}|^\alpha  \exp(-{\rm{tr}} \, \mathbf{V}) \, {}_0F_1\left(\beta; \mathbf{M}_1^{1/2} \mathbf{M}_2 \mathbf{M}_1^{1/2} \mathbf{V} \right) \frac{\mbox{d}\mathbf{V}}{|\mathbf{V}|^{(p+1)/2}} \mathbf{M}_1^{1/2}.
\end{align*}
That is,
$$
f(\mathbf{M}_1,\mathbf{M}_2) = |\mathbf{M}_1|^\alpha \mathbf{M}_1^{1/2} f(\mathbf{I}_p,\mathbf{M}) \mathbf{M}_1^{1/2},
$$
where $\mathbf{I}_p$ denotes the $p \times p$ identity matrix, and $\mathbf{M} = \mathbf{M}_1^{1/2} \mathbf{M}_2 \mathbf{M}_1^{1/2} $.

Now consider the integral,
$$
g(\mathbf{M}) := f(\mathbf{I}_p,\mathbf{M}) = \int_{\mathbf{V}>\mathbf{O}} \mathbf{V} \, |\mathbf{V}|^\alpha  \exp(-{\rm{tr}} \, \mathbf{V}) \, {}_0F_1(\beta; \mathbf{M} \mathbf{V}) \frac{\mbox{d}\mathbf{V}}{|\mathbf{V}|^{(p+1)/2}},
$$
where $\mathbf{M}$ is any $p \times p$ symmetric matrix.  Observe that $g(\mathbf{M})$ is a symmetric $p \times p$ \textit{matrix}, each of whose entries is an integral.  That the integral defining $g(\mathbf{M})$ converges absolutely for all $\alpha > (p-1)/2$ and all $\mathbf{M}$ can be proved using the Poisson integral for the Bessel functions of matrix argument; see Herz (Ann. Math., 61 (1955), 474--523).

Denote by $O(p)$ the group of all $p \times p$ orthogonal matrices.  Note that if $\mathbf{H} \in O(p)$ then
$$
g(\mathbf{H} \mathbf{M} \mathbf{H}^{-1}) = \int_{\mathbf{V}>\mathbf{O}} \mathbf{V} \, |\mathbf{V}|^\alpha \, \exp(-{\rm{tr}} \, \mathbf{V}) \, {}_0F_1\left(\beta; \mathbf{H} \mathbf{M} \mathbf{H}^{-1} \mathbf{V} \right) \frac{\mbox{d}\mathbf{V}}{|\mathbf{V}|^{(p+1)/2}},
$$
Making the substitution $\mathbf{V} \to \mathbf{H} \mathbf{V} \mathbf{H}^{-1}$, using the invariance of the measure $\mbox{d}\mathbf{V}/|\mathbf{V}|^{(p+1)/2}$, and simplifying, we obtain
$$
g(\mathbf{H} \mathbf{M} \mathbf{H}^{-1}) = \mathbf{H} \int_{\mathbf{V}>\mathbf{O}} \mathbf{V} |\mathbf{V}|^\alpha  \exp(-{\rm{tr}} \, \mathbf{V}) \, {}_0F_1(\beta; \mathbf{M} \mathbf{V}) \frac{\mbox{d}\mathbf{V}}{|\mathbf{V}|^{(p+1)/2}} \mathbf{H}^{-1}.
$$
In short, we have shown that, for all $p \times p$ symmetric matrices $\mathbf{M}$ and all $\mathbf{H} \in O(p)$,
$$
g(\mathbf{H} \,\mathbf{M} \,\mathbf{H}^{-1}) = \mathbf{H} \,g(\mathbf{M}) \,\mathbf{H}^{-1}.
$$
This property has appeared earlier in the 1970's and 1980's papers of K. I. Gross and R. A. Kunze on generalized Bessel functions of matrix argument.  Gross and Kunze called it an orthogonal ``covariance'' property to distinguish it from the property of invariance under $O(p)$.

Since every symmetric matrix can be diagonalized by a transformation of the form $\mathbf{M} \to \mathbf{H} \mathbf{M} \mathbf{H}^{-1}$, the covariance property also shows that, in calculating $g(\mathbf{M})$, it suffices to assume that $\mathbf{M}$ is diagonal.

The above covariance property is the best that one can do for general $p$.  For $p=2$, I suggest that you try to evaluate $g(\mathbf{M})$ by using an explicit formula for the ${}_0F_1$ function when $p=2$ (see Muirhead's book, \textit{Aspects of Multivariate Statistical Theory}, Wile, New York, 1982) and then calculating each entry of $g(\mathbf{M})$ term-by-term.

For the case in which $\mathbf{M} = \mathbf{I}_p$ (or a multiple of the identity), there is some hope that the matrix $g(\mathbf{M})$ can be calculated explicitly for general $p$.  For $\mathbf{M} = \mathbf{I}_p$, the covariance property reduces to
$$
g(\mathbf{I}_p) = \mathbf{H} \, g(\mathbf{I}_p) \, \mathbf{H}^{-1}
$$
for all $\mathbf{H} \in O(p)$.  By Schur's lemma, it follows that $g(\mathbf{I}_p) = c \, \mathbf{I}_p$ for some constant $c$.  Therefore, all off-diagonal entries of $g(\mathbf{I}_p)$ are equal to zero.  As for the diagonal entries, by taking traces, we obtain
\begin{align*}
c p = {\rm{tr}} \,(c \,\mathbf{I}_p)
&= {\rm{tr}} \, g(\mathbf{I}_p) \\
&= {\rm{tr}} \, \int_{\mathbf{V}>\mathbf{O}} \mathbf{V} \, |\mathbf{V}|^\alpha \, \exp(-{\rm{tr}} \, \mathbf{V}) \, {}_0F_1(\beta; \mathbf{V}) \frac{\mbox{d}\mathbf{V}}{|\mathbf{V}|^{(p+1)/2}} \\
&= \int_{\mathbf{V}>\mathbf{O}} ({\rm{tr}} \, \mathbf{V}) \, |\mathbf{V}|^\alpha \, \exp(-{\rm{tr}} \, \mathbf{V}) \, {}_0F_1(\beta; \mathbf{V}) \frac{\mbox{d}\mathbf{V}}{|\mathbf{V}|^{(p+1)/2}} \\
&= - \frac{\partial}{\partial t} \int_{\mathbf{V}>\mathbf{O}}|\mathbf{V}|^\alpha \, \exp(- t \, {\rm{tr}} \, \mathbf{V}) \, {}_0F_1(\beta; \mathbf{V}) \frac{\mbox{d}\mathbf{V}}{|\mathbf{V}|^{(p+1)/2}}\Bigg|_{t=1}.
\end{align*}
Expanding the function ${}_0F_1(\beta; \mathbf{V})$ in a series of zonal polynomials $Z_\kappa$, integrating term-by-term using a \href{https://dlmf.nist.gov/35.4.E8}{formula} for the Laplace transform of $Z_\kappa$, and then differentiating term-by-term, one obtains a final result in terms of an infinite series of zonal polynomials.


答:
Not directly related to the way you solved the integral but I thought you might be interested in the Frullani integral which states:
$$\int_0^\infty\frac{f(ax)-f(bx)}{x}dx=\left(f(\infty)-f(0)\right)\ln\frac{a}{b}$$

and in your case:
$$f(x)=e^{-x},a=2,b=1$$
and so:
$$I=\left(0-1\right)\ln\frac{2}{1}=-\ln(2)$$
Hope this helps!


问:
What’s the ratio between the sums of areas of these circles and triangles?
Start with a unit circle. First inscribe it with an equilateral triangle, then  inscribe the equilateral triangle with a smaller circle, repeat the $2$ steps $99$ times, we end up with $101$ circles and $100$ triangles. What’s the ratio between the sums of areas of these circles and triangles?

With 30$^o$ right triangle, I can see the side of the first triangle is $\sqrt{3}$, the radius  of the second circle is 1/2 of the very first one, then?

答: Because of the similarity, the ratios between the shapes stays the same.  If the first triangle has side $\sqrt 3$ and the second circle has radius $\frac 12$, the second triangle has side $\frac 12 \sqrt 3$ and the third circle has radius $\frac 14$.  The areas of either the circles or triangles form a geometric series with ratio $\frac 14$ as the areas scale as the square of the side.


问:
Dice probability, Exemple, having at least three 5 or more, on seven dice throw.
I'm having trouble calculating probability on dice rolls, here's the setup :

(All dices are fair 6 faces classic dice)

It's a D&D Roleplaying kind of game, the rules say this :

For each action, you have a stat, and a skill assigned.
To fight someone on melee, that could be strenght as a stat, and brawling as a skill.

If you have 5 points on strenght, that mean you can throw 5 dices.
Then, let's say you have 4 points on brawl skill, that mean you must land a 4 or higher on a dice to validate it.

Then, the Game Master will set a goal.
If he set it a 3, you must have at least 3 dice that show a 4 or higher to achieve your action.

Get it ?

Exemple :

Me : "I have 5 in strenght and 4 in brawl, I'll punch this guy in the face"

GM : "Ok, but he's skilled at combat, so the goal is at 3"

Then, according to the rules, I can throw 5 dices, and need to have at least 3 showing 4 or more to succesfully punch the said guy.

Here's where I landed thus far :

If I want at least a 4, with five dice, I can apply this formula :

The chance of failing are (3/6)^5

So the chance of succes are 1 - ((3/6)^5), or 0,96875

According to the game rules, this mean I have a goal set to 1.

But if I have a goal set to 3 ?

Is the formula (1 - ((3/6)^5))^3 ? wich shall give 0,9091491699

Or I am missing something ?

I am working on an easy excel, so the GM can more easily calculate the chances of success before setting the goal of every action player might do.

答:
Let's suppose you have $d$ fair six-sided dice, with a target of at least $t$ showing on at least $n$ of the dice.

The probability of this is $$\sum\limits_{j=n}^d {d \choose j}\frac{(6-t+1)^j(t-1)^{d-j}}{6^d}$$

In Excel, you can find ${d \choose j}$ as =COMBIN(d,j)

So in your example of $d=5,t=4,n=3$ you have $10\times \frac{3^3 3^2}{6^5}+5\times \frac{3^4 3^1}{6^5}+1\times \frac{3^5 3^0}{6^5}$ which simplifies to $\frac12=0.5$, as you might expect since a die has probability of $\frac12$ of being $4$ or more, and you want at least half of an odd number of dice to achieve this.


问:
Homeomorphism from $P^2\mathbb{R}$ onto the image of $\mathbb{S}^2$ through the Veronese map
Consider the map $v \colon \mathbb{S}^2 \to \mathbb{R}^6$ defined by
$$ v(x, y, z) = (x^2, y^2, z^2, \sqrt{2}xy, \sqrt{2}yz, \sqrt{2}zx) $$
Furthermore, let $ M := v(\mathbb{S}^2) $ and $P^2\mathbb{R}$ denote the projective plane.

I want to show that the map $v$ defines a homeomorphism $V$ from $P^2\mathbb{R}$ onto $M$.

Well, I know that $P^2\mathbb{R} \cong {\mathbb{S}^2}/{\sim}$ where $(x, y, z) \sim (-x, -y, -z)$. If I somehow manage to prove that $M \cong {\mathbb{S}^2}/{\sim}$, it'd be neat.

答:
No idea why you are complicating this way. $v$ is continuous bijective $\Bbb{S^2/\pm 1}\to v(\Bbb{S}^2)$ and $w:[x:y:z]\to \pm (\frac{x}{\sqrt{x^2+y^2+z^2}},\frac{y}{\sqrt{x^2+y^2+z^2}},\frac{z}{\sqrt{x^2+y^2+z^2}})$ is continuous bijective $\Bbb{P^2(R)\to S^2/\pm 1}$. The composition of both is your homeomorphism:

The inverse of $w$ is obviously continuous, for the inverse of $v$ I would say that $v^{-1}$ gives a continuous function $v(\Bbb{S}^2)\to \Bbb{S^2}/(\pm 1,\pm 1,\pm 1), (a,b,c,d,e,f)\to (\pm\sqrt{a},\pm\sqrt{b},\pm\sqrt{c})$ which implies that $v^{-1}$ is itself continuous.


答: Obviously $v(x,y,z) = v(-x,-y,-z)$. Thus $v$ induces $V : P^2\mathbb R\to \mathbb R^6$  such that $V(P^2\mathbb R ) = M$. Now let us explore what $v(x,y,z) = v(x',y',z')$ means. The first three coordinates show that $x' = (-1)^a x, y' = (-1)^b y, z'= (-1)^c z$ with $a, b, c \in \{0,1\}$. Coordinate no. 4 shows $a = b$, coordinate no. 5 shows $b = c$ and coordinate no. 6 shows $a = c$. Thus $(x,y,z) = \pm (x',y',z')$, i.e. $V$ is injective. Hence $V : P^2\mathbb R \to M$ is a continuous bijection. Since $P^2\mathbb R$ is compact and $M$ is Hausdorff, the map $V$ is a homeomorphism.

Title: Homeomorphism from $P^2\mathbb{R}$ onto the image of $\mathbb{S}^2$ through the Veronese map

Body:
Consider the map $v \colon \mathbb{S}^2 \to \mathbb{R}^6$ defined by
$$ v(x, y, z) = (x^2, y^2, z^2, \sqrt{2}xy, \sqrt{2}yz, \sqrt{2}zx) $$
Furthermore, let $ M := v(\mathbb{S}^2) $ and $P^2\mathbb{R}$ denote the projective plane.

I want to show that the map $v$ defines a homeomorphism $V$ from $P^2\mathbb{R}$ onto $M$.

Well, I know that $P^2\mathbb{R} \cong {\mathbb{S}^2}/{\sim}$ where $(x, y, z) \sim (-x, -y, -z)$. If I somehow manage to prove that $M \cong {\mathbb{S}^2}/{\sim}$, it'd be neat.


答:
When you want to count the number of successes of independent trials where the probability of success is all the same, reach for a Binomial random variable. Your step $(3/6)^5$ as the chance of failure is not quite right. Each roll succeeds or fails with a probability of 1/2. So $(3/6)^5$ is the probability of one particular sequence of 5 rolls, such as $SSFFS$ or $FSFSF$. The question is: in how many of these possible sequences do we exceed the goal set by the GM?

Let $X$ be the random variable for the number of trials out of 5 that succeed (roll a 4 or higher). Then $X \sim \text{Bin}(5, 1/2)$.

The probability that 3 or more of the 5 trials succeed is then
$$P(X \geq 3) = 1 - P(X \leq 2) = P(X = 0) + P(X=1) + P(X = 2)$$

In R you can call 1 - pbinom(2, 5, 0.5) and find that this probability is actually 0.5. In Excel I think you can use 1 - BINOM.DIST(g-1, n, p, TRUE) to do the computation

So in general, you can model these types of probabilities as follows:
$$p = \frac{6 - skill + 1}{6} \text{, probability of success on a single roll}$$
$$g = \text{goal set by the GM}$$
$$n = \text{the stat, the number of rolls you get}$$
$$X \sim \text{Bin}(n, p)$$
$$P(X = k) = \binom{n}{k}p^k (1-p)^{n-k}, \text{the PMF for Binomial dist.}$$
$$P(X \geq g) = 1 - P(X < g) = 1 - \sum_{k=0}^{g-1} \binom{n}{k}p^k (1-p)^{n-k}$$

Title: Dice probability, Exemple, having at least three 5 or more, on seven dice throw.

Body:
I'm having trouble calculating probability on dice rolls, here's the setup :

(All dices are fair 6 faces classic dice)

It's a D&D Roleplaying kind of game, the rules say this :

For each action, you have a stat, and a skill assigned.
To fight someone on melee, that could be strenght as a stat, and brawling as a skill.

If you have 5 points on strenght, that mean you can throw 5 dices.
Then, let's say you have 4 points on brawl skill, that mean you must land a 4 or higher on a dice to validate it.

Then, the Game Master will set a goal.
If he set it a 3, you must have at least 3 dice that show a 4 or higher to achieve your action.

Get it ?

Exemple :

Me : "I have 5 in strenght and 4 in brawl, I'll punch this guy in the face"

GM : "Ok, but he's skilled at combat, so the goal is at 3"

Then, according to the rules, I can throw 5 dices, and need to have at least 3 showing 4 or more to succesfully punch the said guy.

Here's where I landed thus far :

If I want at least a 4, with five dice, I can apply this formula :

The chance of failing are (3/6)^5

So the chance of succes are 1 - ((3/6)^5), or 0,96875

According to the game rules, this mean I have a goal set to 1.

But if I have a goal set to 3 ?

Is the formula (1 - ((3/6)^5))^3 ? wich shall give 0,9091491699

Or I am missing something ?

I am working on an easy excel, so the GM can more easily calculate the chances of success before setting the goal of every action player might do.


问:
About $p$-filters with $p\in\beta X\setminus X$.
I was reading this article and the next underlined part took my atention

The definition of $p$-filter sounds so different from the usual definition of a filter in a topological space. I know that maybe this can be a different definition and maybe the only thing related with filters is the name. Is there another place where I can read about this "$p$-filters? Or is this only a definition concerning this article and nobody worked with that? I am really interesing about to know more  about $p$-filters. I really apreciated any help. Thanks.

答:
If I understand the definitions correctly, $\mathscr{F}$ is not necessarily a filter in the usual sense, but if you close it under finite intersections and then under supersets, you get an ordinary filter on $\pi$ that is still a $p$-filter on $\pi$.

Let $\mathscr{B}=\left\{\bigcap\mathscr{A}:\mathscr{A}\subseteq\mathscr{F}\text{ and }\mathscr{A}\text{ is finite}\right\}$; then $\mathscr{B}$ is closed under finite intersections, and $p\in\operatorname{Cl}\bigcup B$ for each $B\in\mathscr{B}$, so $\mathscr{B}$ is a $p$-filter on $\pi$ that contains $\mathscr{F}$. Finally, let $\mathscr{G}=\{G\subseteq\pi:B\subseteq G\text{ for some }B\in\mathscr{B}\}$; $\mathscr{G}$ is a filter on $\pi$ that is also a $p$-filter containing $\mathscr{F}$. (In fact it is the minimal such filter on $\pi$.)

Say that $U\subseteq X$ is a $\pi$-set if it is the union of members of $\pi$. Then $\left\{\bigcup G:G\in\mathscr{G}\right\}$ is a $\pi$-set filter on $X$ with the property that $p\in\operatorname{Cl}U$ for each $G\in\mathscr{G}$.


问:
Create N points that are spaced as far as possible within a D dimensional sphere
I'm not a big geometry buff, but I was wondering if you're give a dimension $D$ and a number of points $N$ and a radius $R$ of a sphere in that dimension. How would you generate $N$ points on the edge or in the sphere such that their magnitude and cosine similarity between each point is as far as possible? The dimensional space could be something as large as 512.

The ultimate goal is to achieve something like: Assume we're placing planets in a galaxy that is spherical, but we have no idea how big each planet is, so we want to place all the origins of each planet as far as possible from each other in order to avoid overlapping even if the planets are very large.

答:
One way to pick a random point on a $d$-dimensional hypersphere ${\cal S}$ is to generate $n$ Gaussian random variables $x_1, x_2, ..., x_n$ and then the distribution of the vectors

$$ \frac{1}{\sqrt{x_1^2 + x_2^2 + \cdots + x_n^2}} \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}$$

is uniform over the surface ${\cal S}^{d-1}$, as  desribed in

Muller, M. E. "A Note on a Method for Generating Points Uniformly on N-Dimensional Spheres." Comm. Assoc. Computing Machinery 2, 19-20 (Apr. 1959)


问:
Finding $(N,m)$ such that $3 \mid 2^Nm - 1$
Good evening everyone! Here is my "question". Find all pairs of natural numbers $(m,N)$ such that $3| 2^{N}m - 1$. Could I get some tips to get started. I've proven a few special cases using induction.

答:
We should start by noticing that:

$2^N =
\begin{cases}
1 \mod 3, & \text{if $N$ even}\\
-1 \mod 3, & \text{if $N$ odd}
\end{cases}$

Thus,
$ 2^N m =
\begin{cases}
m \mod 3, & \text{if $N$ even}\\
-m \mod 3, & \text{if $N$ odd}
\end{cases}$

and solving $2^Nm = 1 \mod 3$ we conclude that

$
\begin{cases}
m = 1\mod 3, & \text{if $N$ even}\\
m = -1 \mod 3, & \text{if $N$ odd}
\end{cases}$

So all the pairs are $$\{(N,m): [(N = 0 \mod 2) \text{ and } (m = 1 \mod 3)] \text{ or }  [(N = 1 \mod 2) \text{ and } (m = -1 \mod 3)] \}$$


问:
Surjectivity of the Gelfand Transform map
This is a very basic question on understanding the Gelfand Transform...

Let $A$ be a unital commutative Banach Algebra.

Let $0\neq e\neq 1$ be an idempotent in $A$.

Let $\Omega(A)$ be the set of non-zero homomorphisms from $A$ to $\Bbb C$.

Show the map $\hat{e}:\Omega(A)\rightarrow \{0,1\}$ defined by $\phi\mapsto \phi(e)$ is surjective.

I see that we have $\phi(e ) = \phi(e^2) = \phi(e)^2= \cdots = \phi(e^n)= \phi(e)^n$.

Is this enough to show surjectivity? I'm a bit confused by the definitions.

Thank you for your help.

答:
The characters of $A$ (the nonzero homomorphisms $A\to\mathbb{C}$) are in a bijective correspondence with the maximal ideals of $A$. More precisely, the maximal ideals are exactly the kernels of the characters of $A$.

Now, since $e$ is an idempotent we have $e^2=e$, and so $e(e-1)=0$. If $e$ was invertible then we would get $e=1$, a contradiction. Thus $e$ is not invertible, and so it belongs to some maximal ideal $M\subseteq A$. (follows from Zorn's lemma) But this means $e$ belongs to the kernel of some character $\phi$, i.e there is $\phi\in\Omega(A)$ such that $\phi(e)=0$.

Also, since $e\ne 0$, the equality $e(e-1)=0$ implies that $e-1$ also cannot be invertible. So $e-1$ also belongs some maximal ideal, i.e to the kernel of some character $\varphi$. But then we get:

$0=\varphi(e-1)=\varphi(e)-\varphi(1)=\varphi(e)-1$

And so $\varphi(e)=1$ for this $\varphi$. So indeed there is a character $\varphi\in\Omega(A)$ such that $\varphi(e)=1$.

Alright, so your map $\Omega(A)\to\{0,1\}$ is indeed surjective.


问:
Do Dyck Words or Dyck Paths Have Manifestations in Nature?
Working with the Dyck language of well-balanced strings of parentheses, I am led to wonder what, if any, phenomena the concept relates to in the natural world - in the same sense that Fibonacci numbers, for example, have manifestations as patterns in sunflower floret spirals.

Since Dyck words have equivalent Dyck paths, I also included the latter in my question.

I hope this is an appropriate board to ask such a question; if you can suggest a better place, I'll be happy to go there.

答:
From a page in the presentation of the Vancouver (B.C., Canada) science museum Science World:

A fractal is a kind of pattern that we observe often in nature and in art. As Ben Weiss explains, “whenever you observe a series of patterns repeating over and over again, at many different scales, and where any small part resembles the whole, that’s a fractal.”

Not only is it a fractal. It's also a Dyck language, since a Dyck language is an abstraction of precisely the recursive structure described in that paragraph.


问:
Hall $\pi$-subgroup and $HN=G$

Let $\pi$ be any set of prime numbers. A finite group $H$ is a $\pi$-group if all primes that divide $|H|$ lie in $\pi$. If $|G|<\infty$, then a Hall $\pi$-subgroup of $G$ is a $\pi$-subgroup $H$ such that $|G:H|$ is divisible by no prime in $\pi$. Let $\phi$ be a homomorphism defined on $G$.

If $H\subseteq G$ is a Hall $\pi$-subgroup, show that $\phi(G)$ is a $\pi$-group iff $HN=G$, where $N=\text{ker}(\phi)$.

If $HN=G$, then $\phi(H)=\phi(G)$ since $\phi(N)=1$, and since $\phi(H)$ divides $H$, which is a $\pi$-subgroup, we must have that $\phi(G)$ is a $\pi$-group.

But what about the other direction? Assume that $\phi(G)$ is a $\pi$-group. This means all primes dividing $|\phi(G)|$ are in $\pi$, all primes dividing $|H|$ are in $\pi$, and $|G:H|$ is divisible by no prime in $\pi$. We have $|G| = |H||G:H|$, and $|\phi(G)|$ divides $|G|$. How can we prove that $HN=G$ from here?

答: $N\subset HN$, hence $[G:HN]=[\phi(G):\phi(HN)]$. But $[G:HN]$ is "$\pi$-free", hence $[\phi(G):\phi(HN)]$ is "$\pi$-free". But $\phi(G)$ is a $\pi$-group...</p>

Title: Hall $\pi$-subgroup and $HN=G$


问:
Explicit formula of Mobius Transformation that maps non intersecting circles to concentric circles
How to find a Mobius Transformation that maps the circles $C(0,1)$ and $C(4,2)$  to a pair of concentric circles.

I have already discussed this problem in general case in this blog. By the composition of transformation we can do this. But , the idea I have seems burdensome to calculate. Is there any good way of doing this?

答:
An affine transformation $az+b$ may be used to turn one circle into the unit circle (specifically, $(z-a)/r$ if the circle has center $a$ and radius $r$). If the other circle is not contained within it, we may apply $1/x$ to (or $-1/x$ if we want to use $\mathrm{PSL}_2\mathbb{R}$), which fixes the unit circle while swapping its interior and exterior.

The transformations that preserve the real line come from $\mathrm{PSL}_2\mathbb{R}$ and the ones that preserve the unit circle are $\mathrm{PSU}(1,1)$. The intersection of these two is $\mathrm{PSO}(1,1)$ whose elements may be represented by

$$ \begin{bmatrix} \cosh\tau & \sinh\tau \\ \sinh\tau & \cosh\tau \end{bmatrix}. $$

Alternatively, one may say $Mz=\tfrac{az+b}{cz+d}$ and solve $M(1)=1$, $M(-1)=-1$ to get $Mz=\frac{az+b}{bz+a}$, without loss of generality $a^2-b^2=1$ whose solutions ($a>0$) are parametrized $(a,b)=(\cosh\tau,\sinh\tau)$.

The circle in the interior of the unit circle intersects at points $u,v\in(-1,1)$. We may parametrize this interval with $\tanh$, say $u=\tanh(\alpha)$, $v=\tanh(\beta)$ and check that

$$ \begin{bmatrix} \cosh\tau & \sinh\tau \\ \sinh\tau & \cosh\tau \end{bmatrix} \tanh(\sigma)=\tanh(\tau+\sigma) $$

using $\tanh$'s sum formula. Therefore, we may choose $\tau=-\frac{1}{2}(\alpha+\beta)$ and apply this transformation to turn $u,v$ into values $x,-x$ within $(-1,1)$, necessarily intersection points of a concentric circle.

Note the transformation may also be written as

$$ \begin{bmatrix} 1 & \tanh\tau \\ \tanh\tau & 1 \end{bmatrix} $$

and use various hyperbolic trig identities to obtain

$$ \tanh\tau=\pm\sqrt{\frac{1+uv-\sqrt{(1-u^2)(1-v^2)}}{1+uv+\sqrt{(1-u^2)(1-v^2)}}} $$

In the case of $C(0,1)$ and $C(4,2)$, we can apply $-1/x$ so the latter becomes $C(-\tfrac{1}{3},-\tfrac{1}{6})$ with corresponding endpoints $u=-\tfrac{1}{2}$, $v=-\tfrac{1}{6}$. Plugging these into the above, rescaling and using Wolfram|Alpha to denest,

$$ \begin{bmatrix} \sqrt{21}+\sqrt{5} & \sqrt{21}-\sqrt{5} \\ \sqrt{21}-\sqrt{5} & \sqrt{21}+\sqrt{5} \end{bmatrix} $$

preserves $C(0,1)$ and turns $C(-\tfrac{1}{3},-\tfrac{1}{6})$ into $C(0,\tfrac{11-\sqrt{105}}{4})$.


问:
Showing that a solid of revolution is Lebesgue measurable
Let $E \subset \mathbb{R}^2$ be Lebesgue measurable in $\mathbb{R}^2$, bounded and in the right half-plane:  $(r,z) \in E \Rightarrow r >0.$ $R(E)$ is rotating around the $z$-axis.

$$R(E):=\{(x,y,z)\in \mathbb{R}^3 \space| \space(\sqrt{x^2+y^2},z) \in E \}$$

How do I show that $R(E)$ is Lebesgue-measurable in $\mathbb{R}^3$?

I think I have to do something with characteristic functions and the transformation theorem. Then I could choose a smart diffeomoprhism but I'm not really sure how.

答: Consider the map $F:\Bbb{R}\times (0,\infty)\times\Bbb{R}\to\Bbb{R}^3$ defined by $F(\phi,r,z):= (r\cos\phi,r\sin\phi,z)$. Then, $F$ is $C^{\infty}$, hence locally Lipschitz, so it sends Lebesgue measurable sets to Lebesgue measurable sets. What can you deduce about the revolved set $R(E)$ in relation to $F$?


问:
Mobius transformation sending spheres to concentric spheres
In the complex plane it is known that given two disjoint circles we can find a Mobius transformation that maps them to concentric circles.

My question: Given two disjoint $(n-1)$-spheres $Y$ and $Z$ in $\mathbb{R}^n$ is there a Mobius transformation of $\mathbb{R}^n$ that maps them to concentric spheres or is that not true in general?

答:
Without loss of generality, two circles are symmetric across the real axis, and so the Mobius transformation used to make them concentric is also symmetric across the real axis (see here).

This generalizes to spheres by symmetry: we can make spheres concentric using a Mobius transformation which is symmetric across the axis of symmetry (of the two spheres); just pick any plane containing this axis, get a 2D Mobius transformation, then extend to rest of space by symmetry.


问:
Solve system of first order differential equations
I have to solve differential systems like this:
$$
\left\{
\begin{array}{c}
x' = 3x - y + z \\
y' = x + 5y - z \\
z' = x - y + 3z
\end{array}
\right.
$$

Until now I computed the eigenvalues $k = \{2,4,5\}$ by solving the equation resulted from this determinant of this matrix:
$$
\begin{pmatrix}
3 && -1 && 1 \\
1 && 5 && -1 \\
1 && -1 && 3
\end{pmatrix} - kI_3 =
\begin{pmatrix}
3-k && -1 && 1 \\
1 && 5-k && -1 \\
1 && -1 && 3-k
\end{pmatrix}
$$
I don't know what to do next.

NOTE: This is an example but from it I want to learn the method to solve any system of this kind. I learn better from particular examples than directly from generalization.

答:
To find the eigenvalues of: $$A =\begin{bmatrix}3 & -1 & 1\\1 & 5 & -1\\1 & -1 &3\end{bmatrix},$$

we set up $|A - \lambda I| = 0$ and solve the characteristic polynomial, so we have:

$$|A -\lambda I| = \begin{bmatrix} 3-\lambda & -1 & 1\\1 & 5-\lambda & -1\\1 & -1 &3 -\lambda \end{bmatrix} = 0$$

From this, we get the characteristic polynomial as: $$-\lambda^3+11 \lambda^2-38 \lambda+40 = -(\lambda-5) (\lambda-4) (\lambda-2)= 0$$

This gives us three eigenvalues: $λ_1 = 2, λ_2 = 4$  and $λ_3 = 5$

To find the eigenvectors, we set up and solve $[A - \lambda_i I]v_i = 0$, so lets do $\lambda_1 = 2$. We have:

$$[A - \lambda_i I]v_i = \begin{bmatrix} 3-2 & -1 & 1\\1 & 5-2 & -1\\1 & -1 &3 -2 \end{bmatrix}v_1 = \begin{bmatrix} 1 & -1 & 1\\1 & 3 & -1\\1 & -1 & 1 \end{bmatrix}v_1  = 0$$

This leads to the row-reduced-echelon form:

$$\begin{bmatrix} 1 & 0 & \dfrac{1}{2}\\0 & 1 & -\dfrac{1}{2}\\0 & 0 & 0 \end{bmatrix}v_1  = 0$$

This gives us a solution of:

$b = \dfrac{1}{2}c \rightarrow c = 2 \rightarrow b = 1$, and $a = -\dfrac{1}{2} c \rightarrow a = -1$.

Thus the eigenvector for the eigenvalue $\lambda_1 = 2$ is $v_1 = (-1, 1, 2)$.

If we repeat this process two more times for the other two eigenvalues, we end up with the eigenvalue/eigenvector pairs:

$$\lambda_1 = 2, ~v_1 = (-1, 1, 2)$$

$$\lambda_2 = 4, ~v_2 = (1, 0, 1)$$

$$\lambda_3 = 5, ~v_3 = (1, -1, 1)$$

Note that different approaches are needed when we get into repeated eigenvalues, but that is for another problem.

Try to derive those last two eigenvectors and report back if they do not work out.

To write out the solution for the three equation, we would use a linear combination of the the eigenvalues, eigenvectors and some unknown constants, as:

$$X(t) = \begin{bmatrix} x(t) \\ y(t) \\ z(t) \end{bmatrix} = c_1e^{2t} \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} + c_2 e^{4t} \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + c_3 e^{5t} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$$


Title: Finding $(N,m)$ such that $3 \mid 2^Nm - 1$

Body: Good evening everyone! Here is my "question". Find all pairs of natural numbers $(m,N)$ such that $3| 2^{N}m - 1$. Could I get some tips to get started. I've proven a few special cases using induction.

Body: HUGE HINT:  $2 = (3-1)$ so $2^Nm -1 = (3-1)^Nm -1$.


问:
Use differentiation to find the a power series for 1/(x+1)^2
I did see this discussed in another question; however, I did exactly what the answer suggested and still do not seem to get the solution.

The question is to use differentiation to find the representative power series of $1/(x+1)^2$

I could easily see that the derivative of $-1/(x+1)$ is $1/(x+1)^2$.

So, my goal was to find the power series for $-1/(x+1)$ and then take the derivative of that.

I found the power series for $-1/(x+1)$ to be: $\sum_{n=0}^\infty (-1)^\left(n+1\right) x^n$.
Then I simply took the derivative of that power series to get:
$\sum_{n=0}^\infty (n) (-1)^\left(n+1\right) x^\left(n-1\right)$

Can't see where I went wrong.

Book answer:  $\sum_{n=0}^\infty (-1)^n (n+1)x^n$
Makes no sense to me.  Maybe I didn't find the original power series correctly?

答:
It's the same. In your answer, the $n=0$ term is just $0$ because of the multiplication by $n$:
\begin{align}
\sum_{n=0}^{\infty}n(-1)^{n+1} x^n = \sum_{n=1}^{\infty}n(-1)^{n+1}x^n.
\end{align}
Now, change the index of summation, by letting $n=m+1$. Then,
\begin{align}
\sum_{n=1}^{\infty}n(-1)^{n+1}x^n=\sum_{m+1=1}^{\infty}(m+1)(-1)^{m+1+1}x^{m+1}=
\sum_{m=0}^{\infty}(-1)^m(m+1)x^{m+1}
\end{align}
Now, you can relabel the indices however you like:
\begin{align}
\sum_{m=0}^{\infty}(-1)^m(m+1)x^{m+1}=\sum_{n=0}^{\infty}(-1)^n(n+1)x^{n+1}=
\sum_{@=0}^{\infty}(-1)^@(@+1)x^{@+1}=
\sum_{\star=0}^{\infty}(-1)^{\star}(\star+1)x^{\star+1}.
\end{align}


问:
Maps from $\Bbb R^2 \to \Bbb R^2$ that preserve squeeze maps?

What maps from $\Bbb R^2 \to \Bbb R^2$ preserve squeeze maps?

A squeeze map is a type of linear map that preserves Euclidean area regions of the Cartesian plane but is not a rotation or shear mapping.

For a fixed positive real number $a$ the mapping $(x,y)\mapsto (ax,y/a)$ is the squeeze mapping with parameter $a.$

So I'm looking to understand maps that preserve the structure of squeeze maps from $\Bbb R^2$ to itself. That is, if you have a squeeze map on $\Bbb R^2,$ what maps will preserve the squeeze map?

I have thought about this and come up with $3$ examples: $1)$ identity map, $2)$ involution map $3)$ any map that maps involutions to involutions.

答:
I think by "squeeze maps" you mean the $2$-dimensional manifold of symmetric positive-definite matrices $A$ of determinant $1$, which we will denote by $\mathcal{S}$.

You are asking for the set of all matrices $g \in \mathrm{GL}(2,\mathbb{R})$ such that

$$ g \mathcal{S} g^{-1} \subseteq \mathcal{S}$$.

Note that if $g \in O(2,\mathbb{R})$, then $g$ has the property above. Indeed, if $g$ is an orthogonal matrix, then $g^T g = I$, which implies that $g^{-1} = g^T$. Hence

$$ g \mathcal{S} g^{-1} = g \mathcal{S} g^T \subseteq \mathcal{S}.$$

To see why this is so, note that if $A \in \mathcal{S}$, then $gAg^{-1} = gAg^T$ is symmetric. Note also that

$$(gAg^{-1}v, v) = (A g^{-1}v, g^T v) = (A g^{-1}v, g^{-1}v) > 0,$$

for all nonzero $v \in \mathbb{R}^2$. Moreover, $\operatorname{det}(gAg^{-1}) = \operatorname{det}(A) = 1$.

Moreover, if $g = \lambda I$, for some $\lambda \in \mathbb{R}$, $\lambda \neq 0$, then also $g$ has the property we seek.

In short, if $g \in CO(2,\mathbb{R})$ (the conformal group in dimension $2$), then $g$ has the property the OP is interested in.

Now, let us assume that $k \in \mathrm{GL}(2,\mathbb{R})$ such that

$$ k \mathcal{S} k^{-1} \subset \mathcal{S}.$$

Then what can we say about $k$? I think, but I am not sure, that $k \in CO(2, \mathbb{R})$. I will have to think a bit more about it though...

Edit: indeed, my last claim is true. I proved it by taking two special classes for $A$. By taking $A \in \mathcal{S}$ to be diagonal, imposing the condition we are interested in leads to the two columns of $A$ being orthogonal. However, the two columns may, a priori, have $2$ different norms. And then, by taking $A \in \mathcal{S}$ to have $1$s on the diagonal, and after imposing again the condition above, one gets that $k \in CO(2,\mathbb{R})$.

More explicitly, $CO(2,\mathbb{R})$ consists of matrices $g$ of the form

$$ g = \left( \begin{array}{cc} u & -v \\ v & u \end{array} \right),$$

where $u$ and $v$ are real numbers which are not both $0$. Note that $CO(2, \mathbb{R})$ includes for example rotations, reflections and scaling (it is actually generated by these $3$ types of linear transformations).


问:
Basis of vectors Graphically
How do we check the basis of vectors graphically?
Can anyone explain few of the graphs below How would you know whether they are basis or not

答:
Since we're talking about pictures, there are only two cases, two-dimensional, and three-dimensional.

In two dimensions, a basis must consist of exactly two vectors, and these vectors can't be colinear.

In three dimensions, a basis must consist of exactly three vectors, and they cannot all lie in one plane.

Title: Basis of vectors Graphically


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

Thanks in advance for any helpful answers.

答:
We can prove something more general:

If $A$ is an $n\times n$ matrix over any field such that all its row sums and column sums are zero, then its cofactors are identical to each other. That is, all entries of $\operatorname{adj}(A)$ are equal.

Denote by $M_{ij}$ the submatrix obtained by deleting the $i$-th row and $j$-th column of $A$ and let $m_{ij}=\det M_{ij}$. Now suppose $i<n$ and we want to calculate $m_{ik}$ for some index $k$. Add the first $n-2$ rows of $M_{ik}$ to the last row. Since all column sums of $A$ are equal to $0$, the last row becomes $(-a_{i1}, -a_{i2}, \ldots, -a_{i,j-1}, -a_{i,j+1}, \ldots,-a_{in})$. This resulting submatrix is almost identical to $M_{nj}$, except that the $i$-row of $M_{nj}$ is moved to the bottom and negated. It follows that
$$
m_{ij} = (-1)^{n-i}m_{nj}\tag{1}
$$
and this obviously also holds when $i=n$. Similarly, by using column operations, we have $m_{ij} = (-1)^{n-j}m_{in}$. In particular, when $i=n$ we get
$$m_{nj} = (-1)^{n-j}m_{nn}.\tag{2}$$
Combine $(1)$ and $(2)$, we have $(-1)^{i+j}m_{ij}=m_{nn}$. Hence the result.

Title: Question on cofactors


问:
proving uniform convergence and differentiability for any bounded open interval ($a,b$) $\in \mathbb{R}$
I'm trying to prove uniform convergence and differentiability for any bounded open interval ($a,b$) $\in \mathbb{R}$
$$
\sum_{n=1}^{\infty} \left(x-n\sin \left(\frac x{n}\right)\right)
$$

I was trying to use Weierstrass M-test to prove this but I cannot find an $M_n$ that is greater than or equal to the absolute value of the inside function for all of $n$.

Any help, alternative methods or suggestions for $M_n$ or anything else, would be greatly appreciated.

答: Let's say your interval is $(0,b)$ (Take $a = 0$), then: $x- n\sin \left(\frac{x}{n}\right)= n\left(\dfrac{x}{n} - \sin\left(\frac{x}{n}\right)\right)\le \dfrac{n}{6}\cdot \left(\dfrac{x}{n}\right)^3 \le \dfrac{b^3}{n^2}$. Thus you might take $M_n = \dfrac{b^3}{6n^2}$. Note here we use the well-known fact: $x - \sin x \le \dfrac{x^3}{6}, x \in (0,b), b $ is small.


问:
What is the intuition behind the dot product between a vector and a bivector?
From geometric algebra, I know the definition of a bivector. Now suppose we have three 2-d vectors $a$, $b$, $c$. What is the geometrical intuition behind the following formula? (The dot product of a vector and a bivector)

$$
c\cdot(a\wedge b)
$$

If I find the answer to this question, I will understand what is happening in 3-d space, when $c$ is not in the same plane of $a$ and $b$. Or what the value of $(a\wedge b) \cdot c$ is and its relation to the specified formula.

答:
In 2D, let $a \wedge b = B i$, where $i = e_1 e_2$ is the pseudoscalar for the plane, and $B$ is the magnitude of the bivector (i.e. $B = (a \wedge b)/i$).  Then

$c \cdot (a \wedge b) = c i B$.

The vector $c$ is scaled by the magnitude of the bivector $a \wedge b$, and also rotated 90 degrees.  The sense of the rotation is from $e_1$ towards $e_2$ if $B > 0$ (counterclockwise with $e_1$ to the right, and $e_2$ up.)

In dimensions greater than 2, let

$c = c_\parallel + c_\perp$

where $c_\parallel \in \textrm{span}\{a, b\}$, and $c_\parallel \cdot c_\perp = 0$.  None of the components of $c$ that lie outside of the plane $a \wedge b$ contribute to the vector-bivector dot product, leaving

$c \cdot (a \wedge b) = c_\parallel \cdot (a \wedge b).$

This is now a 2D relationship, as above.  We have is a projection of $c$ into the plane spanned by $a$ and $b$.  That projection is by the magnitude of $a \wedge b$, and then rotated by 90 degrees.


问:
Proving that every well orderable set X is equipotent to a unique initial ordinal number
I'm following the proof offered by the book "Introduction to Set Theory" by K. Hrbacek of the claim "every well orderable set X is equipotent to a unique initial ordinal number" (Theorem 1.3 p. 130).

They start saying that, since a well ordered set is isomorphic to an ordinal number, X is equipotent to
some ordinal number, then they talk about the least ordinal number equipotent to $X$; I don't understand how they deduce the existence of such ordinal. Since they emphasized the existence of an ordinal equipotent to X, I guess they are taking the least member of "the set of ordinals equipotent to X"; I know that every set whose elements are ordinals is well ordered (by the membership relation), then I'll agree of doing so but just after we prove the existence of such a set, and I still have no idea of how doing so.

Thank you in advance for your clarifications !

答:
Let $\alpha$ be an ordinal equipotent to $X$. Then we can form the set

$$\{\xi\in\alpha+1:\xi\text{ is an ordinal equipotent to }X\}\,,$$

and since it’s a non-empty set of ordinals, it has a least element. Any ordinal equipotent to $X$ that is not in this set is larger than $\alpha$ and therefore certainly not the least ordinal equipotent to $X$.


问:
Writing an epsilon proof showing $\lim \limits_{x\to \infty} |{\frac{\sin x}{x}|} = 0$
I am trying to write an epsilon proof to show that $$\lim \limits_{x\to \infty} |{\frac{\sin x}{x}|} = 0$$

I'm having a bit of a problem at the moment as to what to include and how to show it properly.
I know I need to show that $$|\frac{\sin x}{x} - L| < \epsilon; \epsilon > 0$$

I don't know if I am allowed to use $L = 0$ in the proof because I have seen people do it before but I don't think your allowed to assume that. Furthermore, I am having trouble removing the magnitude sign with the $| \sin x |$. This is what my proof looks like currently:

For all $\epsilon > 0$, show that $$|\frac{\sin x}{x}- L| < \epsilon$$

Let $$ m = \frac{1}{\epsilon}+1$$
Suppose $$x \ge m$$
Then $$|\frac{\sin x}{x} - L|$$
$$ = |\frac{\sin x}{x}|$$
$$ \le |\frac{1}{x}|$$
$$ < |\frac{1}{\frac{1}{\epsilon}}|$$
$$ = \epsilon$$

Hence $$|\frac{\sin x}{x}- L| < \epsilon$$

Therefore: $$\lim \limits_{x\to \infty} |{\frac{\sin x}{x}|} = 0$$

I would like to know what I have done wrong and what I have missed in this proof. Any answers would be greatly appreciated.

答:
You are allowed to use L=0. However, what you need to prove is:

$\forall \epsilon \gt 0, \exists M>0: x>M\Rightarrow \left |\frac{\sin(x)}{x} \right | \lt \epsilon$

To prove that, show that $\left| \frac{\sin(x)}{x} \right | \le \frac{1}{x} \lt \frac{1}{M}$. Then choose $M=\frac{1}{\epsilon}$ to to get the inequality in $\epsilon$.

What I have proved is:

$\forall \epsilon \gt 0, \exists M>0, M=\frac{1}{\epsilon}:x>M\Rightarrow \left |\frac{\sin(x)}{x} \right | \lt \epsilon$


问:
If $\lim_{n \to \infty} |a_n| = |L|$ then $\lim_{n \to \infty} a_n = L$
This is the question that I am having a bit of trouble with and would like some help and/or advice.

Prove or disprove. Suppose we have a sequence $\{a_n\}_{n = 1}^{\infty}$. If $\lim_{n \to \infty} |a_n| = |L|$, then $\lim_{n \to \infty} a_n = L$.

My thought of approaching this question was to prove the converse of this statement, so what I really want to prove is if $\lim_{n \to \infty} a_n = L$, then $\lim_{n \to \infty} |a_n| = |L|$.

For the limit $\lim_{n \to \infty} a_n = L$, by the definition of the limit: $\forall\varepsilon > 0 \exists N\in\mathbb{N}$ such that if
\begin{equation*}
n > N \hspace{0.5cm} \rightarrow \hspace{0.5cm} |a_n - L| < \varepsilon
\end{equation*}

But we also know that for the limit $\lim_{n \to \infty} |a_n| = |L|$, we have: $\forall\varepsilon > 0 \exists N\in\mathbb{N}$ such that if
\begin{equation*}
n > N \hspace{0.5cm} \rightarrow \hspace{0.5cm} ||a_n| - |L|| < \varepsilon
\end{equation*}

I was wondering how to proceed from here. This is the part where I am having a bit of trouble with. Some advice would be great!

答: You can show a counterexample. Take $a_n=-1$ and $L=1$. We have $lim_{n \to \infty} |a_n| = |L|$, but $lim_{n \to \infty} a_n \neq L$.


答: I am not an expert with sequence, but for this one you can "see through" if you take $a_n = (-1)^n$. Hope this helps.

Title: If $\lim_{n \to \infty} |a_n| = |L|$ then $\lim_{n \to \infty} a_n = L$

Body:
This is the question that I am having a bit of trouble with and would like some help and/or advice.

Prove or disprove. Suppose we have a sequence $\{a_n\}_{n = 1}^{\infty}$. If $\lim_{n \to \infty} |a_n| = |L|$, then $\lim_{n \to \infty} a_n = L$.

My thought of approaching this question was to prove the converse of this statement, so what I really want to prove is if $\lim_{n \to \infty} a_n = L$, then $\lim_{n \to \infty} |a_n| = |L|$.

For the limit $\lim_{n \to \infty} a_n = L$, by the definition of the limit: $\forall\varepsilon > 0 \exists N\in\mathbb{N}$ such that if
\begin{equation*}
n > N \hspace{0.5cm} \rightarrow \hspace{0.5cm} |a_n - L| < \varepsilon
\end{equation*}

But we also know that for the limit $\lim_{n \to \infty} |a_n| = |L|$, we have: $\forall\varepsilon > 0 \exists N\in\mathbb{N}$ such that if
\begin{equation*}
n > N \hspace{0.5cm} \rightarrow \hspace{0.5cm} ||a_n| - |L|| < \varepsilon
\end{equation*}

I was wondering how to proceed from here. This is the part where I am having a bit of trouble with. Some advice would be great!


问:
Floating point division resulting in a value exceeding 1 but should be equal to 1...maybe!
I am computing the apparent magnitude of comets and minor planets using data from the Minor Planet Center.  The formula I use has a division and that result is passed to the arccosine function (in Python).

Occasionally the result of the division is slightly greater than 1, for example

6.272370848320445 / 6.2723697617591405 = 1.0000001732297912

I assume the result is not actually greater than 1; rather the result is an artefact of floating point arithmetic, particularly given the number of zeros after the decimal point.

If my assumption is correct, I want to catch this sort of situation and treat the result as if the division yielded exactly 1 (and so can safely be passed to arccosine).

If I make an arbitrary choice of say five decimal points and truncate the division result, then if the result is 1 I'm good, otherwise, drop the result.

I'm not sure if what I'm proposing is correct mathematically, so looking for guidance please.  I have assumed this issue arises due to floating point arithmetic, but if that assumption is wrong, then I'm not sure what else I can do.  It is possible the source data from the Minor Planet Center contains spurious data which is causing the actual error (and thus I shouldn't be trying to work around it but just let it fail).

The formula can be found here (using the H,G model).

答:
In your example the numerator is slightly larger than the denominator (look at the fifth decimal place) so that quotient should be greater than $1$.

Pasting the division into the google search bar gives
$1.00000017323$.

Your program is calculating the quotient correctly. If it should be no larger than $1$ the problem is upstream where the data comes from.


答:
I have experienced exactly the same situation with round-off (in calculation of finite rotations in C#). You theoretically know that the division equals maximally to 1 ergo you can implement it as follows:

$\max(-1.0, \min(1.0, x/y))~$ instead of $~x/y$

and your code will be safe.

Title: Floating point division resulting in a value exceeding 1 but should be equal to 1...maybe!

Body:
I am computing the apparent magnitude of comets and minor planets using data from the Minor Planet Center.  The formula I use has a division and that result is passed to the arccosine function (in Python).

Occasionally the result of the division is slightly greater than 1, for example

6.272370848320445 / 6.2723697617591405 = 1.0000001732297912

I assume the result is not actually greater than 1; rather the result is an artefact of floating point arithmetic, particularly given the number of zeros after the decimal point.

If my assumption is correct, I want to catch this sort of situation and treat the result as if the division yielded exactly 1 (and so can safely be passed to arccosine).

If I make an arbitrary choice of say five decimal points and truncate the division result, then if the result is 1 I'm good, otherwise, drop the result.

I'm not sure if what I'm proposing is correct mathematically, so looking for guidance please.  I have assumed this issue arises due to floating point arithmetic, but if that assumption is wrong, then I'm not sure what else I can do.  It is possible the source data from the Minor Planet Center contains spurious data which is causing the actual error (and thus I shouldn't be trying to work around it but just let it fail).

The formula can be found here (using the H,G model).


答:
Let consider derivative of function $f(x)=x-n\sin \left(\frac x{n}\right)$, then we have $f'(x)=1-\cos\frac {x}{n}\geqslant 0$, which gives, that $f$ is increasing and it's supremum for any $(a, b)$ is $f(b)=b-n\sin \left(\frac {b}{n}\right)$.

Knowing $b-n\sin \left(\frac {b}{n}\right) = n \left(\frac {b}{n}-\sin \left(\frac {b}{n}\right)\right) = n \left(\frac {b}{n}- \left[\frac {b}{n}- \frac {1}{6}\left(\frac {b}{n}\right)^3 + O\left(\left(\frac {b}{n}\right)^5\right)\right]  \right) =O\left( \frac {1}{n^2}\right)$ we have Weierstrass M-test.

For differentiability we should take look at $ \sum\limits_{n=1}^{\infty} \left(1-\cos\frac {x}{n}\right) = \sum\limits_{n=1}^{\infty} 2\sin^2 \frac {x}{2n}$ which uniform convergence let me leave to you

Title: proving uniform convergence and differentiability for any bounded open interval ($a,b$) $\in \mathbb{R}$

Body:
I'm trying to prove uniform convergence and differentiability for any bounded open interval ($a,b$) $\in \mathbb{R}$
$$
\sum_{n=1}^{\infty} \left(x-n\sin \left(\frac x{n}\right)\right)
$$

I was trying to use Weierstrass M-test to prove this but I cannot find an $M_n$ that is greater than or equal to the absolute value of the inside function for all of $n$.

Any help, alternative methods or suggestions for $M_n$ or anything else, would be greatly appreciated.


问:
Intuitive explanation of Cauchy's Integral Formula in Complex Analysis
There is a theorem that states that if $f$ is analytic in a domain $D$, and the closed disc {$  z:|z-\alpha|\leq r$} contained in $D$, and $C$ denotes the disc's boundary followed in the positive direction, then for every $z$ in the disc we can write:
$$f(z)=\frac{1}{2\pi i}\int\frac{f(\zeta)}{\zeta-z}d\zeta$$

My question is:
What is the intuitive explanation of this formula? (For example, but not necessary, geometrically.)

(Just to clarify - I know the proof of this theorem, I'm just trying to understand where does this exact formula come from.)

答:
As Chandru1 pointed out, a beautiful particular case of Cauchy's integral formula gives us a very particular winding number. For the sake of simplicity, let us look only at closed paths on the boundary of the unit disc, $\gamma : [0,1] \longrightarrow S^1$. Then, the winding number of $\gamma$ is the number of times $\gamma$ goes round the circumference. That is, if we write the complex number $\gamma (t)$ in its exponential form

$$
\gamma (t) = e^{i\theta (t)} \qquad \qquad \qquad  [1]
$$

then you can prove that there is a continuous choice for the argument $\theta (t)$ of $\gamma (t)$; that is, a continuous function $\theta : [0, 1] \longrightarrow \mathbb{R}
$ such that [1] holds for all $t$. In Algebraic Topology, $\theta$ is called a "lifting" of $\gamma$. Moreover, any two such continuous "liftings" of $\gamma$ differ necessarily by a constant integer multiple of $2\pi$: if $\widetilde{\theta}(t)$ is another (continuous) lifting of $ \gamma$, then there is a constant integer $k$, not depending on $t$, such that $\widetilde{\theta}(t) = \theta (t) + 2k\pi$ for all $t$. (This is called the "lifting lemma" in Algebraic Topology. For a proof in Complex Analysis, see theorem 7.1, in Stewart and Tall.) So the number of times $\gamma $ goes round the circumference (its winding number) is well-defined as

$$
w (\gamma , 0) = \frac{\theta (1) - \theta (0)}{2\pi} \ .
$$

Then you can easily prove (see op.cit, section 7.5), that

$$
w (\gamma , 0 ) = \frac{1}{2\pi i}\int_\gamma \frac{1}{\zeta}d\zeta \ .
$$

So, Cauchy's integral formula for the constant function $f \equiv 1$, $z = 0$ (and $\gamma (t) = e^{i2\pi t}$, hence $\theta (t) =2 \pi t$) tells us that

$$
\frac{1}{2\pi i}\int_\gamma \frac{1}{\zeta}d\zeta = 1 \ .
$$

That is, if you go round the circumference just one time, well, you are going round the circumference indeed exactly once. Isn't math amazing? :-)

ContentLicense: CC BY-SA 2.5


问:
Motivation for the Mapping Class Group
Question: What is the motivation for studying the mapping class group?  In particular, what types of questions does it attempt to answer and what kind of invariant is it?

Motivation for this Question: Recently I've seen a number of references to things which assume knowledge of the mapping class group.  I've attempted to page through Dan Margalit and Benson Farbs' book on the subject as well as check out the wikipedia page for it, but both sources give the motivation as just "what it is" as opposed to, bluntly, "why should I care about this."

答:
The mapping class group (in genus $g$) is the fundamental group of the moduli space of compact Riemann surfaces of genus $g$.  Indeed, the latter space
is the quotient of a contractible space (Teichmuller space) by the mapping class
group.  Therefore a lot of the geometry of this moduli space is encoded in the
mapping class group.  </p>

<p>This is explained in the first one or two pages of the first chapter of (version 5.0 of) Farb and Margalit's book.  </p>

Title: Motivation for the Mapping Class Group

Body:
The mapping class group (in genus $g$) is the fundamental group of the moduli space of compact Riemann surfaces of genus $g$.  Indeed, the latter space
is the quotient of a contractible space (Teichmuller space) by the mapping class
group.  Therefore a lot of the geometry of this moduli space is encoded in the
mapping class group.

This is explained in the first one or two pages of the first chapter of (version 5.0 of) Farb and Margalit's book.


问:
Communicating mathematics
As a TA, my ratings have been fairly mediocre.  There's nothing to suggest I'm a problem, but I have room to improve.

What's more disturbing to me is that I even struggle to communicate math with people who know a lot more than I do.  It's no surprise that they struggle to communicate with me.  They know more, probably forgot what it's like to know less, and generally do not give the details on everything, sometimes even neglecting to mention that a particular detail needs to be filled in.  But when I communicate, I try to leave out as few details as possible and clearly enunciate when I am leaving them out, and to minimize the number of mistakes I make.  I am extremely nitpicky about this.

So there must exist other obstacles to communication other than just misrepresentations.  I suspect that what I'm lacking is a pedagogical element, which would also explain my interaction with students.  How can I go about finding out what is lacking, and then fixing it?  What are some common teaching mistakes made by people with the attitude that mathematics should be precise and airtight?  It's not that I'm such an adherent of my own philosophy that I refuse to "dumb things down."  That used to be the problem, but it appears that when I attempt to convey vague ideas, people do not understand me.

答:
A fairly random example of a great pedagogical technique for mathematics - examples first.

For example, suppose I was trying to explain to you what a ring is in algebra.  I could start by telling you that a ring is a set $R$ together with two binary operations $+$ and $.$ such that:

$r+s=s+r$ and $(r+s)+t$=$r+(s+t)$ for all $r,s,t\in R$

There exists $0\in R$ such that $r+0=r$ for all $r\in R$

etc. ...

However, you are much more likely to understand what a ring is if I give you some concrete examples.  For example, I could say, 'Rings are like $\mathbb{Z}$.  If we consider the integers, then we have two important operations $+$ and $\times$, which have the following interesting properties...'  Then I could point out that other interesting sets like the integers $\mod n$, which also have two operations which behave in a similar way.  Only then would I give you the formal axiomatic definition of a ring.  The next step would be showing that many interesting things that are true for the integers are true for all rings, and that many other interesting things about the integers (like unique factorization) are true for rings with certain properties.  Then you would gain some understanding of what rings are and why they are worth introducing.

Similarly, I don't think anyone would find the following definition at all meaningful if they hadn't studied topology before:

A topological space is a set $X$ together with a collection $\tau$ of subsets of $X$ (called the open sets) such that:

$\varnothing,X\in\tau$.

$\tau$ is closed under taking unions: for all (possibly infinite) collections of sets $(U_\alpha)_{\alpha\in A}$ with $U_\alpha\in\tau$, the union $\bigcup_{\alpha\in A}U_\alpha$ is in $\tau$.

$\tau$ is closed under taking finite intersections: for all $A, B\in\tau$, $A\cap B\in\tau$ (and therefore all intersections of finitely many members of $\tau$ are contained in $\tau$).

Much better is to start off by introducing the more concrete idea of metric spaces (by first showing that a lot of concepts in real analysis, like convergence and continuity, can be expressed entirely in terms of the distance between points, and showing that more abstract ideas of distance can be useful) and then showing that the definition of a continuous function between metric spaces can be expressed entirely in terms of the open sets in that space, and then introducing topology from there.

Having given you some examples, I'll now state the general principle: if you want to explain some mathematics to someone, always start by telling them some motivating examples.  This serves two purposes.  Firstly, it shows them why they should care about the idea you're introducing, since all mathematical concepts were originally created because they were interesting in some way.  Secondly, it helps them to understand the ideas that you're telling them about, since they have some concrete examples to link them to in their mind.

Of course, that's just one useful pedagogical technique, and it's not necessarily useful in all contexts.  But it is often very useful.

For a much better exposition of this idea, see this blog post by Timothy Gowers.

Title: Communicating mathematics