问:
Arbitrarily large sized gaps in the $L$ hierarchy
Question: Assume $V=L$. Let $\xi < \omega_1$ be arbitrary. Then there is $\alpha < \omega_1$ such that $L_\alpha \cap P(\omega) = L_{\alpha+\xi} \cap P(\omega)$.

Here's my idea:

When $V = L$, $L_{\omega_2} \vDash ZFC-PowerSet$ (since when $V=L$ the $L$ and $H$ hierarchy coincide for uncountable cardinals (basically the proof of $GCH$)).

Since $L_{\omega_2} \vDash ZFC-Powerset$ it can develop the notions of formulas, definability, constructibility, ordinal addition, etc, and will moreover be correct about it (i.e, it's absolute). And since  $P(\omega) \subseteq L_{\omega_1} \subseteq L_{\omega_2}$, we have that

$L_{\omega_2} \vDash L_{\xi + \omega_1} \cap P(\omega) = L_{\omega_1} \cap P(\omega)$

Then take a countable $X \prec L_{\omega_2}$ such that $trcl(\{\xi\}) \subseteq X$ and apply the Mostowski collapse on $X$. By the condensation theorem this is $L_\beta$ for some $\beta \in \omega_2$. $\beta \in \omega_1$ since $L_\beta$ is countable. Moreover $\pi(\xi) = \xi \in L_\beta$ because $X$ contains the transitive closure. Finally $\omega_1$ is definable in $L_{\omega_2}$ (as the least uncountable ordinal) and thus $\omega_1 \in X$. Let $\alpha = \pi(\omega_1) \in L_\beta$. $\alpha < \omega_1$ So we then have:

$L_\beta \vDash L_{\xi + \alpha} \cap P(\omega) = L_{\alpha} \cap P(\omega)$

And again $L_\beta \vDash ZFC-Powerset$ and is absolute about the $L$-hierarchies, etc. So then we have that $L_{\xi + \alpha} \cap P(\omega) = L_\alpha \cap P(\omega)$ as desired.

I think this is mostly fine except for the absoluteness/correctness parts which I handwaved. I am usually uncomfortable and unsure when I appeal to absoluteness (feels too handwavey) like above so I just wanted to see if I made any mistakes. I think I'm mostly fine about the absoluteness of the definability and constructibility stuff. However, I'm not sure about the complexity and absoluteness of ordinal addition. Is it absolute?

答:
Yes, your proof is right. Ordinal addition is absolute: the ordinal successor operation is absolute, unions are absolute, and a transfinite recursion defined from absolute operations is absolute.

A couple extra things:

Your question about absoluteness of ordinal addition suggests a generalization, since this is pretty much the only aspect of ordinal addition you used. Let $F(\alpha,\xi)$ be any absolute operation on the ordinals with $F(\alpha, \xi)\ge \alpha$. Then for any $\xi<\omega_1$ there is an $\alpha<\omega_1$ such that $L_{F(\alpha,\xi)}\cap P(\omega) = L_{\alpha}\cap P(\omega).$

Even though this is under the assumption of V=L, it can be rephrased without it and the proof is similar: For any $\xi < \omega_1^L$ there is an $\alpha < \omega_1^L$ such that $L_{\alpha+\xi}\cap P(\omega) = L_\alpha\cap P(\omega).$

Also note that by just increasing how many ordinals we put in the elementary submodel, we can replace "there is an $\alpha<\omega_1$" with "there are unboundedly many $\alpha < \omega_1$".


问:
An approximation for $\left(1-{1\over x}\right)^{2y} - \left(1-{2\over x}\right)^y$
For large $x$ it is common to use the approximation $$\left(1-{1\over x}\right)^{x} \approx e^{-1}$$ leading to $$\left(1-{1\over x}\right)^{y} \approx e^{-y/x}$$ and so $$\left(1-{2\over x}\right)^y\approx e^{-2y/x}\approx \left(1-{1\over x}\right)^{2y}$$

But is there a simple approximation for the difference
$$\left(1-{1\over x}\right)^{2y} - \left(1-{2\over x}\right)^y \approx \,?$$

When $x=y$ the difference is about $\frac{e^{-2}}{y}$ but the pattern seems less obvious when they are not equal.

答:
We have

$$\left( 1+ \frac{a}{x}\right)^x = e^a \left( 1 - \frac{a^2}{2x} +O(x^{-2}) \right)$$

Using this, assuming $y/x = a$ (constant) we get

$$\left(1-{1\over x}\right)^{2y} - \left(1-{2\over x}\right)^y \approx e^{-2 a} \frac{a}{x}=e^{-2 y/x} \frac{y}{x^2} $$

ContentLicense: CC BY-SA 4.0


问:
Tychonoff theorem motivation
I am studying for a master's degree and I have to choose a topic and finally I decided to choose. Tychonoff Theorem in General Topology.
Here I studied about Tychonoff Theorem and also I found some easy proof of this theorem. But I am looking for motivation behind this theorem and what is the idea of  this theorem. If I would found some interesting application regarding masters level then it'll be great.
Currently I am following Munkres Topology , Ryszard Engelking and Armstrong basic topology.

Any help will be appreciated.

Thanks in advance!

答:
It would seem that what led Tychonoff to his product theorem was primarily the question of exactly which (Hausdorff) spaces admit nice compactifications.

It is this question which is the focus of his 1930 paper Über die topologische Erweiterung von Räumen where a partial answer was supplied. The main theorem of the paper loosely reads as follows.

For each cardinal $\tau$ there exists a compact Haudorff space $R_\tau$ of weight $\tau$ with the property that any normal space of weight $\leq\tau$ embeds into $R_\tau$. When $\tau=\aleph_0$, the space $R_{\aleph_0}$ is homeomorphic to the Hilbert cube.

He points out that this implies a special case of the Urysohn Metrisation Theorem amongst other things.

The point is of course that the compact space $R_\tau$ is a $\tau$-weighted product of unit intervals, and the hard part of the paper appears in $\S2$, where Tychonoff proves that this product is indeed compact. This would be the very first and rather special case of what would later be called the Tychonoff Product Theorem.

It is amusing that the result here is of such secondary importance to the paper that I cannot even find a lemma or proposition statement to extract to highlight it. Neither is there any mention whatsoever of das Auswahlaxiom, despite the enthusiasm of other commenters here.

Something else which strikes me as odd is that Tychonoff seems to be constructing Stone-Čech type compactifications seven years before either of Stone's or Čech's papers were to be published. Indeed, it is this 1930 paper in which Tychonoff introduced the notion of a completely regular space, and his second main result is that

A space embeds as a subspace of a compact Hausdorff space if and only if it is completely regular.

So, it is all of this which seems to have been the motivation behind Tychonoff's Product Theorem.

The history from here I have not been so successful at tracking down. According to wikipedia Tychonoff pointed out in a 1935 paper of his that the construction given in Über die topologische Erweiterung von Räumen goes through to show that arbitrary products of compact spaces are compact. This would of course be his famous result. I have not been able to find the paper in question, so will decline to comment.

Rest assured, however, that a proof of Tychonoff's Theorem did finally emerge. Actually in the 1937 paper On Bicompact Spaces by Eduard Čech. It is on page 830 in the line starting The Cartesian product.... Here it is used by Čech to construct compactifications of completely regular spaces, and Tychonoff is accredited accordingly.

It would truly seem that the mathematics community did not come to recognise the importance of the Tychonoff Product Theorem until long after it had been accepted into the subject.

Title: Tychonoff theorem motivation


问:
If $N$ is a normal subgroup of semidirect product group of $N$ and $H$ where $H$ is generated by $a$ and $b$, is $ab=ba$?
I need help to get the idea of the presentation of semidirect products groups. I got this question from an example the $A_4$ is a semidirect product of $N=\langle a,b\rangle$ and $\langle (1 2 3)\rangle$.

If $N$ is a normal subgroup of semidirect product group of $N$ and $H$ where $N$ is generated by $a$ and $b$, is $ab=ba$? If so why?

I know if $H$ and $N$ are both normal and their intersection is trivial, then $hn=nh$ for all $h$ in $H$ and $n$ in $N$, but not sure why in semidirect case.

regards

答:
In the interest of adding a proper answer so that future users know how to spend their time:

The symmetric group $S_5 = C_2 \ltimes A_5$ is a counterexample, since
$A_5 = N = \langle(1,2,3), (1,2,3,4,5)\rangle$ yet is nonabelian.

I hope this helps ^_^


问:
Applying Ito's product rule to $X = tW$ and $Y = e^{W(t)}$
I am given $X = tW$ and $Y = e^{W}$ and am asked to compute $d(\frac{X}{Y})$. To me, this seems to be an application of Ito's product rule which I know to be:

$$d(X_1 X_2) = X_2 dX_1 + X_1dX_2 + G_1 G_2 dt$$ where $dX_1 = F_1dt + G_1dW$ and $dX_2 = F_2dt + G_2dW$

So following this, I write $$d(tW \cdot e^{-W}) = tW \cdot d(e^{-W}) + e^{-W}d(tW) +G_1G_2 dt$$

However, I am confused about computing $d(e^{-W})$ and $d(tW)$. The examples in my book give my a function $u(X,t)$, assume $X(\cdot) = W(\cdot) \implies dX = dW \implies F\equiv 0, G\equiv 1$, and apply Ito's chain rule:$$du(X,t) = (u_t + u_x F + \frac{1}{2}u_{xx}G^2)dt + u_x G dW$$ from $dX = Fdt + GdW$

Is $d(e^{-W}) = \frac{de^{-W}}{dt} + \frac{de^{-W}}{dW} = 0dt + -e^{-W}dW$? Or do we use the above chain rule somehow?

答:
Two facts we can use are, first we have the product rule given by $$d(f(t)G_t) = f(t)dG_t + G_tdf(t) + df(t)dG_t.$$ In this case of the rule $f(t)$ is a deterministic function. Second from Ito's formula we know if $F_t = f(W_t)$ we have $$dF_t = \frac{1}{2}f''(W_t)dt + f'(W_t)dW_t.$$

Now given $X_t = tW_t$ and $Y_t = exp(W_t)$. We want to find $dX_t$ and $dY_t$.

Using the product rule for $X_t$ we have
$$dX_t = W_tdt + tdW_t + dtdW_t = W_tdt + tdW_t.$$

Then using the fact from Ito's formula we have
$$dY_t = exp(W_t)dW_t + \frac{1}{2}exp(W_t)dt.$$


问:
Lebesgue measurable simple function and its Borel measurable counterpart
Q) Suppose that $f$ is a simple Lebesgue measurable function on $R^n$. Show that
there exists a Borel measurable function $g$ such that $f(x) = g(x)$ for almost every $x \in R^n$.

I know that a simple Lebesgue measurable function takes finitely many values on Lebesgue measurable sets. But I'm not sure how I can find an appropriate Borel measurable function $g$, i.e. function wherein the pre-image of every open set is Borel measurable, which is the same as $f$ except atmost on a countable set?

答:
Here is a proof for arbitrary Lebesgue measurable $f$. As the other answer shows, $A=F\cup N$ for some Borel set $F$ and a null set $N$. Let $U\in \tau_{\mathbb R}.$ Then the fact that $f$ is Lebesgue measurable implies that $f^{-1}(U)=V$ is Lebesgue measurable and so has the form $F\cup N$, as above.

Let $\{I_n\}_{n\in \mathbb N}$ be an enumeration of the collection of intervals with rational endpoints, which form a base for $\tau_{\mathbb R}.$
Then, there are Borel sets $\{F_n\}_{n\in \mathbb N}$ and null sets $\{N_n\}_{n\in \mathbb N}$ such that $f^{-1}(I_n)= F_n\cup N_n.$ Furthermore, $U=\bigcup^\infty_{k=1}I_{n_k}$ for some subsequence. Then, $f^{-1}(I_{n_k})=F_{n_k}\cup N_{n_k}.$

Define $g:\mathbb R^n\to \mathbb R$ by $g(x)=f(x)$ if $x\in F_n$ for some $n$  and $g(x)=0$ otherwise. Then, $g$ agrees with $f$ except perhaps on $\bigcup_n N_n$, which is null. And $g^{-1}(U)=g^{-1}(\bigcup^\infty_{k=1}I_{n_k})=\bigcup^\infty_{k=1}F_{n_k}\cup g^{-1}(\{0\}),$ which is a Borel set. It follows that $g$ is Borel measurable.

Remark: if you know that $\mathscr B(\mathbb R)$ is generated by the intervals $\{(-\infty, r):r\in \mathbb Q\}$, then you can show that it suffices to prove the claim for these intervals only. Then the proof is much easier: For each $r\in \mathbb Q,$ choose a Borel set $B_r\subseteq f^{-1}((-\infty,r))$ such that $f^{-1}((-\infty,r))=B_r\cup N_r$ for some null set $N_r$. Now define $g(x)=f(x)$ if $x\in B_r$ for some $r$ and $0$ otherwise. Then, $g^{-1}((-\infty,r))=B_r\cup g^{-1}(\{0\}).$


问:
A Sequence is Square Summable if and only if....
Let $\{z_k\}_{k=0}^{\infty}$ be a sequence of complex numbers. Is it true that $\sum_{k=0}^{\infty} \frac{2k^2}{4k^2 + 8k +3} |z_k|^2 < \infty$ if and only if $\sum_{k=0}^{\infty} |z_k|^2 < \infty$. I'm not really sure how to approach this problem? I was able to reduce a way more complicated problem to determining whether this is true, but I don't remember much about series.

EDIT:

I ran the sum through Wolframalpha for $z_k = \frac{1}{k}$ and $z_k = \frac{1}{2^k}$, and it converges in both cases.

答: Hint: You can check that $$\dfrac{2}{15} \le \dfrac{2k^2}{4k^2+8k+3} \le \dfrac{1}{2}$$ for all integers $k \ge 1$. Try using this fact along with the comparison test.


问:
Simple application of Dominated convergence theorem
Q) Let $\mu$ be a positive measure defined on the Borel sigma algebra of $\mathbb{R}$. Show that if $f\in L^1(\mu)$, then

$$\int_{\mathbb{R}} fd\mu = \lim_{R\to \infty}\int_{[-R,R]} fd\mu$$

My attempt: I mean $$\int_{[-R,R]} fd\mu = \int f.1_{[-R,R]}d\mu$$

and somehow the limit is taken inside. $f.1_{[-R,R]}$ need not be dominated by $f$ so I don't know how to find an integrable function that dominates $f.1_{[-R,R]}$ and not sure how to use that $\mu$ is a positive measure?

答:
Hint:

You have the exact right idea! Remember what the DCT says

If $f_n \to f$ and for some $g \in L^1$, each $|f_n| \leq g$, then
$\lim \int f_n d\mu = \int f d\mu$

Now we use your idea of taking $f_n = f \cdot \chi_{[-n,n]}$. All that's left is to show that $|f_n| \leq g$ for some $g \in L^1$. Do you see why?

Of course, we know that $|f_n| \leq |f|$... Can you show that $|f| \in L^1$? Can you finish the proof from here?

I hope this helps ^_^


问:
$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$ Proof
Does anyone know hot to prove this inequality?

Having: $a, b, c \gt 0$

$$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$$

I tried with the AM-GM inequality but I couldn't get any improvement. I'm on a still point and I don't know how to continue.

Furthermore, I don't know how to get rid of the known term. I tried with the AM-HM inequality but I'm still not getting any results.

Also, just some hints would be appreciated, thanks

答:
Use general Schur's inequality to get (if $a\geqslant b\geqslant c$)
$$
\sum_{cyc}{\frac{(b+a)(b-a)}{c+a}} = \sum_{cyc}{\frac{(b-a)(b-c)}{c+a}} \geqslant 0
$$
For the other case (assuming $c$ is minimal), make substitution
$p$ for $b$,
$q$ for $a$ and
$r$ for $c$. Rest expressions are the same.


问:
Is my Expectation for $Y = \frac{1}{1+ e^{aX}}$ for $X\sim\exp (1)$ correct?
If $X\sim \exp(1)$, for $x \in \{0,+\infty \}$, and $Y = \frac{1}{1 + e^{aX} }$, where $a>0$ is a constant. (So the limit of Y becomes $y \in \{0, 1/2\}$) - Is it correct?

CDF $F_Y(y)$ of Y is :

$$F_Y(y) = \mathbb{P}(Y\leq y)= \Big(\frac{1}{1 + e^{aX}} \leq y \Big)$$

after mathematical manipulations, I get

$$F_Y(y) = \frac{1}{(y-1)^{1/a}}$$

So the PDF is:

$$f_Y(y) = \frac{1}{a}(y-1)^{-\frac{a-1}{a}}$$

The expected value of Y is now

$$\mathbb{E}[Y] = \int_0^{1/2} y\times\frac{1}{a}(y-1)^{-\frac{a-1}{a}} dy $$

$$\mathbb{E} = \dfrac{a\left(2a\mathrm{e}^\frac{\ln\left(-\frac{1}{2}\right)}{a}+\left(1-2a\right)\mathrm{e}^\frac{\ln\left(-1\right)}{a}\right)\mathrm{e}^{-\frac{\ln\left(-\frac{1}{2}\right)}{a}-\frac{\ln\left(-1\right)}{a}}}{2\left(a-1\right)}$$

答:
It is more direct to compute $$\begin{align}
\operatorname{E}[Y] &= \operatorname{E}[(1+e^{aX})^{-1}] \\
&= \int_{x=0}^\infty \frac{1}{1+e^{ax}} f_X(x) \, dx \\
&= \int_{x=0}^\infty \frac{e^{-x}}{1+e^{ax}} \, dx \\
&= \int_{x=0}^\infty e^{-x} \sum_{k=1}^\infty (-1)^{k-1} e^{-k a x} \\
&= \sum_{k=1}^\infty (-1)^{k-1} \int_{x=0}^\infty e^{-(ka+1) x} \, dx \\
&= \sum_{k=1}^\infty \frac{(-1)^{k-1}}{ka+1},
\end{align}$$ for $a > 0$.  If $a < 0$, then the series expansion is modified accordingly.


问:
Solve $\sin(z) = z$ in complex numbers
Show that $\sin(z) = z$ has infinitely many solutions in complex numbers.
Little Picard theorem should help, but using big Picard theorem is undesirable.
Thanks a lot!</p>

答: Using a non trivial result by Picard: $\sin(z) - z$ has an essential singularity at $\infty$.  Therefore it attains all values infinitely many times with at most one exception.  Since $$\sin(z + 2\pi) - (z+2\pi) = \sin(z) - z - 2\pi$$ one can shift the image over multiples of $2\pi$.  By Picard's theorem there cannot be any value (including $0$) that is only attained a finite number of times since only a single exception is allowed.

ContentLicense: CC BY-SA 3.0

Title: Solve $\sin(z) = z$ in complex numbers


问:
Orthogonal complement and shortest distance on $L^{2}[-1,1]$
Consider the subset $A\subset L^{2}[-1,1]$ defined by
$$
  A=\left\{f\in L^{2}[-1,1]:\int_{-1}^{1}f(t)dt=0\right\}
$$
I want to find the set $A^{\perp}=\{g\in L^{2}[-1,1]:(f,g)=0\}$ for all $f\in A$ and $(,)$ is the inner product on $L^{2}$. Also I want to know what is the shortest distance between $t^{2}$ and $A$.

My attempt to find $A^{\perp}$: Let $f\in A$ fixed and let $g\in L^{2}[-1,1]$, then, using Cauchy-Schwarz inequality I get
$$
 0\leq (f,g)\leq \|f\|\|g\|
$$
So I conclude that $A^{\perp}=\left\{g\in L^{2}[-1,1]:\int_{-1}^{1}|g(t)|^{2}dt=0\right\}$, am I correct?

I have troubles finding the shortest distance between $t^{2}$ and $A$, how can I calculate it?

答:
First notice that any constant function is in $A^\perp$.

Take any $g \in A^\perp$. Then $(f,g) = 0$ for all $f \in A$. Set $\overline g = \frac  12 \int^1_{-1} g(t) dt$, and take $$f(x) = g(x) -  \overline g, \,\,\,\,x\in [-1,1].$$ Then $f \in A$ (you should check this), and so $$0 = (f,g) = (f,g) - (f,\overline g) = (f,g-\overline g) = (g-\overline g, g-\overline g) = \| g - \overline g\|_2^2.$$ Thus $g = \overline g$, and so $g$ is constant. Thus $A^\perp$ is precisely the constant functions.

To find the closest constant $c$ to $h(t) = t^2$ simply minimize $$\phi(c) = \int^1_{-1} (t^2 - c)^2 dt = \frac{2}{5} - \frac{4c}{3} + 2c^2$$ using a calculus argument: $$\phi'(c) = 4c - 4/3 \,\,\,\,\, \implies \,\,\,\,\, c = 1/3.$$ This isn't a coincidence, the best constant approximation to $h(t) = t^2$ is the average value $\overline h = \frac 1 2\int^1_{-1} h(t) dt = 1/3$. The distance is then $$\sqrt{\phi(1/3)} = \sqrt{\frac 8 {45}}.$$


问:
Find the steady state matrix for Markov Chains
Question: Let $P$ be a transition matrix.  Suppose that

$$P^{k} =\begin{bmatrix}0.37 & 0.37 & 0.37\\0.33&0.33&0.33\\ 0.3&0.3&0.3\end{bmatrix}$$

As $k$ approaches infinity, if $x_k$ are state vectors and $x_0$ is a probability vector, then what does $x_k$ converges to?

My attempt: Since $Px=x$, we know that $(P - I_3)=x$. By subtracting $P$ with the identity matrix, I got:

$$(P - I_3) = x = \begin{bmatrix}-0.63&0.37&0.37\\ 0.33&-0.67&0.33\\ 0.3&0.3&-0.7\end{bmatrix}$$

For the next step, I generally find the null space of the matrix, which provide me with the info to find $x_1$, $x_2$, and $x_3$. But in this case, solving for the reduced row echelon form of the matrix is nearly impossible and I'm not really sure how I should approach this problem.

Can someone please show me how this is done? Thanks in advance

答: The steady state distribution is $(0.37,0.33,0.3)$ because for any distribution $x$, $Px$ is this distribution.


问:
Recursive sequence depending on the parameter

For the given parameter $\mathbb R\ni t\geq 1$, the sequence is
  defined recursively: $$a_1=t,\;\;a_{n+1}a_n=3a_n-2$$ $(a)$ Let $t=4$.
  Prove the sequence $(a_n)$ converges and find its limit.

$(b)$ Which parameters $t\geq 1$ is the sequence $(a_n)$ increasing
  for?

My attempt:

Bolzano-Weierstrass:A sequence converges if it is monotonous and
   bounded

$$a_{n+1}a_n=3a_n-2\implies a_{n+1}=3-\frac{2}{a_n}$$
$(a)$

First few terms: $a_1=4,a_2=\frac{5}{2},a_3=\frac{11}{5}$

Assumption: the sequence is decreasing

Proof by induction:
the basis (n=1) is trivial: $\frac{5}{2}<4$

Assumption: $a_n<a_{n-1},\;\forall n\in\mathbb N$

Step: $$a_n<a_{n-1}\implies\frac{1}{a_n}\geq\frac{1}{a_{n-1}}\Bigg/\cdot(-2)$$
$$\iff-\frac{2}{a_n}\leq-\frac{2}{a_{n-1}}\iff \underbrace{3-\frac{2}{a_n}}_{a_{n+1}}\leq\underbrace{3-\frac{2}{a_{n-1}}}_{a_n}$$
The limit: $$L=3-\frac{2}{L}\implies L^2-3L+2=0$$
I take into account only $2$ because the parabola is convex and $$a_n\to L^-.$$
Then I have to prove: $a_n\geq 2\;\forall n\in\mathbb N$ after the formal computing: $a_{n+1}\geq 3-\frac{2}{2}=2$
$\underset{\implies}{\text{Bolzano-Weierstrass theorem}}(a_n)\to 2$

$(b)$ Since the sequence doesn't have to be convergent, only increasing:
$$a_2=3-\frac{2}{t}\geq t\implies t\in[1,2]$$
Then, it should follow inductively,analogously to $(a)$, this time it is increasing.
Is this correct?

答:
If you have already proven $a_n > 2, \forall n$ , then

Let $b_n=a_n-2$, $b_n>0$. $a_{n+1} = 3 - \frac{2}{a_n} \Rightarrow b_{n+1} + 2=3-\frac{2}{b_n+2} \Rightarrow b_{n+1} = \frac{b_n}{2+b_n} < \frac{b_n}{2}$.

Therefore as $n\to \infty, b_n \to 0, a_n \to 2.\blacksquare$

The next is pure hindsight based on dan_fulea's solution but I believe it can be useful when the two fixed points are distinct.

$a_{n+1} - 1 = 2-\frac{2}{a_n} = \frac{2(a_n - 1)}{a_n}$

$a_{n+1} - 2 = 1-\frac{2}{a_n} = \frac{a_n - 2}{a_n}$

Therefore $\frac{a_{n+1}-1}{a_{n+1}-2} = 2\cdot \frac{a_n-1}{a_n-2}$

$\frac{a_n-1}{a_n-2}$ is a geometric sequence with initial value $\frac{3}{2}$,

so $\frac{a_n-1}{a_n-2} = 2^{n-1} \frac{3}{2} = 1+\frac{1}{a_n-2} \Rightarrow a_n = 2+ \frac{1}{2^{n-1}\frac{3}{2}-1} = \frac{6\cdot 2^{n-1}-2}{3\cdot 2^{n-1}-2}.\blacksquare$

In general if there are two distinct fixed points $r$ and $s$ then the ratio $\frac{a_n-r}{a_n-s}$ is a geometric sequence.


问:
Interpreting a lemma on coset representatives
I have been reading through "Number Fields" by Marcus and I came across the following lemma on page 115:

Let $f:G \rightarrow G'$ be a homomorphism of abelian groups and let $S$ be a subgroup of $G$ which is carried isomorphically onto a subgroup $S'$ in $G'$. Suppose that $D'$ is a set of coset representatives of $S'$ in $G'$. Then its total inverse image $D = f^{-1}(D')$ is a set of coset representatives for $S$ in $G$.

I am a bit stuck on trying to prove this myself and the primary reason is that I am having trouble understanding the statement. I have interpreted this lemma to mean that the pre-image of $D'$ through the function $\overline{f}:G/S \rightarrow G'/H'$ via $\overline{f}(gS) = f(g)S'$ is the total inverse image mentioned and that every element of the pre-image of $D'$ lies in $G/S$. If this is the case then the result appears to be clear (although perhaps I ought to prove that $\overline{f}$ is well-defined which shouldn't be difficult) since $\overline{f}$ is a function defined on a set of coset representatives of $S$ in $G$.

Explicit questions - Have I correctly interpreted the lemma? How would one prove this lemma?

答:
The “preimage of $D’$” is just the usual pre-image, no need to factor through any quotient or induced function. That is,
$$D = \{d\in G\mid f(d)\in D’\}.$$

I’ll use additive notation, since the groups are abelian.

To show $D$ is a complete set of coset representatives for $S$ in $G$, let $g\in G$. Then $f(g)$ is in a coset of $S’$, so there exists $d’\in D’$ such that $f(g)\in d’+S’$. therefore, there exists $s’\in S’$ such that $f(g)=d’+s’$. Since $f$ maps $S$ isomorphically into $S’$, there exists a (unique) $s\in S$ such that $f(s)=s’$. Thus $f(g)=d’+f(s)$, so $d’=f(g)-f(s)=f(g-s)$. Thus, $g-s\in D$. And clearly, $g\in (g-s)+S$. Thus, every element of $G$ is in a coset of the form $d+S$ for some $d\in D$.

Now assume that $d_1,d_2\in D$ are such that $d_1+S = d_2+S$. Then there exists $s\in S$ such that $d_1=d_2+s$. That means that $f(d_1) = f(d_2)+f(s)\in f(d_2)+S’$, hence $f(d_1)+S’=f(d_2)+S’$. But that means that $f(d_1)=f(d_2)$ (since both $f(d_1)$ and $f(d_2)$ are in $D’$, which is a complete set of coset representatives for $S’$). Therefore, $f(s)=0$ (since $f(d_1)=f(d_2)+f(s)$). Since $f$ is one-to-one when restricted to $S$, we must have $s=0$, so $d_1=d_2+s=d_2$. Thus, distinct elements of $D$ correspond to distinct cosets of $S$ in $G$.

Hence, every element is in a coset of the form $d+S$ with $d\in D$, and distinct elements of $D$ are incongruent modulo $S$. Thus, $D$ is a complete set of coset representatives for $S$ in $G$, as claimed.


问:
Prove $\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}} \ln \tan x\,\,dx= - \frac{2G}{5}$
Context: This question
asks to calculate a definite integral which turns out to be equal to $\displaystyle 4 \, \text{Ti}_2\left( \tan \frac{3\pi}{20} \right) -
4 \, \text{Ti}_2\left( \tan \frac{\pi}{20} \right),$ where $\text{Ti}_2(x) = \operatorname{Im}\text{Li}_2( i\, x)$ is the
Inverse Tangent Integral function.

The source for this integral is this question on brilliant.org.

In a comment, the OP
claims that the closed form can be further simplified to $-\dfrac\pi5 \ln\left( 124 - 55\sqrt5 + 2\sqrt{7625 - 3410\sqrt5} \right) + \dfrac85 G$.

How can we prove that?

I have thought about using the formula $$\text{Ti}_2(\tan x) = x \ln \tan x+ \sum_{n=0}^{\infty} \frac{\sin(2x(2n+1))}{(2n+1)^2}. \tag{1}$$
but that only mildly simplifies the problem.

Equivalent formulations include:

$$\, \text{Ti}_2\left( \tan \frac{3\pi}{20} \right) -
 \, \text{Ti}_2\left( \tan \frac{\pi}{20} \right) \stackrel?= \frac{ \pi}{20} \ln \frac{  \tan^3( 3\pi/20)}{\tan ( \pi/20)}   + \frac{2 G}{5} \tag{2}$$

$$ \sum_{n=0}^{\infty} \frac{\sin \left(\frac{3\pi}{10}(2n+1) \right)- \sin \left(\frac{\pi}{10}(2n+1)\right)}{(2n+1)^2} \stackrel?=\
\frac{2G}{5} \tag{3}$$
$$\int_{\pi/20}^{3\pi/20} \ln \tan x\,\,dx \stackrel?= - \frac{2G}{5} \tag{4}$$

A related similar question is this one.

Body:
Let $J(a)=\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}}\tanh^{-1}\frac{2a\cos2x}{1+a^{2}}dx$ and evaluate

\begin{align}
J’(a) &= \int_{\frac\pi{20}}^{\frac{3\pi}{20}}\frac{2(1-a^2)\cos2x}{a^4+1-2a^2\cos4x}dx=\frac1{2a}{\left.\tan^{-1}\frac{2a\sin2x}{1-a^2}\right|_{\frac{\pi}{20}}^{\frac{3\pi}{20} } }\\
&=\frac1{2a}\tan^{-1}\frac{a-a^5}{1+a^6}
= \frac1{2a}(\tan^{-1}a - \tan^{-1}a^5)
\end{align}
where $\sin\frac{3\pi}{10}-\sin\frac{\pi}{10}=\frac12$ and $\sin\frac{3\pi}{10}\sin\frac{\pi}{10}=\frac14$ are recognized. Then
\begin{align}
\int_\frac\pi{20}^\frac{3\pi}{20}\ln(\tan x)~dx
&= -\int_\frac\pi{20}^\frac{3\pi}{20}\tanh^{-1}(\cos2x)dx =-J(1)
=-\int_0^1 J’(a)da \\
& =-\frac12 \int_0^1\left(\frac{\tan^{-1}a}{a}\right.
-\underset{a^5\to a}{\left.\frac{\tan^{-1}a^5}{a}\right)}da\\
&=-\left(\frac12-\frac1{10}\right) \int_0^1\frac{\tan^{-1}a}{a}da=-\frac25 G
\end{align}


问:
Is natural projection a natural transformation?
Given a group $G$ and its normal subgroup $N$, there is a natural projection $\pi: G\to G/N,\ g\mapsto gN$.

Question: Is 'natural projection' a kind of natural transformation?

I know determinant is a natural transformation between $GL_n(-)$ and $(-)^*$, the functors $\bf\text{CRing} \to \bf\text{Grp}$. There is also a natural transformation between the identity functor of $\bf\text{Grp}$ and the factor-commutator functor $(-)/[-,-]$. However, I can't figure out which functors the 'natural projection' relates to.

I have just begun learning category theory in order to gain a better understanding to some concepts in algebra, and I'm totally unfamiliar with the common techniques in category theory.

答:
Yes. It’s a natural transformation between the functors $G$ and $G/N$, whose source is the category whose

objects are pairs $(G, N)$ consisting of a group and a normal subgroup, and whose

morphisms are commutative squares.

The target is the category of groups.


问:
If all lines on a sphere converge, what happens to latitude lines?
I mean, on an image like this, how do latitude lines not converge? Everywhere i read says that there can be no parallel lines on a sphere. Is this related to the different definition of 'line' in spherical geometry?

Thanks in advance.

答:
All straight lines indeed converge on a sphere. But contrary to longitude lines for example, which are straight lines (great circles) and do converge at the poles, latitude lines are a different kind of line: small circles, meaning they aren't actually straight (great circle) lines (except the Equator).

If you were to walk along a latitude line which isn't the Equator, you would have to turn in order to follow the line. Taking this example to the extreme, imagine you're walking near the North Pole at 89°N. To keep at 89°N and follow the latitude line eastward, you'd have to slowly turn left, essentially circling the pole.

So this special geometry of latitude lines allow them to be parallel to each other, because they aren't great circles or straight lines.


问:
Problem related with a linear transformation and its rank,nullity...
I am stuck on the following problem:

Suppose $W$ be  a vector space of dimension $m \geq 2.$ Let $F \colon W \to W$ be a linear transformation such that $F^{n+1}=0$ and $F^n \ne 0$ for some $n \ge 1.$ Then I have to check which of  the following statement(s) is/are true?

1.$\text{trace}(F) \ne 0 $

2.$\text{Rank}(F^n) \le \text{Nullity}(F^n)$

3.$n=m$

4.$F$ is diagonalizable.

I do not know how to progress with the problem. Any kind of hints with some explanation will be helpful.

答:
Everything here can be solved with the use of Jordan normal form of a matrix.

So, if we want a non-zero nilpotent matrix $F$, we must write it as
$$F=T^{-1}JT,
$$
$T$ - some matrix with $det T\ne 0$. $J$ consists of Jordan cells, each of them has the view
$$J_l=\begin{pmatrix}\lambda&1&0&\dots\\0&\lambda&1&\dots\\\vdots&\ddots&\ddots&\ddots\\0&\dots&0&\lambda \end{pmatrix},$$
$\lambda$ is an eigenvalue.

It's easy to compute the powers of Jordan blocks and hence the powers of matrices. In particular, $(J_l)^k$ has $\lambda^k$ on its diagonal. Thus, in order to this block to be nilpotent, one must take $\lambda=0$. If we now consider the whole matrix $J$, it's easy to see that $tr(F)=0$, since it's invariant under changes of basis.

As Jordan normal form is unique (up to some changes, cf. wiki), the diagonal form of the matrix is its Jordan form. Since all eigenvalues are zero, diagonalisability would mean that the whole matrix is zero. Contradiction.

Next, it's quite easy to see that if the order of a Jordan block $J_l$ with zero eigenvalue is equal to $r$, then $(J_l)^r=0$; it's easy to conclude on (3).

Finally, $F^{n+1}=0$ and $F^{n}\ne 0$ mean that the maxim order of Jordan cells in Jordan decomposition of $F$ is equal to $n+1$. Let's say that we have $s$ cells of the order $n+1$ and the sum of dimensions of all other cells is $t$. Apparently, $$m=s(n+1)+t.$$
It's easy to see that $\text{rank}(F^n)=s$, $\text{nullity}(F^n) = ns+t$, which allows to conclude on (2).


问:
Maximal area under projectile trajectory
Problem:

A projectile is fired with an initial velocity of $v$ at an angle of $\theta$ from the ground. Then its trajectory can modeled by the parametric equations
\begin{align*}
x &= vt \cos \theta, \\
y &= vt \sin \theta - \frac{1}{2} gt^2,
\end{align*}where $t$ denotes time and $g$ denotes acceleration due to gravity, forming a parabolic arch.

Then the maximal area under the trajectory is given by $k \cdot \frac{v^4}{g^2}$ for some constant $k.$ Find $k.$

My Work:
I substituted the first equation for t into the second equation getting:
$$y = \frac{\sin\theta x}{\cos\theta} - \frac{gx^2}{2v^2(\cos\theta)^2}.$$
How do I proceed?

答:
Find the range. Set $y=0$. One solution is $x=0$, divide by $x$, and solve for the other root.

Integrate to find the area. $$A=\int _0^{x_{max}} y dx$$

Take the derivative with respect to $\theta$ and set it to $0$:
$$\frac{dA}{d\theta}=0$$

For the given $\theta$ calculate the area.


问:
Equivalent definition of Lebesgue integral
We define $\int f d \mu=\sup \{\mathcal{L}(f, P): P \text { is an } \mathcal{S} \text { -partition }\}$ where $\mathcal{L}(f, P)=\sum_{i=1}^{n} \mu\left(A_{i}\right) \cdot \inf _{A_{i}} f$ for an $\mathcal{S}$-partition $P=A_{1}, \ldots, A_{n}$.

I am looking to show that when $\mu(X)< \infty$ and $f$ is bounded then this definition is equivalent to using the upper sum $\mathcal{U}(f, P)=\sum_{i=1}^{n} \mu\left(A_{i}\right) \cdot \sup _{A_{i}} f$ with $\int f d \mu=\inf \{\mathcal{U}(f, P): P \text { is an } \mathcal{S} \text { -partition }\}$.

It was easy to prove that $\int f d \mu \leq \inf \{\mathcal{U}(f, P): P \text { is an } \mathcal{S} \text { -partition }\}$ by using the fact that $\mathcal{L}(f, P) \leq \mathcal{U}(f, P)$.

I am now left to prove $\int f d \mu \geq \inf \{\mathcal{U}(f, P): P \text { is an } \mathcal{S} \text { -partition }\}$.
I think this can be done by first proving it for when $f$ is simple and then generalizing but I am struggling. Any help would be much appreciated. Thanks in advance

答:
Note that $\mathcal{U}(f,P) = \sum_{i=1}^n \sup_{A_i}f\cdot \mu(A_i) = \int\psi_P$ where $\psi_P = \sum_{i=1}^n \sup_{A_i}f\cdot \chi_{A_i}$ is a simple function.

Since $\psi_P \geqslant f$, we have $\inf \{\int\psi:\psi \geqslant f, \psi  \text{ simple} \} \leqslant \mathcal{U}(f,P)$ and, it follows that taking the infimum over all partitions $P$, we have $\inf \{\int\psi:\psi \geqslant f, \psi  \text{ simple} \} \leqslant \inf_P\mathcal{U}(f,P)$.

On the other hand, if $\psi = \sum_{j=1}^m b_j \chi_{B_j}$ is the canonical representation of a simple function $\psi \geqslant f$, then $b_j \geqslant  \sup_{B_j} f$ and $\int \psi =  \sum_{j=1}^m b_j \mu(B_j) \geqslant  \sum_{j=1}^m \sup_{B_j} f\cdot \mu(B_j) = \mathcal {U}(f,P_\psi)  $ for some partition $P_\psi$. This implies that $\inf \{\int\psi:\psi \geqslant f, \psi  \text{ simple} \}  \geqslant \inf_P \mathcal{U}(f,P).$

Hence, $\inf \{\int\psi:\psi \geqslant f, \psi  \text{ simple} \}  = \inf_P \mathcal{U}(f,P)$ and by a similar argument we also have $\sup \{\int\varphi:\varphi \leqslant f, \varphi  \text{ simple} \}  = \sup_P \mathcal{L}(f,P)$.

Since $f$ is bounded, for any $\epsilon>0$ there exists, by the simple approximation lemma, simple functions $\varphi_\epsilon$ and $\psi_\epsilon$ such that $\varphi_\epsilon \leqslant f \leqslant \psi_\epsilon$ and $\psi_\epsilon - \varphi_\epsilon < \epsilon$.

Thus,

$$\int \varphi_\epsilon \leqslant \sup \{\int\varphi:\varphi \leqslant f, \varphi  \text{ simple} \}  = \sup_P \mathcal{L}(f,P) \\ \leqslant \inf_P \mathcal{U}(f,P) = \inf \{\int\psi:\psi \geqslant f, \psi  \text{ simple} \} \leqslant \int\psi_\epsilon,$$

and, for all $\epsilon > 0$,

$$0 \leqslant \inf_P \mathcal{U}(f,P) -\sup_P \mathcal{L}(f,P) \leqslant \int \psi_\epsilon - \int \varphi_\epsilon < \epsilon \mu(X)$$

Therefore,

$$\inf_P \mathcal{U}(f,P) = \sup_P \mathcal{L}(f,P) = \int f $$


问:
Diophantine equations of a certain form
Let us fix a rational number $c=\frac{a}{b}$ where (a) the square root of $c$ is irrational, and (b) $a$ and $b$ are both positive integers no larger than say, 10. [The integer $b$ may be 1.]

Does the diophantine equation

$$n^2 = cm^2 + 1 $$

have an infinite number of solutions i.e., an infinite number of pairs $(m,n)$ so that the above equation holds--for any such choice for $c$? I'd be really surprised if there were such a $c$ that yielded an infinite number of solutions, but I cannot prove or disprove either way.

If this is too broad a question, I would be quite happy if one could prove or disprove whether

$$n^2 = \frac{4m^2}{3} + 1 $$

has a bounded number of solutions.

答:
There’d be an infinite number of (integer) solutions once there is a single non-trivial (i.e., nonzero) solution, say $(m_0,n_0)$. Note that necessarily, in this case we must have $b|m_0^2$ (assuming $a$ and $b$ are coprime). To see this observe that you have
$$1=(n_0^2-cm_0^2)^2=(n_0^2+cm_0^2)^2-c(2m_0n_0)^2\,,$$
which gives another (larger, magnitude-wise) solution $$m_1:=2m_0n_0\,,\,~\,~\,~ n_1:= n_0^2+cm_0^2 \,.$$ Note that $n_1$ is an integer.

(Edit: It turns out that a nontrivial solution can be found by first simply solving the Pell equation $$n^2-abm^2=1\,.$$ Note that $ab$ cannot be a square. Granted a nontrivial solution to this, say $ (m’,n’)$, then a nontrivial solution to your problem becomes $(m_0,n_0):=(bm’,n’)$.))

For instance, for the example you gave—$3n^2=4m^2+3$—we have the non-trivial solution $$(m_0,n_0)=(6,7)\,,$$
which quickly leads to the infinitude of solutions via the above construction
$$(6,7)\,,(97,84)\,,(18817,16286)\,,\cdots\,.$$

Title: Diophantine equations of a certain form


问:
If $N$ is a normal subgroup of semidirect product group of $N$ and $H$ where $H$ is generated by $a$ and $b$, is $ab=ba$?
I need help to get the idea of the presentation of semidirect products groups. I got this question from an example the $A_4$ is a semidirect product of $N=\langle a,b\rangle$ and $\langle (1 2 3)\rangle$.

If $N$ is a normal subgroup of semidirect product group of $N$ and $H$ where $N$ is generated by $a$ and $b$, is $ab=ba$? If so why?

I know if $H$ and $N$ are both normal and their intersection is trivial, then $hn=nh$ for all $h$ in $H$ and $n$ in $N$, but not sure why in semidirect case.

regards

答:
Let $n\ge3$.

$S_n=\langle(12),(123\dots n)\rangle$.

And any index two subgroup is normal.

So $S_n\rtimes\Bbb Z_2$ will do as a counterexample.


问:
An approximation for $\left(1-{1\over x}\right)^{2y} - \left(1-{2\over x}\right)^y$
For large $x$ it is common to use the approximation $$\left(1-{1\over x}\right)^{x} \approx e^{-1}$$ leading to $$\left(1-{1\over x}\right)^{y} \approx e^{-y/x}$$ and so $$\left(1-{2\over x}\right)^y\approx e^{-2y/x}\approx \left(1-{1\over x}\right)^{2y}$$

But is there a simple approximation for the difference
$$\left(1-{1\over x}\right)^{2y} - \left(1-{2\over x}\right)^y \approx \,?$$

When $x=y$ the difference is about $\frac{e^{-2}}{y}$ but the pattern seems less obvious when they are not equal.

答:
In the spirit as @leonbloy, we can make the problem more general considering
$$z=\left(1-\frac{k}{x}\right)^{b y}-\left(1-\frac{b k}{x}\right)^y$$ with $y=a x$. Using the same Taylor expansion, this would give
$$z=\frac{a (b-1) b k^2 e^{-a b k}}{2 x}\left(1-\frac{(b+1) k (3 a b k-8)}{12 x}+O\left(\frac{1}{x^2}\right)\right)$$ and, back to $y$, this becomes
$$z=\frac{b(b-1)k^2}{2}  \,\frac{y }{x^2}\exp\left({-b k\frac{ y}{x}}\right)+\cdots$$

Edit

As @Henry pointed out in a comment, it would be simpler to let $\frac k x = \frac 1 X$ and write
$$z=\left(1-\frac{1}{X}\right)^{b y}-\left(1-\frac{b }{X}\right)^y$$ with $y=a X$. For large $X$
$$z=\frac{a (b-1) b e^{-a b}}{2 X}\left(1-\frac{\left(b+1\right) (3 a b-8)}{12 X}+O\left(\frac{1}{X^2}\right) \right)$$ and, back to $y$,
$$z=\frac{b(b-1)}{2}  \,\frac{y }{X^2}\exp\left({-b \frac{ y}{X}}\right)+\cdots$$


答:
Note that the projectile’s flying time in the air is $T=\frac{2v\sin\theta}g$. The area under the trajectory is

\begin{align}
A =\int_0^T y(t)x’(t)dt=\int_0^T
(vt \sin \theta - \frac{1}{2} gt^2)v\cos\theta dt=\frac{2v^4}{3g^2}\cos\theta\sin^3\theta
\end{align}

Then, set $dA/d\theta=0$ to get the optimal angle $\theta =\frac\pi3$. Thus, the maximal area is

$$A_{max}=\frac{2v^4}{3g^2}\cos\frac\pi3\sin^3\frac\pi3=\frac{\sqrt3v^4}{8g^2}
$$
and
$k=\frac{\sqrt3}8$.

Title: Maximal area under projectile trajectory

Body:
Problem:

A projectile is fired with an initial velocity of $v$ at an angle of $\theta$ from the ground. Then its trajectory can modeled by the parametric equations
\begin{align*}
x &= vt \cos \theta, \\
y &= vt \sin \theta - \frac{1}{2} gt^2,
\end{align*}where $t$ denotes time and $g$ denotes acceleration due to gravity, forming a parabolic arch.

Then the maximal area under the trajectory is given by $k \cdot \frac{v^4}{g^2}$ for some constant $k.$ Find $k.$

My Work:
I substituted the first equation for t into the second equation getting:
$$y = \frac{\sin\theta x}{\cos\theta} - \frac{gx^2}{2v^2(\cos\theta)^2}.$$
How do I proceed?


问:
Mean and Variance for Random Variables X>0
I am looking to understand mean and variance for random variables a little better. Can someone help me with some examples or guide me in the right direction?

I am interested in seeing the differences between finite and infinite so I would like help with coming up with examples of random variables X>0 with:

Finite mean and finite variance

Finite mean but infinite variance

Infinite mean and infinite variance.

答:
Simple examples:  density function $f_n(x)=\frac{c_n}{1+x^n}$ for $x\ge 0$ and $f_n(x)=0$ otherwise.  For $n=2$ mean and variance are infinite, for $n=3$ finite mean and infinite variance, for $n=4$ mean and variance finite.

$c_n$ defined so that $\int_0^\infty f_n(x)dx=1$.


问:
Double sampling distribution
One bag contains $n_1$ white and $n_2$ red balls. Another bag contains $m_1$ white and $m_2$ red balls.
$N$ balls are drawn at random from the first bag and transferred to the second.
Then $M$ balls are drawn at random from second bag.

What is the probability that

exactly $x$ balls are white?

$x$ balls or more are white?

答:
Let $n=n_1+n_2$ denote the total number of balls in the first bag and $m=m_1+m_2$ the total number of balls in the second bag. Let $N_1$ denote the number of white balls transferred to the second bag and $N_2$ the number of red balls transferred to the second bag. Let $M_1$ denote the number of white balls drawn from the second bag and $M_2$ the number of red balls drawn from the second bag. Hence $N_1$, $N_2$, $M_1$ and $M_2$ are random but $N=N_1+N_2$ and $M=M_1+M_2$ are deterministic.

Conditionally on $(N_1,N_2)$, $M_1$ is the number of white balls in a subset of $M$ balls chosen from $m_1+N_1$ white balls and $m_2+N_2$ red balls, hence
$$
P(M_1=x|N_1=y)={m_1+y\choose x}{m_2+N-y\choose M-x}{m+N\choose M}^{-1}.
$$
Likewise,
$$
P(N_1=y)={n_1 \choose y}{n_2\choose N-y}{n\choose N}^{-1}.
$$
Hence
$$
P(M_1=x)=\sum_y{n_1\choose y}{n_2\choose N-y}{m_1+y\choose x}{m_2+N-y\choose M-x}{n\choose N}^{-1}{m+N\choose M}^{-1}.
$$
The expectation is simpler, since one can get directly
$$
E(M_1)=\frac{M}{n}\frac{nm_1+n_1N}{m+N}.
$$

Title: Double sampling distribution


问:
For what values of $a$ does $1/(1+1/x) = a$ have no solution for $x$? Similarly, $ (6x-a)/(x-3) = 3$?

For what values of $a$ does this equation have no solution in $x$?
$$\frac1{\;1+\dfrac1x\;} = a$$

Similarly, for what value of $a$ does this equation have no solution in $x$?
$$\frac{6x-a}{x-3} = 3$$

I get only the value $a = 1$ for the first, and $a = 18$ for the second with my method.

I basically delineate which values of $x$ would be unacceptable, such as $x=-1$ for the first equation, and then solve for $x$ in terms of $a$. If that results in an expression with $a$ that calls for excluding a value of $a$ that would result in a denominator of $0$, I exclude that.

I also solve for the value of $a$ that would make $x$ equal a prohibited $x$ value and exclude that.

I'm not sure that's the correct method, so I would appreciate if anyone could run through their solutions. Thanks!

答:
Imagine there is a solution $(a,x)$. If $x\neq0$, the equation is equivalent to $\frac{x}{x+1}=a$. Now if $x\neq-1$, the equation is equivalent to $x=a(x+1)=ax+a$. From which, $$x(1-a)=a\tag{1}$$ And now if $a\neq1$, $$x=\frac{a}{1-a}\tag{2}$$
So as long as $a\neq1$ and also $a$ is not such that equation (2) implies $x=0$ or $x=-1$, there is a solution. Is there a value of $a$ where the equation (2) implies $x=0$? Yes: $a=0$. Is there a value of $a$ where equation (2) implies $x=-1$? No, because equation (1) would be $-1+a=a$, equivalent to $-1=0$.

So the values of $a$ that lead to no solution are $1$ and $0$.

Imagine there is a solution $(a,x)$. If $x\neq3$, the equation is equivalent to $6x-a=3(x-3)=3x-9$. From which, $$x=\frac{a-9}{3}\tag{3}$$ So as long as $a$ is such that equation (3) does not imply $x=3$, there is a solution. Is there a value of $a$ such that equation (3) implies $x=3$? Yes, $a=18$. There are no other considerations. So $a=18$ is the only value for $a$ for which there is no solution for $x$.


问:
Lower bound on number of disjoint spherical caps that can be packed on hypersphere
Let $S^n = \{x \in \mathbb{R^{n+1}: \ ||x||_2=1}\}$ be the L2 unit sphere in $\mathbb{R^{n+1}}$.

I saw the following result (https://ocw.mit.edu/courses/brain-and-cognitive-sciences/9-520-statistical-learning-theory-and-applications-spring-2006/lecture-notes/class22.pdf ,  page 28):

You can pack $(\frac{1}{\epsilon})^{n+1}$ disjoint spherical caps of radius $\epsilon$ on the surface of $S^n$.

The statement does not include a proof, so I assume it shouldn't be too hard to derive.

Thanks for the help!

答:
it shouldn't be too hard to derive.

I tried not to hard to derive it, but failed.  Moreover, I think that the proposed bound is wrong. Below are my calculations.

Note that when $\tfrac 1\varepsilon$ is not an integer then there is no sense to pack $\tfrac 1{\varepsilon^n}$-many caps, so we try to pack at least so many caps.

Since for $n\ge 2$ it is hard to provide explicit coordinates of the centers of packed caps, we try to obtain the required bound comparing ($n$-dimensional) areas. Unfortunately, an observation that an area of a cap is less that  $\tfrac 1m$ area of a sphere should not imply that we can pack $m$ copies of the cap on the sphere, so we should use the following trick.

We assume that a radius of a spherical cap its a distance from its center on a sphere to its boundary point.

Given $\varepsilon$, let $X$ be any maximal subset of $S^n$ such that the distance between any two distinct points of $X$ is at most $\color{red}2\varepsilon$. Then (open) spherical caps of radius $\varepsilon$ centered at points of $X$ are pairwise disjoint. So we have to show that $|X|\ge  \tfrac 1{\varepsilon^{n+1}}$.

The maximality of $X$ implies that a union of closed spherical caps of radius $2\varepsilon$ centered at points of $X$ covers $S^n$. We have that $n$-dimensional area of $S^n$ equals $A_{n+1}=\frac{2\pi^{(n+1)/2}}{\Gamma\left(\tfrac {n+1}2\right)}$. The formula for the area $C_{n+1}$ of the spherical cap of a given radius is even more complicated. In particular, it involves a regularized incomplete beta function. In order to esimate it, we use a formula $C_{n+1}=\frac{2\pi^{n/2}}{\Gamma\left(\tfrac n2\right)}\int_0^\phi\sin^{n-1} \theta d\theta$ of Li’s paper, where, as I understood, $2\phi$ is the angle spanned by the cap, which follows $\varepsilon =\sin \phi/2$. The area comparison provides us a bound $|X|\ge A_{n+1}/C_{n+1}$. But for fixed $n$ and small $\varepsilon$ (and so $\phi$ and $\theta$), we have $\sin\theta\approx\theta$, so $$\int_0^\phi\sin^{n-1} \theta d\theta\approx \int_0^\phi \theta^{n-1} d\theta=\frac{\phi^n}n\approx \frac{(2\varepsilon)^n}n .$$

Since $\Gamma\left(\tfrac n2\right)\le \Gamma\left(\tfrac {n+1}2\right)$ for $n\ge 3$, we have

$$\frac {A_{n+1}}{C_{n+1}}\approx \frac{2\pi^{(n+1)/2}}{\Gamma\left(\tfrac {n+1}2\right)} \cdot \frac {\Gamma\left(\tfrac n2\right)} {2\pi^{n/2}}\cdot \frac n{(2\varepsilon)^n}\le \frac {n\sqrt{\pi}}{(2\varepsilon)^n},$$
which is smaller than the required bound $\tfrac 1{\varepsilon^{n+1}}$, provided $\varepsilon\le \frac{2^n}{n\sqrt{\pi}}$.

Moreover, replacing in the above calculations $2\varepsilon$ by $\varepsilon$, we see that we even cannot pack on $S^n$ more than  $\approx \frac {n\sqrt{\pi}}{\varepsilon^n}$ caps, which is smaller
than the required bound $\tfrac 1{\varepsilon^{n+1}}$, provided $\varepsilon\le \frac{1}{n\sqrt{\pi}}$.


问:
Construct a finite field of 128 elements.
By using Galois' Theorem I know I need to find an irreducible polynomial of degree 7 in $\mathbb{Z}_2$. However I'm not sure how to find one efficiently, without showing that it is not the product of a polynomial of degree 5 and another of degree 2, or one of degree 3 and another of degree 4.

答:
As Lubin explained, a bit of experimenting will take you the distance sooner rather than later. Shaking my bag of tricks and examining what falls out.

We need an irreducible septic $f(x)\in\Bbb{F}_2[x]$. Clearly $f(x)$ must contain the terms $x^7$ and $1$ for otherwise $x\mid f(x)$. The binomial $x^7+1$ is divisible by $x+1$ so we need a trinomial at the very least. Let's try
$$
f(x)=x^7+x^a+1
$$
for some $a\in\{1,2,\ldots,6\}$.

Such a trinomial has no linear factors for $f(0)=f(1)=1$.

Could it have a quadratic factor? The sole irreducible quadratic, $x^2+x+1$, has a primitive third root $\omega$ as a root. We have
$$f(\omega)=\omega^7+\omega^2+1=\omega+\omega^a+1.$$
We see that this is zero if and only $a\equiv2\pmod3$. This rules out $a=2$ and $a=5$.

But all the other choices for $a$ actually work! This is because from the construction of the field $\Bbb{F}_8$ we know that all the irreducible cubic polynomials of $\Bbb{F}_2[x]$ are factors of $x^8-x=x(x^7+1)$ and hence also of $x^7+1$. So if $p(x)$ is an irreducible cubic, we have
$$f(x)\equiv x^a\pmod{p(x)}.$$
This immediately tells us that $p(x)\nmid f(x)$ (any common factor would be a factor of $x^a$ also etc).

So
$$
x^7+x+1,x^7+x^3+1, x^7+x^4+1\quad\text{and}\quad x^7+x^6+1
$$
all work.

A few concluding remarks for extra credit:

You may have spotted that the irreducible septic trinomials come in reciprocal pairs. If $f(x)$ is an irreducible septic then so is its reciprocal $\tilde{f}(x)=x^7f(\frac1x)$. I invite you to prove this in a more general case as a fun exercise. Basically if $p(x)$ is the minimal polynomial of an element $\alpha\neq0$, algebraic over a field $K$, then $\tilde{p}(x)$ is the minimal polynomial of $1/\alpha$.

Lubin hinted at repeated squaring. We can do it here quickly for example with the choice $a=1$. So let $\beta$ be a zero of $x^7+x+1$. Therefore $\beta^7+\beta+1=0$,
implying $\beta^7=\beta+1$ and thus also $\beta^8=\beta^2+\beta$. Then start squaring. Remember Freshman's dream, $(a+b)^2=a^2+b^2$ when in characteristic two. Behold.
$$
\begin{aligned}
\beta^{16}&=\beta^4+\beta^2\\
\beta^{32}&=\beta^8+\beta^4=\beta^4+\beta^2+\beta\\
\beta^{64}&=\beta^8+\beta^4+\beta^2=\beta^4+2\beta^2+\beta=\beta^4+\beta\\
\beta^{128}&=\beta^8+\beta^2=2\beta^2+\beta=\beta.
\end{aligned}
$$
So $\beta$ is an element of $\Bbb{F}_{128}$. As it is not an element of the prime field and the extenstion degree $7$ is a prime, we can deduce that $\Bbb{F}_{128}=\Bbb{F}_2(\beta)$.

Oh, one last thing occurred to me while looking at that last calculation. Namely, it may remind you of the calculation of writing consecutive powers $\alpha^i$, $i=1,2,\ldots,7$, where $\alpha$ is a zero of $x^3+x+1$, see here. You may have done that as an exercise when constructing the field $\Bbb{F}_8$. Notice how similar it is. The difference is that as we there reduce powers $\alpha^j$ to linear combinations of $\alpha^0,\alpha^1$ and $\alpha^2$, here we, in each step, reduce $\beta^{2^j}$ to the exact same linear combination of $\beta^{2^0}$, $\beta^{2^1}$ and $\beta^{2^2}$. This is no coincidence! It reflects the relation of the polynomial $$m(x)=x^3+x+1=\sum_j b_jx^j$$ and its linearized associate $$L_m(x)=\sum_j b_j x^{2^j}=x^8+x^2+x.$$ This may be beyond the first course in finite fields, but because $3,7$ and $127$ are all Mersenne primes, it follows that we can repeat this process and show that $L_{L_m/x}(x)/x=x^{127}+x+1$ is also irreducible over $\Bbb{F}_2$. See this old answer of mine for more details. This is further developed when viewing extension fields of $\Bbb{F}_2$ as modules over the polynomial ring $\Bbb{F}_2[\tau]$ with the indeterminate $\tau$ acting via the Frobenius automorphism.


问:
Prove $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\gt1$
Prove that $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\gt1$, where a, b, c >0.

I tried,
$\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} = \frac{a}{\sqrt{a}\sqrt{a+b}}+\frac{b}{\sqrt{b}\sqrt{b+c}}+\frac{c}{\sqrt{c}\sqrt{c+a}} \ge \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{\sqrt{a}\sqrt{a+b}+\sqrt{b}\sqrt{b+c}+\sqrt{c}\sqrt{c+a}}$.

Then $\sqrt{a}\sqrt{a+b} \le \sqrt{2}/2 (2a+(a+b))$.

This only gave me $\sqrt{2}/2 + \sqrt{2}(\frac{ab+bc+ca}{a+b+c})$.

答: Hint: $\dfrac{a}{a+b} > \dfrac{a}{a+b+c}$


问:
"Let" in math texts
I'm trying to do the following exercise for my Real Analysis class:

Let $p$ be a given natural number. Give an example of a sequence $\left(x_{n}\right)$ that is not a Cauchy sequence, but that satisfies $\lim \left|x_{n+p}-x_{n}\right|=0$

However, I am in constant doubt in regard to the "let" word in math texts. Can I choose, say $p = 1$, or when one says "let" I am supposed to stick with $p \in \mathbb{N}$ and nothing more? What kind of "control" do I have over $p$?

For instance, when one says "Given $\epsilon \gt 0$", I usually see things like: Let $\epsilon = \frac{\epsilon}{2}$ so one can finish a certain argument.

Can someone help me? I need to solve this doubt once and for all, it bothers me very often.

答: No, you cannot take $p=1$, in order to prove the statement you need to show that it holds for arbitrary $p \in \mathbb{N}$. Taking $\epsilon' = \frac{\epsilon}{2}$ is allowed because this still works for an arbitrary choice of $\epsilon$. If I take $\epsilon = 0.1$ or if take it to be $\epsilon =0.9$, the proof still works since $\epsilon'$ is defined relatively, but if you only prove something for say $p=1,2$, then you have not proven the claim generally, and your proof might break down for $p=4$ say

Title: "Let" in math texts


问:
upper bound for the mean of logistic normal distribution
Assume that ${\bf x} \in R^{K \times 1}$ is a multivariate normal distribution with mean of $\mu $ covariance matrix of $\Sigma $. I want to compute the following term.

$y_k=E_x(\frac{e^{x_k}}{\sum_{k'=1}^{K}e^{x_{k'}}})$

I know that there is no analytical solution for this. But, is there any upper bound for $y_k$?

答:
If $X_1, \ldots, X_n$ are identically distributed, then
$$E(X_1/\sum_{j=1}^n X_j) = 1/n$$
to see this, note that
$$1 = E(\sum_{j=1}^n X_j / \sum_{k=1}^n X_k ) = \sum_{j=1}^n E(X_j/\sum_{k=1}^n X_k) = nE(X_1/\sum_{j=1}^n X_j)$$
When $\mu_1 = \ldots = \mu_K$ and the $\Sigma_{ii} = \Sigma_{jj}$ for every $1 \leq i,j \leq K$, then the $e^{x_k}$ are identically distributed so you can use the above to get $y_k = 1/K$.

You might be able to get the general case by writing each normal distribution as $\sigma Z + \mu$, where $Z \sim N(0,1)$, but I'm not sure if the calculation will work out.


问:
Regular open set whose boundary has nonzero volume.
I found this question quite interesting, but its answers were disappointingly non-geometric. I'd be interested to know whether there exists a geometric example.

To be precise about what I mean by a geometric example, recall that an open set $U\subset \mathbb{R}^n$ is called regular if it is equal to the interior of its closure.  Here is my question:

Question: Does there exist a regular open set $U\subset \mathbb{R}^n$ whose boundary has nonzero Lebesgue measure?

I would guess that the answer is no in $\mathbb{R}^1$, but it ought to be yes in $\mathbb{R}^2$.  Part of why this is interesting is that it seems to me that the Mandelbrot set might fall into this class (although according to this MathOverflow post, it is an open question whether the boundary of the Mandelbrot set has positive Lebesgue measure).

答:
Osgood constructed an example of a Jordan curve of positive area in Trans. Amer. Math. Soc. 4 (1903), 107-112. Knopp's construction of such a curve is the content of a demonstration on Wolfram|Alpha.

By the Schoenflies theorem we can extend such a Jordan curve $\gamma$ to a homeomorphism $\mathbb{C} \to \mathbb{C}$ such that the interior and exterior of the unit circle are sent to the interior and exterior of $\gamma$, respectively. Since homeomorphisms preserve regular open sets, the interior and exterior of such a Jordan curve will be examples of regular open sets whose boundary has nonzero volume.

Added: Brian M. Scott shows in Cantor set is boundary of regular open set how to get a Cantor set as  boundary of a regular open set. The argument appears to apply just as well for a fat Cantor set, so it is even possible in one dimension.


答:
Well, usually the word "let" it's use for defining the context of a variable and also for letting fixed its value. However, it's usual to redefine the value of certain variable as in your example "$\epsilon = \frac{\epsilon}{2}$" strictly speaking, it shouldn't happen, but it's a convention, in order to avoid choosing from the beginning the value $\epsilon/2$ or others, which works but which could be seen as non intuitive, and to avoid rewrite the whole proof  from the start.

All you know about $p$, it's that is a natural number, could be any, but you cannot choose one at your convenience, unless you will go into all of them, for example considering $p$ as an odd number, then as an even number, and so on...

Title: "Let" in math texts

Body:
I'm trying to do the following exercise for my Real Analysis class:

Let $p$ be a given natural number. Give an example of a sequence $\left(x_{n}\right)$ that is not a Cauchy sequence, but that satisfies $\lim \left|x_{n+p}-x_{n}\right|=0$

However, I am in constant doubt in regard to the "let" word in math texts. Can I choose, say $p = 1$, or when one says "let" I am supposed to stick with $p \in \mathbb{N}$ and nothing more? What kind of "control" do I have over $p$?

For instance, when one says "Given $\epsilon \gt 0$", I usually see things like: Let $\epsilon = \frac{\epsilon}{2}$ so one can finish a certain argument.

Can someone help me? I need to solve this doubt once and for all, it bothers me very often.


答:
To attempt an answer to your question, any time you read an instance of "Let $x$ be an element of $S$," you are being asked to consider an element $x$ whose only salient property as far as you are concerned is that it is a member of $S$. Until further assumptions are made about the specific nature of $x$, you should only use properties of $x$ that it has as an element of $S$ (and any supersets that $S$ is contained in, depending on the context). Some examples:

Let $x$ be a real number. (Or "Let $x$ be an element of $\mathbb R$" to match the language I used above.)

This means to only consider the properties that $x$ has as a real number, or possibly also as a complex number, for example depending on the situation.

Let $n$ be a given natural number.

This means that for our discussion, $n$ is a natural number, but that is all we know about it. Depending on what we talk about next, we may or may not also use other properties of $n$, like that it is a rational number, or that it is a complex number, for instance.

Let $A$ be a subset of $\mathbb R$.

This is similar to the first example. All we want to do at this point is consider the properties of $A$ we know it has as a subset of $\mathbb R$. It may be the collection of natural numbers, it may be the empty set, it may be $\{\pi, \pi^2\}$, and so on.

Part of the point of making statements like this is that if we then go on to prove something involving this arbitrary element $x$, then we can substitute the value of $x$ with any specific element of $S$ and obtain a "new" theorem. For your example, if you can prove the statement you are given assuming nothing other than $p$ is a natural number, then you know, for example that there is a sequence $\{x_n\}$ in $\mathbb R$ that is not Cauchy, with the property that
$$
\lim_{n\to\infty}(x_{n+314159}-x_n) = 0,
$$
which looks pretty impressive, but is really nothing other than a special case where $p = 314159$ of the theorem stated in terms of the otherwise arbitrary $p$. We get to know that this "$314159$ theorem" is true for free if the only property of $p$ we use in our proof is that $p$ is a natural number since $314159$ is also a natural number.

Edit: "The $\epsilon$ issue." In analysis, we often want to prove theorems that have the form "For all $\epsilon > 0$, $P(\epsilon)$ is true." Where $P(\epsilon)$ is a statement involving $\epsilon$. For example,

$P(\epsilon) = $ there exists $\delta > 0$ so that $|x^2-100| < \epsilon$ if $|x-10| < \delta$.

$P(\epsilon) = $ there exists $N\in\mathbb N$ so that for all $n,m\ge N$, $|x_n-x_m| < \epsilon$.

When you think about these theorems in this way, I think it becomes a little bit clearer why in order to establish the truth of $P(\epsilon)$ for all $\epsilon > 0$, it suffices to establish the truth of $P(\epsilon/2)$ for all $\epsilon > 0$. Essentially, $\epsilon$ is a place-holder for "all the positive real numbers," and for that purpose, if $C$ is any constant like $2$ or $\frac12$, then $C\epsilon$ is also a place-holder for all the positive real numbers if $\epsilon$ is a place-holder for all positive real numbers. So it suffices to show that for all $\epsilon > 0$, $P(C\epsilon)$ is true. (It is critical however that $C$ does not depend on $\epsilon$! For instance, if we tried to let $C$ depend on $\epsilon$, then we could "cheat" and prove $P(\frac1\epsilon \epsilon)$ is true for all $\epsilon > 0$, but this only shows that $P(1)$ is true...)

Edit to the edit: I wrote in parentheses at the end of my last edit that $C$ is not allowed to depend on $\epsilon$, but this is not quite correct. So long as $C\epsilon$ is a place-holder for all the positive real numbers, $C$ can itself be a function of $\epsilon$. Here are two examples and one non-example of when $C\epsilon$ is a place-holder for all the positive real numbers when $\epsilon$ is itself a place-holder for all the positive real numbers (you can try to think of more examples for fun!):

$C = (1+\epsilon^2)$. Then $C\epsilon = \epsilon + \epsilon^3$ is still a place-holder for all the positive real numbers.

$C = \frac{1}{\sqrt{\epsilon}}$. Then $C\epsilon = \sqrt{\epsilon}$ is still a place-holder for all the positive real numbers.

$C = \frac{1}{\epsilon}$. Then $C\epsilon = 1$ is not a place-holder for all the positive real numbers. This was the kind of "malicious" dependence on $\epsilon$ that I had in mind when I said that $C$ should not depend on $\epsilon$ in the parenthetical sentence at the end of my first edit.

Sometimes it can be convenient to allow $C$ to depend on $\epsilon$. I know of at least one example where the bound I had by the end of the proof was $\epsilon + \epsilon^3$, which motivates my first example I wrote here where $C = 1 + \epsilon^2$.

Title: "Let" in math texts


问:
Get explicit expression of $f(m,n)$
$$
f( m,n) =\left\{ \begin{array}{l}
 1&\text{ if } m=n=0;\\
 0&\text{ if } m\ne 0,\ n=0;\\
 \displaystyle\sum_{t\in \left\{ -1,0,1 \right\}}{f(m+t,n-1)} &\text{ if }\ n>0.\\
\end{array} \right.
$$
This is a question about convolution. I can calculate the value of specific $f(m_0,n_0)$ by writing a brute force computer program, but I want to know how to get the explicit expression.

答:
Let $g:\Bbb Z \longrightarrow \Bbb Z$ be such that $g\equiv 0$ except that $g(-1) = g(0) = g(1) = 1$.
Then it's easy to verify that for $n>0$ we have

$$f(m, n) = \sum_{j\in\Bbb Z}f(j, n-1) \cdot g(m-j).$$

Let $f_n(m) := f(m,n)$, and observe that $g = f_1$.
We can then rewrite for $n>0$

$$f_n(m) = \sum_{j\in\Bbb Z}f_{n-1}(j) \cdot g(m-j) = (f_{n-1}*f_1)(m).$$

Of course, we can further expand and obtain that for $n>0$

$$f_n(m) = (\underbrace{f_1 * f_1 * \dots * f_1}_{n \text{ times}})(m),$$

At this point, one can proceed using the convolution theorem. Do you think you can take it from here?

Let $\mathfrak D[f]$ denote the Fourier transform of $f$. The convolution theorem states that

$$\mathfrak D[f*g] = \mathfrak D[f] \cdot \mathfrak D[g].$$

We can couple this with the inverse fourier tranform to turn the problem of calculating $n$ convolutions into the problem of calculating $n$ multiplications.

Indeed, by the theorem we have $\mathfrak D[f_n] = {\left(\mathfrak D[f_1]\right)}^n$.
Then, with the inverse fourier transform, we can recover

$$f_n(m)
= \frac 1{2\pi}\int_0^{2\pi} \mathfrak D[f_n](\omega)\, e^{i\omega m} \, d\omega
= \frac 1{2\pi}\int_0^{2\pi}{\big(\mathfrak D[f_1](\omega)\big)}^n e^{i\omega m} \, d\omega.\tag{1}$$

Now, $\mathfrak D[f_1]$ is a power series of $e^{-i\omega}$, and it's a good exercise to check that, for this kind of function, the integral $(1)$ yields the coefficient of $e^{-i\omega m}$.

With $x = e^{-i\omega}$, we hence have $\mathfrak D[f_1](\omega) = x + 1 + \frac1x,$ and therefore

$$f_n(m) = \left[x^m\right]\left(x + 1 + \frac1x\right)^n. \tag{2}$$

As a side note, you can easily see here the connection with the solution you found.
Indeed, we have that
$$\begin{align}\left[x^m\right]\left(x + 1 + \frac1x\right)^n
&= \left[x^m\right]\left(\frac{x^2 + x + 1}x\right)^n
\\&= \left[x^m\right]\left(\frac{\left(x^2 + x + 1\right)^n}{x^n}\right)
\\&= \left[x^{m+n}\right]\left(x^2 + x + 1\right)^n
\end{align}$$
I do prefer the formulation in $(2)$; I find it better highlights the problem's symmetry about $m=0$.

Now, calculating this is a matter of combinatorics.

In each of the $n$ factors of the product $\left(x + 1 + \frac1x\right)^n$, we must choose one of $x$, $1$ or $1/x$, and the overall product must be $x^m$.
Let $a_{-1}$ be the number of $x^{-1}$ we choose, and similarly for $a_0$ and $a_1$.
Then, these numbers must satisfy the equations

$$a_{-1}+a_0+a_1 = n \quad\text{ and }\quad a_1 - a_{-1} = m.$$

Subtracting them, we find $a_0 + 2a_{-1} = n - m$, so that $a_0 \equiv n - m \pmod 2$ and $a_0 \leqslant n - m$.
Choosing a value for $a_0$ that satisfies these constraints then determines the values of $a_{-1}$ and $a_1$:

$$\begin{align}
a_{-1} &= \frac{n-m-a_0}2,
\\a_{1} &= \frac{n+m-a_0}2.
\end{align}$$

Accounting also for the different orders in which we can choose the factors, we hence have

$$f_n(m) = \sum_{\substack{0\,\leqslant\, t \,\leqslant\, n-m\\t\,\equiv\, n-m\pmod 2}} \frac{n!}
{t!{\left(\frac{n+m-t}2\right)}!{\left(\frac{n-m-t}2\right)}!}.$$

You can specialize the formula depending on the parity of $n-m$ to get a 'traditional' index on the summation, but this is the gist of it.

Title: Get explicit expression of $f(m,n)$


问:
interchange double summation
I am confused I am looking at my professor's note but I don't know how he changes the following double summation. Can someone explain it to me?

$$
\sum_{k = 0}^{\infty}\,\,
\sum_{x = k + 1}^{\infty}\operatorname{f}\left(x\right) = \sum_{x = 1}^{\infty}\,
\sum_{k = 0}^{x - 1}
\operatorname{f}\left(x\right)
$$

答:
There are obvious convergence issues to be worried about with such an interchange. But assuming everything converges uniformly, we can make the interchange as follows (let me change $x$ to $n$ since $x$ will be a little awkward in the discussion that follows). This is hard to describe but easy to picture. If you get confused, my suggestion is to draw everything out.

The value of
$$\sum_{k=0}^\infty \sum_{n=k+1}^\infty f(n)$$
is geometrically simple. Think of the pairs $(k,n)$ as integer points in $\mathbb{R}^2$, say. Then the sum for $k=0$ corresponds to the sum over all integer points with $x$-coordinate $0$ and $y$-coordinate greater than or equal to $1$, which is an infinite column of points. Likewise, the sum for $k=1$ corresponds to the sum over all integer points with $x$-coordinate $1$ and $y$-coordinate greater than or equal to $2$, another infinite column. In general, the sum will be over all integer points with non-negative $x$-coordinate and $y$-coordinate satisfying $n \ge k+1$. This is a triangular region (draw it out), and the first sum corresponds to summing over the triangle column by column.

Of course, another equivalent way of summing over this triangle is not column by column, but row by row. In this case, the first row is at $y=1$, and consists of only the point with $x=0$. The second row is at $y=2$, and consists of the points with $x=0$ or $x=1$. In general, the $n$th row has $y$-coordinate $y=n$ and $n$ total points: $x=0,1,\cdots ,n-1$. Summing this triangle row by row gives us
$$\sum_{n=1}^\infty \sum_{k=0}^{n-1}f(n).$$
The two sums are equal since they sum over the exact same set of points, just in two different orderings.

Title: interchange double summation


问:
Expectation of Gaussian CDF
I have $X_i$ from $n$ i.i.d draws of N($\theta$,1). By the SLLN, I know $\bar{X_n}$ converges almost surely to $\theta$. I would like to find the limiting distribution of $\phi(x-\bar{X_n})$ for fixed $x$ where $\phi$ is the CDF of a standard normal. Is it valid to say $E\phi(x-\bar{X_n}) = \phi(x-E\bar{X_n}) = \phi(x-\theta)$ as $n$ approaches $\infty$? If not, how can I find the expectation of $\phi(x-\bar{X_n})$?

答:
You need to use a combination of the Central Limit Theorem and the Delta method. By the CLT, we know that
$$
\sqrt{n}(\bar{X}_n - \theta) \implies N(0,1),
$$
where $\implies$ denotes convergence in distribution. Then, the delta method tells us that for any function $g$ such that $g'(\theta)$ exists and is not equal to zero, that
$$
\sqrt{n}(g(\bar{X}_n) - g(\theta)) \implies N(0, [g'(\theta)]^2)
$$
Now, for your case, you can take (for a fixed $x$):
$$
g_x(z) = \phi(x-z),
$$
which has first derivative with respect to $z$:
$$
g'_x(z) = (x-z)\phi(x-z)
$$
So we get
$$
\sqrt{n}(\phi(x-\bar{X}_n) - \phi(x-\theta))
\implies
N(0, (x-\theta)^2 \phi^2(x-\theta))
$$

An important thing to note here is that this breaks down for $x=\theta$, since the variance becomes zero and this is no longer a limiting distribution. To get around this, we can use the second order delta method. You can read more about all of this on the wiki page.

edit:
To differentiate $\phi(x-z)$ with respect to $z$, we note that
$$
\phi(x-z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{(x-z)^2}{2}}
\implies
\phi'(x-z) = (x-z) \frac{1}{\sqrt{2\pi}} e^{-\frac{(x-z)^2}{2}}= (x-z) \phi(x-z)
$$

Title: Expectation of Gaussian CDF


答:
The short answer is: It depends on the situation.

Here, you can't choose $p=1$, as it is given that $p$ is a fixed natural number.

However, when it comes to working with $\epsilon$'s, you are required to show a result to be true for all $\epsilon > 0 $.
Hence, given an $\epsilon > 0$, its a convention for authors to use, "choose $\epsilon = \frac{\epsilon}{2}$ ", when they actually mean to say, they are trying to prove the result for this fixed $\epsilon$, by choosing some
${\epsilon}' = \frac{\epsilon}{2} $, and thus, they are proving the result for all $\epsilon > 0$, as this fixed $\epsilon$ was arbitrary.

Title: "Let" in math texts


问:
Which complexity problem is this related to?
Define a universe U containing N elements. We are given N sets, each of which is a set.

For example, U = {1, 2, 3, 4} and
sets

```<code>S1 = {{1}, {2, 4}},
S2 = {{2}, {1, 3}},
S3 = {{3}, {2, 4}},
S4 = {{4}, {1, 3}}
</code>```

The goal is to find the smallest subset of U that contains at least one element from each of the N sets. So for example, the subset {1,3} is a correct answer while the subset {1,2} is not because it does not contain any set in S3, S4.

I tried to formulate the above as an instance of the hitting set problem (because it seemed closer to it in spirit), but failed to do so. One way I managed to cleanly reduce the problem is by expanding each set to include all supersets. Then the desired answer is the smallest sized set in the intersection of the expanded sets. But this approach is undesirable as the expanded set size grows exponentially.

Any thoughts on connections to a known complexity problem are much appreciated.

答:
Thinking of the hitting set problem is the right idea. This problem (let's call it CONTAINMENT) is NP complete. It is clearly in NP, and

Theorem: $\text{CONTAINMENT}$ is NP-hard.

Proof: We reduce the minimal vertex cover problem$^1$ (known to be NP-hard) to an instance of $\text{CONTAINMENT}$. Given a graph $G$ with vertices $V$ (WLOG assume they are numbered from $1$ to $|V|$) and edges $E$, for each edge $e_i = (u,v) \in E$, we create the set $S_i = \{\{u\}, \{v\}\}$. Since you seem to require that the number of sets $S_i$ is the same as the number of elements in $U$, we can do the following. Set $U = \{1, 2, \cdots, \max(|V|, |E|)\}$. If $n = |E|$, then last $|E| - |V|$ elements will never appear in a set that is an element of any $S_i$ (they are in essence irrelevant to the reduction and are only present to satisfy the demands for the input of your problem). If $n = |V|$, then in addition to the $S_i$ we created above (there are $|E|$ such sets) we can add $|V| - |E|$ extra sets $S_i$ where $S_i = \{ \emptyset \}$ (the set where its only element is the empty set). (The condition for these $S_i$ will be trivially satisfied.) Note that the output of this reduction has size $O(\max(|E|, |V|))$.

Running $\text{CONTAINMENT}$ on this reduction will return a set of vertices corresponding  to minimal vertex cover of $G$. A minimal vertex cover must cover at least one of the singleton sets in each of the $S_i$, which is exactly what your problem demands. $\square$

Footnotes:

The minimum vertex cover problem asks of a graph $G = (V, E)$ to find $U \subseteq V$, with $|U|$ as minimal as possible, such that every edge in $E$ is incident to a vertex in $U$.


问:
Bounded component of the complement of a compact set
Suppose $A_1$ and $A_2$ are two compact sets in $\mathbb{R}^n$ for $n>1$. Suppose these sets are disjoint. It seems correct that if the complement of their union has a bounded component $C$, and $\overline{C}\cap A_1\not=\emptyset$, then $C$ must be a bounded component of $A_1$. Is this correct? How to prove it?

答: It's not true as you can show in this picture. For example the complement of union has two bounded component and the closure of the ''exterior ring to $A_2$'' satisfyng that your closure intercept $A_1$ and are not a component of $A_1$.


问:
Simplifying $\cos^{-1}x +\cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$
A question has this equation: $$f(x) = \cos^{-1}x + \cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$$ and you're supposed to simplify it and find $f\left(\frac{2}{3}\right)$ and $f\left(\frac{1}{3}\right)$.

By taking $\cos\alpha  = x$, the equation on the right can be simplified to $\cos^{-1}\left(\cos\left(\frac {\pi}{3} - \alpha\right)\right)$. Finally, you get $\frac{\pi}{3}$ as the final answer.

But the answers are $\frac{\pi}{3}$ and  $2\cos^{-1}\left(\frac{1}{3}\right)-\frac{\pi}{3}$. How does it work out that way?

答:
$$
f(\cos\alpha)=\cos^{-1}(\cos\alpha)+\cos^{-1}\bigg(\frac{1}2\cos\alpha+\frac{\sqrt3}2\sin\alpha\bigg)\\
= \cos^{-1}(\cos\alpha) +\cos^{-1}\bigg(\cos\frac \pi 3\cos\alpha+\sin\frac{\pi}3\sin\alpha\bigg)\\
=\cos^{-1}(\cos\alpha)+\cos^{-1}\bigg[\cos\bigg(\frac\pi3-\alpha\bigg)\bigg]\\
= \pm\alpha+2\pi k_1 \pm\ (\frac\pi3-\alpha)+2\pi k_2 \\
=\pm\alpha  \pm\ (\frac\pi3-\alpha)+2\pi k
$$
where $k,k_1,k_2 $ are integers.
For the first case we get the prescribed solution by taking the $+$ signs in both and $k=0$. In the second case, if we take the $+$ sign on the first and the $-$ sign on the second, and $k=0$ we get:
$$
2\alpha-\frac\pi3=2\cos^{-1}\big(\frac13 \big)-\frac\pi3
$$


问:
Does intersection mean set theoretic point of view for independent events?
It is evident that when two events $A$ and $B$ are independent, then $P(A\cap B)=P(A)P(B)$
A good example is, "Tossing a coin and rolling a die". What is the probability that heads occurs on coin and even number occurs on die.

We have event $A:$ Heads occur on coin

an event $B:$ $2,4$ or $6$ occurs on die

We write the probability as $P(A\cap B)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

But when we write sample space we get $A=\left\{H\right\}$
and $B=\left\{2,4,6\right\}$
But $A\cap B=\phi$
So why is this contradiction?

答:
You haven't defined a common probability space for your coin and your die.  You are basically talking about two outcomes that occur in separate universes, and so talking about their intersection is meaningless.

To fix that, let's build a common space for these two experiments. Define
$$
\Omega:=\{H,T\}\times\{1,2,3,4,5,6\}.
$$
Then the outcome of your coin is just the first coordinate projection, and the outcome of your die is just the second.  And in this case, the formal translations of your two events would be
$$
A:=\{(H,1),(H,2),(H,3),(H,4),(H,5),(H,6)\}
$$
and
$$
B:=\{(H,2),(H,4),(H,6),(T,2),(T,4),(T,6)\}.
$$

Then you have
$$
A\cap B:=\{(H,2),(H,4),(H,6)\}
$$
Now, assuming you define your probability measure $P$ on $\Omega$ to be the uniform distribution, you can verify that:

The coin and die each have the correct (marginal) distribution

The events $A$ and $B$ have exactly the probabilities you would expect ($\frac{1}{2}$)

The intersection has probability $\frac{1}{4}$, just as it should for independent events $A$ and $B$.


问:
How to find the angle which is in a trapezoid formed by two triangles?
The problem is as follows:

The figure from below represents a quadrilateral $ABCE$. Using the
information, $BE=BC$,  $\angle\,BAC=\angle\,ADB=60^{\circ}$ and
$DC=10\,m\,AE=10\,m$. Find the angle $x$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&125^{\circ}\\
2.&140^{\circ}\\
3.&130^{\circ}\\
4.&120^{\circ}\\
\end{array}$

I couldn't really find a relationship other than spot an isosceles triangle on $\triangle BEC$. This would mean that $\angle BEC= \angle BCE$. I also could spot that the $\triangle ABD$ is an equilateral. But other than that I'm stuck. What sort of identity or construction here is required to solve this problem?.

Can somone help me?. Can this be solved relying only in Euclidean postulates?. Please an answer must include a drawing because in this figure I can't find exactly where to look for, but I think that it is related with congruence or maybe similarity.

答:
Since $\angle BAD = \angle BDA = 60^\circ$, $\triangle ABD$ is equilateral, as you have observed.

We have $AB = BD$, $AE = DC$, $BE = BC$.

By SSS, $\triangle ABE \cong \triangle DBC$.

Now $\angle BAE = \angle BDC$. What is $\angle BDC$?


答:
The actual space of outcomes from these two events is not what you suggest, but is instead the set
$\Omega = \{H, T\} \times \{1, \cdots, 6\}$. Now we have two events. Getting a heads is getting an outcome that is in the set
$$\mathcal{H} = \{(H, 1), (H, 2), \cdots, (H, 6)\}$$
while getting an even number on the die is getting an outcome that is in the set
$$\mathcal{E} = \{(H, 2), (T, 2), \cdots, (H, 6), (T, 6)\}$$
so when we ask what is the probability that we get both, we are asking what the probability of getting an outcome in $\mathcal{H} \cap \mathcal{E}$. We say that the two events are independent if and only if
$$P(\mathcal{H}) \cdot \mathcal{P}(\mathcal E) = P(\mathcal{H} \cap \mathcal{E})$$
where $P(S) = \sum_{s \in S} P(s)$. Now of course, for any event $s \in \Omega$, $P(s)$ will change depending on modelling assumptions. So in some cases rolling a die and flipping a coin might not be independent. Of course, examples where this happens in the real world will mostly be contrived/unrealistic, but this is absolutely allowed in theory! (Math doesn't distinguish between a coin, a die, etc. What's relevant is the space of outcomes and the respective probabilities of obtaining each outcome.) The only way to know truly if two events are independent is to apply the definition above.

Title: Does intersection mean set theoretic point of view for independent events?

Body:
It is evident that when two events $A$ and $B$ are independent, then $P(A\cap B)=P(A)P(B)$
A good example is, "Tossing a coin and rolling a die". What is the probability that heads occurs on coin and even number occurs on die.

We have event $A:$ Heads occur on coin

an event $B:$ $2,4$ or $6$ occurs on die

We write the probability as $P(A\cap B)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

But when we write sample space we get $A=\left\{H\right\}$
and $B=\left\{2,4,6\right\}$
But $A\cap B=\phi$
So why is this contradiction?


问:
Prove that the equivalence class $\mathbb R^3/V$ is isomorphic to $\mathbb{R}$

Let, $V$ be a subspace of $\mathbb{R}^3$ such that $V=\{(a,b,c)\in\mathbb{R^3}|a=b\}$. Show that $\mathbb R^3/V$ is isomorphic to $\mathbb{R}$.

I can understand that two 3 tuples $(a,b,c)$ and $(x,y,z)$ are related ie belong to the same equivalence class if $(a-x,b-y,c-z)\in V$ ie. if $a-x=b-y$.

What should I do after that?

答:
We have $V=span\{(1,1,0),(0,0,1)\}$ now we know that
$$
\dim\frac{\mathbb{R}^3}{V}=3-\dim V=1
$$

so, if $v$ it's a generator of the quocient, the map $v\mapsto 1$ extends to a linear isomorphism between $\frac{\mathbb{R}^3}{V}$ and $\mathbb{R}$.


问:
How to show that $AA^*+B^*B$ is invertible if $U\stackrel{A}\to V\stackrel{B}\to W$ is exact?
Let $U, V,$ and $W$ be finite dimensional vector spaces with inner products. If $A: U \rightarrow V$ and $B: V \rightarrow W$ are linear maps with adjoints $A^{*}$ and $B^{*},$ define the linear map $C: V \rightarrow V$ by
$$
C=A A^{*}+B^{*} B
$$
If $U \stackrel{A}{\longrightarrow} V \stackrel{B}{\longrightarrow}  W$ is exact , then my question is how to show $C: V \rightarrow V$ is invertible?

Here is my thought. We obviously have $W \stackrel{B^*}{\longrightarrow} V \stackrel{A^*}{\longrightarrow}  U$ is exact. Then $V=\text{Im}(B^*)\oplus \text{Im}(A)=\text{Ker}(A^*)\oplus\text{Ker}(B)$ thanks to it. Hence,

$C(V)=AA^*(\text{Ker}(B))+B^*B(\text{Ker}(A^*))=AA^*(\text{Im}(A))+B^*B(\text{Im}(B^*)),$ but what should I do next? I still wonder how to solve this problem by the matrix method.

Any help would be appreciated. Thanks a lot!

答:
As you observed, there is a decomposition $V=E\oplus F$ where $E=\ker(B)=\operatorname{im}(A)$ and $F=\ker(A^*)=\operatorname{im}(B^*)$.  Note that $AA^*$ vanishes on $F$ and maps $E$ to itself.  Moreover, $\ker(AA^*)=\ker(A^*)=F$ (if $AA^*v=0$ then $0=\langle AA^*v,v\rangle=\langle A^*v,A^*v\rangle$ so $A^*v=0$), so $AA^*$ is injective when restricted to $E$.  This means that the restriction of $AA^*$ to a map $E\to E$ is invertible.  Similarly, $B^*B$ vanishes on $E$ and restricts to an invertible map $F\to F$.

So, writing $C$ as a block matrix with respect to the decomposition $V=E\oplus F$, it is $\begin{pmatrix} AA^* & 0 \\ 0 & B^*B\end{pmatrix}$ where both diagonal entries are invertible (as maps $E\to E$ and $F\to F$).  Thus $C$ is invertible.


Title: Prove that the equivalence class $\mathbb R^3/V$ is isomorphic to $\mathbb{R}$

Body:
Let, $V$ be a subspace of $\mathbb{R}^3$ such that $V=\{(a,b,c)\in\mathbb{R^3}|a=b\}$. Show that $\mathbb R^3/V$ is isomorphic to $\mathbb{R}$.

I can understand that two 3 tuples $(a,b,c)$ and $(x,y,z)$ are related ie belong to the same equivalence class if $(a-x,b-y,c-z)\in V$ ie. if $a-x=b-y$.

What should I do after that?

Body: $V=\{(a,a,c)|a,c\in\Bbb R\}=\rm{span}\{(1,1,0),(0,0,1)\}\cong\Bbb R^2$.  And $\Bbb R^3/\Bbb R^2\cong\Bbb R$.


问:
Cutting out submanifolds with "orthogonal" functions
Let $Z$ be an embedded manifold in some $\mathbb{R}^M$.  Then locally $Z$ is cut out by independent functions $(g_1, \ldots, g_l): \mathbb{R}^M \to \mathbb{R}^l$, where $l = \operatorname{codim} Z$.  That is, for every $z \in Z$, there exists a neighborhood $\tilde{U}$ of $z$ in $\mathbb{R}^M$ such that $\tilde{U} \cap Z = \bigcap_{i=1}^l g_i^{-1}(0)$.  (See for example page 24 in Guillemin and Pollack.)  Here, independent means that $d(g_1)_z, \ldots, d(g_l)_z$ are linearly independent in a neighborhood of $z$.  My question is:

Can the functions $g_1, \ldots, g_l$ be chosen so that $d(g_1)_z, \ldots, d(g_l)_z$ are actually orthogonal?

My idea is that we can first apply Gram-Schmidt to $\{d(g_i)_z\}$ first to obtain orthogonal vectors $\{\widetilde{d(g_i)_z}\}$, and then take $\tilde{g}_1, \ldots, \tilde{g}_l$ to be the corresponding linear combinations of $\{g_i\}$ with differentials $\{\widetilde{d(g_i)_z}\}$.  Then I believe $\{\tilde{g}_i\}$ still cuts out $Z$.  However, being orthogonal at one point $z$ generally does not mean that the differentials are orthogonal in a neighborhood of $z$.

In fact, I'm interested in the special case where $Z$ is actually a submanifold of some other embedded manifold $Y \subseteq \mathbb{R}^M$.  Then there exists independent functions $(g_1,\ldots,g_l)$ locally cutting out $Z$ such that $(g_{k+1},\ldots,g_l)$ cut out $Y$ locally.  In this case, is it possible to make $\{d(g_1)_z, \ldots, d(g_k)_z\}$ orthogonal to $\{d(g_{k+1})_z, \ldots, d(g_l)_z\}$?

答: What a great question! As you realized, you can easily arrange this at any given particular point. But to achieve what you desire locally will require examining the proof of the Tubular Neighborhood Theorem (see Guillemin and Pollack, p. 76, #16). You want to impose some differential geometric conditions that the diffeomorphism from the normal bundle to a neighborhood of the submanifold preserve orthogonality along the submanifold. I believe that if you check the details of the proof Guillemin and Pollack outline, you'll end up with that condition. (Otherwise, it can be achieved by following geodesics in your ambient manifold $Y$.)</p>


问:
Prove Set is disjoint
If $$n \in \mathbb{Z}, \geq3$$
$${A_{n}}=\left \{ {an+1: a \in \mathbb{Z}} \right \}$$

$${B_{n}}=\left \{ n|(b+1): b \in \mathbb{Z} \right \}$$

How do we prove that the set is disjoint? So This is easy to choose a number, an integer. Then I prove that An is not a subset of Bn and same time, Bn is not a subset of An. I think I need some help.

答: The definition of $B_n$ does not make sense. If we interpret it as $B_n=\{b \in \mathbb Z: n|(b+1)\}$ then statement is correct: Suppose $m \in A_n \cap B_n$. Then $m=an+1$ and $m=kn-1$ for some integer $k$. But then  $(k-a)n=2$ which is  contradiction since $n$ divides then left hand side but not the right hand side.

Title: Prove Set is disjoint


答:
Using Principal values

$$-\dfrac\pi3\le\cos^{-1}x-\dfrac\pi3\le\pi-\dfrac\pi3 $$

$$\cos^{-1}x-\dfrac\pi3=\begin{cases} \cos^{-1}\left(\dfrac x2+\dfrac{\sqrt{3(1-x^2)}}2\right) &\mbox{if } \cos^{-1}x-\dfrac\pi3\ge0\iff  x\le\cos\dfrac\pi3 \\
-\cos^{-1}\left(\dfrac x2+\dfrac{\sqrt{3(1-x^2)}}2\right) & \mbox{if }   x>\cos\dfrac\pi3\end{cases}$$

Now observe that $\dfrac13<\cos\dfrac\pi3=\dfrac12<\dfrac23$

Title: Simplifying $\cos^{-1}x +\cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$

Body:
A question has this equation: $$f(x) = \cos^{-1}x + \cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$$ and you're supposed to simplify it and find $f\left(\frac{2}{3}\right)$ and $f\left(\frac{1}{3}\right)$.

By taking $\cos\alpha  = x$, the equation on the right can be simplified to $\cos^{-1}\left(\cos\left(\frac {\pi}{3} - \alpha\right)\right)$. Finally, you get $\frac{\pi}{3}$ as the final answer.

But the answers are $\frac{\pi}{3}$ and  $2\cos^{-1}\left(\frac{1}{3}\right)-\frac{\pi}{3}$. How does it work out that way?


问:
Prove that $2\sqrt{pq}$ is irrational, with $p$ and $q$ being primes and $p \neq q$
I'm trying to prove that $2\sqrt{pq}$, with $p$ and $q$ primes and $p \neq q$, is irrational.
I have something like this:

Let's suppose that $2\sqrt{pq}$ is rational. Then, we have that $2\sqrt{pq} = \frac{m}{n}$, with $m,n \in \mathbb{Z^{+}}$, and $(m,n) = 1$.

Now, I need to prove that $m \neq 1$ and $n \neq 1$. When I try to prove that $m \neq 1$ I get:

If $m = 1$, then:
\begin{align*}
2\sqrt{pq} &= \frac{m}{n}\\
2n\sqrt{pq} &= 1\\
\end{align*}
This implies $2 \vert 1$, wich is false. Thus, $m\neq1$.

But I don't really know how to prove that $n \neq 1$. The most I can get is:

If $n = 1$, then:
\begin{align*}
2\sqrt{pq} &= \frac{m}{n}\\
2\sqrt{pq} &= m\\
\end{align*}
And this implies $2\sqrt{pq}$ is an integer.

But I don't know how to continue after this.

答:
If $2\sqrt {pq} = a/b$ with $(a,b) = 1$ and $p,q$ distinct primes, then $4pq = a^2/b^2$ so that $b^24pq = a^2$ which implies that $p | a$, and $a=pk$. Then $4pqb^2 = p^2k^2$. Then $p | 4q$ and if $p > 2$ then $p | q$, a contradiction. You might consider the case where $p = 2$ and $q = 3$ in the following way:

If $2\sqrt6 = a/b$ with $(a,b) = 1$ then $24b^2=a^2$ and $3|a$. Then $24b^2 = 9k^2$ so that $8b^2 = 3k^2$ which implies that $3|b$, a contradiction.


答:
When you used the replacement $\cos \alpha = x$, the expression in the bracket became

$$\frac{\cos \alpha}{2} + \frac{\sqrt{3} \sin \alpha}{2} = \cos \frac{\pi}{3} \, \cos \alpha + \sin \frac{\pi}{3} \, \sin \alpha$$

but note that this can be written as both

$$\cos \left( \frac{\pi}{3} - \alpha \right) \quad \text{or} \quad \cos \left( \alpha - \frac{\pi}{3} \right)$$

In order for $\cos^{-1} ( \cos \theta) = \theta,$ we want to avoid $\,\theta \,$ being negative.

When evaluating $f(\frac{2}{3})$, we have that $\,\cos \alpha = \frac{2}{3} \,$, so $\, 0 < \alpha < \frac{\pi}{3}$; but when evaluating $f(\frac{1}{3})$, we have that $\,\cos \alpha = \frac{1}{3} \,$, so $\, 0 < \frac{\pi}{3} < \alpha$.

Thus

$$\begin{align}
f(\frac{2}{3}) &= \alpha + \cos^{-1} \left( \cos \left[ \frac{\pi}{3} - \alpha \right] \right) \\
\\
&= \alpha + \left( \frac{\pi}{3} - \alpha \right) = \boxed{\frac{\pi}{3}}
\end{align}$$

but

$$\begin{align}
f(\frac{1}{3}) &= \alpha + \cos^{-1} \left( \cos \left[ \alpha - \frac{\pi}{3} \right] \right) \\
\\
&= \alpha + \left( \alpha - \frac{\pi}{3} \right) \\
\\
&= 2 \alpha - \frac{\pi}{3} = \boxed{2 \cos^{-1} \left( \frac{1}{3} \right) - \frac{\pi}{3}}
\end{align}$$


问:
Sine and Cosine Models
This is a general question about modeling the seasons using sine and cosine functions; I am trying to use sine and cosine to model cyclic behavior in sales due to the seasons (spring, summer, fall and winter); I have the following seasonal factors so to speak:-

$t = 0$ Start of year

$t = 3$ Spring - $\frac{\sin (\pi t)}{6} $

$t = 6$ Summer - $\frac{\sin(\pi t)}{12}$

$t = 9$ Fall -  $\frac{\sin(\pi t)}{18}$

$t = 12$ Winter -  $\frac{\cos(\pi t)}{6}$

I want to get sales to be max in each period for a given product. For example I want sales for sandals to max in the spring and sales for cardigans to max in the fall.

I Would appreciate advice and guidance as am not sure if these are correct. But looking at the graphs, they seem reasonable.

P.S. this is not a homework problem, this is just me trying out different modeling techniques in sales forecasting.

答:
Consider an arbitrary sinusoid:

$$A\cos\left(\frac{2\pi}{T} t+\phi\right)$$

Where $A$ is the amplitude (the peak price), $T$ is the period (the amount of time it takes for the function to repeat itself), and $\phi$ is the phase shift (the "starting point" of the cycle). You want your period to be 12 months, so the function will be

$$A\cos\left(\frac{\pi}{6}t+\phi\right)$$

Keep in mind though that this function will drop below zero (and I'm sure you don't want to pay your customers to take the merchandise off your hands on off seasons!), so in order for it to vary from $A$ to $0$ instead, cut the amplitude in half and add half the amplitude to the function:

$$\frac{A}{2}\cos\left(\frac{\pi}{6}t+\phi\right)+\frac{A}{2}$$

The phase shift $\phi$ can be calculated by multiplying the time where the peak should occur by the angular frequency (the coefficient of $t$). So for example, if you wanted the price to peak every spring, $\phi=3(\frac{\pi}{6})=\frac{\pi}{2}$.

Title: Sine and Cosine Models


问:
Lp space Inequality with L Infinity
I am currently studying $L^p$ spaces and am trying to prove the following inequality, which I just can't seem to work out:

Let $1 \leq p < r < \infty$. Then if $$f \in L^p(\mathbb{R}) \cap L^\infty(\mathbb{R}), \,\,\|f\|_r \leq (\|f\|_p)^\frac{p}{r}(\|f\|_\infty)^{1 - \frac{p}{r}}.$$

I have tried proving this via the generalized Hölder inequality to no avail, and I also tried just straight-out expanding everything out. Since $p < r < \infty$ and $f \in L^p(\mathbb{R}) \cap L^\infty(\mathbb{R})$, we have that $f \in L^r(\mathbb{R})$ as well.
$$\|f\|_r = \left(\int_\mathbb{R}|f(x)|^r \, dx \right)^\frac{1}{r}, \,\, (\|f\|_p)^\frac{p}{r} = \left(\int_\mathbb{R} |f(x)|^p \, dx\right)^{\frac{1}{p} \cdot \frac{p}{r}} = \left( \int_\mathbb{R} |f(x)|^r \, dx\right)^\frac{1}{p}$$
Now this implies that $\|f\|_r \leq (\|f\|_p)^\frac{p}{r}$, so all I have to do is show that $(\|f\|_\infty)^{1 - \frac{p}{r}} \geq 1$, then this would be proven. But I am not sure where to go here. I tried using the infimum definition of $L^\infty$ without success, and I also tried using the fact that in this case we can write $\|f\|_\infty = \lim_{n \to \infty} \|f\|_n$, again with no success.

Am I approaching this the wrong way, or have I made a mistake? Any hints, corrections, or tips would be appreciated! Thanks!

答: This is much simpler. $\int |f|^{r} =\int |f|^{p} |f|^{r-p} \leq \|f\|_{\infty}^{r-p} \int |f|^{p}$ Just rise both side to power $\frac 1 r$ to finish.

Title: Lp space Inequality with L Infinity


问:
$x\mapsto x^k$ is an automorphism on $C_n$ iff $\gcd(n,k)=1$
Let $G = \langle g\rangle$ be a cyclic group of order $n$ (with multiplication as its operation). Prove that $f(x) = x^k$ is an automorphism on $G$ if and only if $\gcd(n, k) = 1$.

答:
I am going to use additive notation for the cyclic group here. Let $C_n : = \{1,2,3,\dots,n-1\}$ be the cyclic group of order $n$

Forward case: assume $f : C_n \to C_n$ where $f(x) = kx$ is an automorphism. Then, it must be the case that
$$f(\langle x \rangle) = \langle kx \rangle = \langle x \rangle$$
This is only true if $\gcd(n,k)=1$, hence the desired reult.

Backwards case: assume $\gcd(k,n)=1$ and $f : C_n \to C_n$ where $f(x)=kx$. We need to show that $f$ is an isomorphism (and hence an automorphism). First, let us show it is a homomorphism. Pick any $x,y \in C_n$. Then:
$$f(x+y) = k(x+y) = kx+ky = f(x)+f(y)$$
hence $f$ is a homomorphism. Next, to show it is an isomorphism, we need to show it is bijective. Since obviously $|C_n|=|C_n|$, we just need to show $f$ is injective to show it is bijective. Choose any $x,y \in C_n$ such that $f(x)=f(y)$. Then,
$$f(x)=f(y) \implies kx=ky$$
Since $gcd(k,n)=1$, it must be the case that $x=y$, so $f$ is an injection, and hence bijective, so $f$ is an automorphism.


问:
Where does $\frac {\pi}{4}$ come from in this inverse sine graph? $y=-3\sin\frac {x\pi}{4})$
I understand that the vertical shift is $0$ that is why the graph starts at $(0,0)$. Also I understand that the amplitude is $3$ because the maximum y value is $3$ and the minimum y value is $-3$. Last but not least the graph is inverse sine but it seems to be a period and a half. I am not sure where the  $\frac {\pi}{4}$ comes from. I need help understanding that part. Photo of problem attached below. Thanks!

https://imgur.com/gallery/coL4IUw

答:
$y = \sin(x)$ gets a value $1$ at $x=\frac{\pi}{2}$

$ \Rightarrow y = -3\sin(x)$ gets a value $-3$ at $x=\frac{\pi}{2}$

$ \Rightarrow y = -3\sin(ax)$ gets a value $-3$ at $ax=\frac{\pi}{2}$

We have $y=-3$ at $x = 2$.

So, $ax = \frac{\pi}{2} \Rightarrow 2a = \frac{\pi}{2} \Rightarrow\boxed{a = \frac{1}{4}\pi}$

The equation is $\boxed{ y = -3\sin(ax)=-3\sin\left(\frac{1}{4}\pi x\right)}$


问:
Find the elementary matrix which performs a row operation
For a homework problem, I am required to find an elementary matrix E_-2R_1+R_3 which performs the row operation R_3 = -2R_1+R_3 on a given augmented matrix A = $$ \left[
\begin{array}{cccc|c}
  2&6&2&0&0\\
  4&6&1&0&0\\
  -1&0&3&2&0
\end{array}
\right] $$ when it multiplies from the left side. I am also required to show methods for deriving it and and to give row operation notation for the reverse process.

Does this mean that I need to reduce the matrix to RREF to get E? To find the inverse using the reverse process, will I need to involve an identity matrix in any way?

Thank you in advance for suggestions.

答:
Take an identity matrix and perform the operation.  Get $\begin{pmatrix}1&0&0\\0&1&0\\-2&0&1\end{pmatrix}$.

To get the reverse matrix, calculate the inverse.  Or, apply $2R_1+R_3$ to the identity, getting $\begin{pmatrix}1&0&0\\0&1&0\\2&0&1\end{pmatrix}$.


问:
How to find the distance covered by a light in a flat crystal?
The problem is as follows:

The figure from below shows a lactose crystal which is about to be
studied for its refraction index, for this purpose an orange light
beam is focused by means of a lens over a LED. The crystal was cut as
a thin plate. The plate is a rectangle $ABCD$. The measured
$AB=15\overset{\circ}{A}$ and $FH= 6\,A^{\circ}$. Using this
information find the distance $DE$ in Angstrom.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&3\overset{\circ}{A}\\
2.&5\overset{\circ}{A}\\
3.&4\overset{\circ}{A}\\
4.&6\overset{\circ}{A}\\
\end{array}$

Although it is not specified in the problem. I'm assuming that the path which will take the light will be on all the blue lines of the figure, which it would make this into a find the distance $DE$ problem.

However I'm not very sure on how to find such distance. What I thought to do was to use similarity between the triangles:

$$\triangle ABC ~ \triangle FHC$$

But I am still stuck with that part. I have no idea how to continue or what to do then. Can someone help me here?.

I think there may be needed some sort of construction or something, but I don't know. Does it exist a way to solve this relying only in Euclidean postulates?.

Please include a drawing in the answer this part is important because I don't know very well how to spot the right relations here to find the requested distance.

答:
$\triangle CAB\sim\triangle CHF$ so $CH/CA=FH/AB=2/5$.

$\triangle AHF\sim\triangle ACE$ so $FH/CE=(AC-CH)/AC=1-2/5=3/5$.

This gives $CE=5FH/3=10$ and $DE=CD-CE=15-10=5.$


问:
Intuition for straight lines in polar coordinates
In cartesian coordinates, it's very intuitive to get an idea of how a graph of a straight line comes an equation. With polar coordinates, it's not as easy. How can I start to understand how the graph of a line comes from a polar equation without converting it to cartesian form? Thanks.</p>

答: Let $\ell$ be a line and $A(a,\alpha)$ be the point on $\ell$ that is nearest to the pole. For a point $P(r,\theta)$ on $\ell$, $OP=r$ and $\angle POA=|\theta-\alpha|$. Note that $OA=OP\cos\angle POA$. The equation of $\ell$ is $a=r\cos(\theta-\alpha)$.