Math Challenge - August 2019

[nuuskur]

Why is my formatting disappearing all the time? This is the second time this happened to #33

 

[Pi-is-3]

Why is my formatting disappearing all the time? This is the second time this happened to #33

I don't know a lot about $\LaTeX$ on this site, but writing the entire answer in a word and then copy pasting helps me a lot. And @fresh_42 advised me to use windows hot keys. That helps a lot. Best part of word is that formatting is preserved.

 

[nuuskur]

How do you deduce that you have $\mathcal{P}(\mathbb{N})$ elements?

Edited #33 with explanation.

I don't know a lot about $\LaTeX$ on this site, but writing the entire answer in a word

oh lord please have mercy, I don't have the mental capacity required to type mathematical text in word

 

[Pi-is-3]

Edited #33 with explanation.

oh lord please have mercy, I don't have the mental capacity required to type mathematical text in word

I'm sorry. I advised it because when I came to this site, I learned that copy pasting latex causes some problems in this website. So I found out alternative ways. It also had the replace function, so I would type shortcuts, and then find and replace them. It also preserved the formatting.

You can also do it in a $\LaTeX$ editor if you want. That's what I did initially.

 

[member 587159]

Spoiler: Problem 4

Let $\Sigma$ be an infinite sigma algebra on a set $X$ and pick pairwise different $A_n\in\Sigma, n\in\mathbb N$. Put $S_n := \bigcup _{k=1}^nA_k\in\Sigma,n\in\mathbb N$. The sequence $S_n,n\in\mathbb N$ is increasing. Put
$$T_1 := S_1 \quad\mbox{and}\quad T_n := S_n\setminus \bigcup _{k=1}^{n-1}S_k\in \Sigma, \quad n\geq 2.$$
The $T_n$ are pairwise disjoint, thus $\Sigma$ contains at least $\mathcal P(\mathbb N)$ many elements, thus $\Sigma$ is uncountable. Indeed, the map
$$\mathcal P(\mathbb N) \to \Sigma, \quad A\mapsto \begin{cases} \emptyset, &A=\emptyset \\ \bigcup_{x\in A}T_x, &\emptyset\subset A\subset\mathbb N \\ X, &A=\mathbb N \end{cases}$$
is injective.

I'm not convinced this map is injective. It seems that it can happen that $T_x=\emptyset$ and then multiple things get mapped to $\emptyset$, for example.

For example, if $A_1 = X$ and $A_2\in \Sigma,A_2\neq X$. Then $S_1 = S_2 = X$ and $T_2 = \emptyset$. Thus $\{2\}$ and $\emptyset$ get mapped to $\emptyset$, which violates injectivity.

 

[nuuskur]

I'm not convinced this map is injective. It seems that it can happen that $T_x=\emptyset$ and then multiple things get mapped to $\emptyset$, for example.

For example, if $A_1 = X$ and $A_2\in \Sigma,A_2\neq X$. Then $S_1 = S_2 = X$ and $T_2 = \emptyset$. Thus $\{2\}$ and $\emptyset$ get mapped to $\emptyset$, which violates injectivity.

Mhm, good catch. Picking pairwise different $\emptyset \subset A_n\subset X$ gives injectivity.

 

[member 587159]

Mhm, good catch. Picking pairwise different $\emptyset \subset A_n\subset X$ gives injectivity.

Still won't work. Take for example $A_1=\{x\}$ and $A_2 =X\setminus \{x\}$ (assuming these sets are in the sigma algebra, but this can happen take for example the finite complement sigma algebra). Then $S_2 =X$ and any choice of $A_3$ will lead to $S_3=X$ so $T_3=\emptyset$.

 

[nuuskur]

Still won't work. Take for example $A_1=\{x\}$ and $A_2 =X\setminus \{x\}$ (assuming these sets are in the sigma algebra, but this can happen take for example the finite complement sigma algebra). Then $S_2 =X$ and any choice of $A_3$ will lead to $S_3=X$ so $T_3=\emptyset$.

Arghh, I hate these technicalities. We make a decreasing sequence instead inside of a proper subset. The point is that when we have a countable subset of disjoint elements, their generated sub sigma algebra is isomorphic to $\mathcal P(\mathbb N)$ (the $T_n$ correspond to the singletons)

I Changed the picking scheme in #33. You can't pick complements now to make some pathological cases.

 

[fresh_42]

I don't know a lot about $\LaTeX$ on this site, but writing the entire answer in a word and then copy pasting helps me a lot. And @fresh_42 advised me to use windows hot keys. That helps a lot. Best part of word is that formatting is preserved.

https://www.physicsforums.com/help/latexhelp/ (short instruction help for PF)
http://detexify.kirelabs.org/symbols.html (additional symbols, some available, some not)
Google: TeX editor (for longer or separate texts) and HotKey (for keyboard programming, e.g. Alt+F for \frac{}{})

 

[nuuskur]

Spoiler: Problem 9

If I get this right the situation is something like this

Spoiler: triangles

As time passes the triangle inside gets smaller and smaller until it vanishes. Call the side length $L(t)$. After $\delta t$ seconds has passed, the new side length would be $L(t+\delta t)$. Apply the cosine law to the triangle $EAF$ which yields
$$|FE|^2 = |AF|^2 + |AE|^2 - 2\cos \frac{\pi}{3} |AF||AE| = |AF|^2 + |AE|^2 - |AF||AE|.$$
i.e
$$L(t+\delta t)^2 = (v\delta t)^2 + (L(t) - v\delta t)^2 - (v\delta t)(L(t) - v\delta t).$$
Rearranging we get
$$L(t+\delta t)^2 - L(t)^2 = (3v\delta t - 3L(t))v\delta t$$
Divide by $\delta t$ and take the limit as $\delta t \to 0$, thus
$$\frac{d}{dt}L(t)^2 = -3vL(t).$$
Apply the chain rule and we get an initial value problem:
$$L'(t) = -\frac{3}{2}v, \quad L(0) = L.$$
Hence $L(t) = L-\frac{3}{2}vt$ and at $t=\frac{2L}{3v}$ the triangle vanishes.

 

[fresh_42]

Yes, and the flight path (parametrized by time)?

 

[archaic]

Spoiler: Problem 9

If I get this right the situation is something like this

Spoiler: triangles

View attachment 247484

As time passes the triangle inside gets smaller and smaller until it vanishes. Call the side length $L(t)$. After $\delta t$ seconds has passed, the new side length would be $L(t+\delta t)$. Apply the cosine law to the triangle $EAF$ which yields
$$|FE|^2 = |AF|^2 + |AE|^2 - 2\cos \frac{\pi}{3} |AF||AE| = |AF|^2 + |AE|^2 - |AF||AE|.$$
i.e
$$L(t+\delta t)^2 = (v\delta t)^2 + (L(t) - v\delta t)^2 - (v\delta t)(L(t) - v\delta t).$$
Rearranging we get
$$L(t+\delta t)^2 - L(t)^2 = (3v\delta t - 3L(t))v\delta t$$
Divide by $\delta t$ and take the limit as $\delta t \to 0$, thus
$$\frac{d}{dt}L(t)^2 = -3vL(t).$$
Apply the chain rule and we get an initial value problem:
$$L'(t) = -\frac{3}{2}v, \quad L(0) = L.$$
Hence $L(t) = L-\frac{3}{2}vt$ and at $t=\frac{2L}{3v}$ the triangle vanishes.

Another way of doing this is to notice that since the triangle is always equilateral, the speed in the direction of the center is $v\cos{30}$ and the length to the center is $l/\sqrt{3}$ so $t=\frac{v\cos{30}}{l/\sqrt{3}}$.
Or you could also think of the distance between two successive points that varies by $\Delta d \approx -(v\Delta t+v\cos{60}\Delta t)$

 

[Pi-is-3]

Another way of doing this is to notice that since the triangle is always equilateral, the speed in the direction of the center is $v\cos{30}$ and the length to the center is $l/\sqrt{3}$ so $t=\frac{v\cos{30}}{l/\sqrt{3}}$.
Or you could also think of the distance between two points that varies by $\Delta d \approx -(v\Delta t+v\cos{60}\Delta t)$

I didn't write the answer for this one, because I didn't really come up with the solution. It was an example question in my physics textbook, and it had 2 methods of solving it. One of them was similar to yours and the other was similar to that of @nuuskur gave.

 

[nuuskur]

Oops, I forgot about the trajectory. Well, in short, it's ugly. Actually, this problem is much better in polar coordinates..

Spoiler: Problem 9 - trajectory

Suffices to find explicitly one trajectory. Denote position of plane by $(r(t), \varphi (t))$, where at $t=0$ we have $r(0) = L$ and $\varphi (0) = 0$ (apply translation or reverse traversion later to line up with some initial parameters if necessary).

The plane that is being followed has position $(r(t), \varphi (t) + \alpha)$, where $\alpha = \frac{2\pi}{3}$. So we get
$$\frac{(r\sin \varphi)_t}{(r\cos \varphi )_t} = \frac{r(\sin (\varphi +\alpha) - \sin\varphi)}{r(\cos (\varphi +\alpha) - \cos\varphi)}$$
which yields (after a while)
$$r'\sin\alpha + r\varphi '(\cos\alpha -1) = 0,$$
i.e
$$r' - \sqrt{3}r\varphi ' =0\tag{1}$$
The flight speed is constant so we have
$$v^2 = (r\sin\varphi)_t^2 + (r\cos\varphi)_t^2 = r'^2 + r^2\varphi '^2 \tag{2}$$
From (1) and (2) we get $r' = \frac{\sqrt{3}}{2}v$ sooo I think I have calculated incorrectly somewhere. Let's roll with it for now. Anyway, the angle is given by (1) and (2) as
$$\varphi ' = \frac{v}{2r} = \frac{v}{2L - \sqrt{3}vt} \implies \varphi (t) = \int \frac{v}{2L-\sqrt{3}vt}dt = - \frac{\sqrt{3}}{3} \ln |2L-\sqrt{3}vt| + C$$
where $C = \frac{\sqrt{3}}{3} \ln 2L$

I think I made a calculation error somewhere, too tired to (double check) $^{10}$ right now.

 

[fresh_42]

Oops, I forgot about the trajectory. Well, in short, it's ugly. Actually, this problem is much better in polar coordinates..

Spoiler: Problem 9 - trajectory

Suffices to find explicitly one trajectory. Denote position of top plane by $(r(t), \varphi (t))$, where at $t=0$ we have $r(0) = L$ and $\varphi (0) = 0$ (apply translation or reverse traversion later to line up with some initial parameters if necessary).

The plane that is being followed has position $(r(t), \varphi (t) + \alpha)$, where $\alpha = \frac{2\pi}{3}$. So we get
$$\frac{(r\sin \varphi)_t}{(r\cos \varphi )_t} = \frac{r(\sin (\varphi +\alpha) - \sin\varphi)}{r(\cos (\varphi +\alpha) - \cos\varphi)}$$
which yields (after a while)
$$r'\sin\alpha + r\varphi '(\cos\alpha -1) = 0,$$
i.e
$$r' - \sqrt{3}r\varphi ' =0\tag{1}$$
The flight speed is constant so we have
$$v^2 = (r\sin\varphi)_t^2 + (r\cos\varphi)_t^2 = r'^2 + r^2\varphi '^2 \tag{2}$$
From (1) and (2) we get $r' = \frac{\sqrt{3}}{2}v$ sooo I think I have calculated incorrectly somewhere. Let's roll with it for now. Anyway, the angle is given by (1) and (2) as
$$\varphi ' = \frac{v}{2r} = \frac{v}{2L - \sqrt{3}vt} \implies \varphi (t) = \int \frac{v}{2L-\sqrt{3}vt}dt = - \frac{\sqrt{3}}{3} \ln |2L-\sqrt{3}vt| + C$$
where $C = \frac{\sqrt{3}}{3} \ln 2L$

I think I made a calculation error somewhere, too tired to (double check) $^{10}$ right now.

I think it's ok more or less. I don't have time now to check in detail either.
In my version, the ugly parts are simply initial conditions and then it doesn't look ugly at all:

To get the flight path we decompose $\vec{v}(t)$ in components parallel and perpendicular to $\vec{r}(t).$ The perpendicular component is $|v_\perp| = v\cdot \sin \psi = \dfrac{v}{2}$ so we have the angular velocity $\dot{\omega}(t)=\dfrac{v_\perp (t)}{r(t)}.$ We parameterize the motion by cylindric coordinates $\vec{r}(t)=(r(t)\cos \varphi(t)\, , \,-r(t)\sin \varphi(t)\, , \,0)^\tau$ and receive the momentary rotation angle by the integration
\begin{align*}
\varphi(t)&=\varphi(0) + \int_0^t \omega(t')\,dt'\\
&=\varphi(0)+ \dfrac{v}{2} \int_0^t \dfrac{1}{r(t')}\,dt'\\
&=\varphi(0)+ \dfrac{v}{2} \int_0^t \dfrac{1}{\frac{L}{\sqrt{3}}-v\frac{\sqrt{3}}{2}t'}\,dt'\\
&= \varphi(0)+ \int_0^t \dfrac{1}{\dfrac{2L}{\sqrt{3}v}-\sqrt{3}t'}\,dt'\\
&=\varphi(0)+ \dfrac{1}{\sqrt{3}}\int_0^t \dfrac{1}{\frac{2L}{3v}-t'}\,dt'\\
&=\varphi(0)+ \dfrac{1}{\sqrt{3}} \log\left( \dfrac{\frac{2L}{3v}}{\frac{2L}{3v}-t} \right)\\
&=\varphi(0)+\dfrac{1}{\sqrt{3}} \log\left( \dfrac{r(0)}{r(t)} \right)
\end{align*}
so the flight path is the logarithmic spiral with
$$
r(t) = r(0)\cdot e^{-\sqrt{3}\left( \varphi(t)-\varphi(0) \right)}
$$
The distance towards the center decreases by a factor of $e^{-2\pi \sqrt{3}}\approx 1.88 \cdot 10^{-5}$ with every complete turn.

 

[nuuskur]

Spoiler: Problem 8 - an idea

Making use of induction. Suppose $a,b,c$ is a nonzero solution. Due to $x^2\equiv 0,1\pmod{3}$ it must hold that $a^2,b^2$ are multiples of three. Putting $d:= (a,b)$, if $d\mid c$ holds, then $\left (\frac{a}{d},\frac{b}{d},\frac{c}{d}\right )=:(a',b',c')$ would be a solution with $(a',b')=1$ which is impossible.

Could we show $d\mid c$?

 

[Pi-is-3]

Spoiler: Problem 8 - an idea

Making use of induction. Suppose $a,b,c$ is a nonzero solution. Due to $x^2\equiv 0,1\pmod{3}$ it must hold that $a^2,b^2$ are multiples of three. Putting $d:= (a,b)$, if $d\mid c$ holds, then $\left (\frac{a}{d},\frac{b}{d},\frac{c}{d}\right )=:(a',b',c')$ would be a solution with $(a',b')=1$ which is impossible.

Could we show $d\mid c$?

In this method, $a^2=9x^2, b^2=9y^2$ .

This implies $c^2$ is a multiple of 3. Then you can continue with the same method I did. Using contradiction to prove that solutions don't exist.

However, I'm not sure how to prove that d divides c (if I'm not wrong, d is the gcd of a,b).

 

[member 587159]

Sure. First I'll explain why x is even. Since l is less than equal to m and n, we can divide both the sides by $2^{2l}$ . That in any scenario gives either x is even or x and y are even or all x,y,z are even (when l=m=n) by taking mod 4 both sides. A similar reasoning is followed for z.

As to why it is okay to consider only two cases instead of three (by including the case of m being smaller), considering the case for x is the same as considering the case for y. If it hadn't been the same, I would have to consider 3 cases. I don't know how to explain this one any better. You could think that if (a,b,c) satisfy the equation, then (b,a,c) will also satisfy.

I'm satisfied with your explanations. Well done!

In fact, there is a shorter solution.

You got to $a^2+b^2 +c^2\equiv 0 \bmod 4$, implying that $2|a,b,c$. Now, if we take a minimal positive solution $(a,b,c)$ this procedure yields the solution $(a/2,b/2,c/2)$, contradicting minimality.

I think this is in the lines of what @nuuskur was trying.

 

[fresh_42]

I was thinking about it in a "give a counter-example" fashion. The question then is rather easy don't you think? I could put forward any function $f$ where $f(0) \neq f(1)$ for example.

... plus $f'(x)\neq 0$ for all $x$.

If you have three conditions which are needed to draw a single conclusion, then the violation of any of the conditions together with the opposite of the conclusion is a counterexample. The best solution would have been to list three examples, violating one condition after the other. And a fourth example where the conclusion holds without any of the conditions.

 

[nuuskur]

I think this is in the lines of what @nuuskur was trying.

Indeed, formally it's known as method of infinite descent - stems from induction (equivalent, too, I think). By the way, I revised #33.

 

[nuuskur]

Spoiler: Problem 7a

Suppose $g\mapsto g^{-1}$ is an endomorphism and take $g,h\in G$. Then $gh \mapsto (gh)^{-1}$. Due to compatibility $gh \mapsto g^{-1}h^{-1}$, thus
$$(gh)^{-1} = g^{-1}h^{-1} = (hg)^{-1} \iff gh = hg.$$
Suppose $G$ is abelian then for every $g,h\in G$
$$gh \mapsto (gh)^{-1} = h^{-1}g^{-1} = g^{-1} h^{-1}.$$

Spoiler: Problem 7b

The map $g\mapsto g^{-1}\sigma (g)$ on $G$ is injective, because if $g^{-1}\sigma (g) = h^{-1}\sigma (h)$, then $hg^{-1} = \sigma (hg^{-1})$. The only fixpoint is $e$, thus $h=g$. The map is also surjective, because $G$ is a finite set.

Now, pick $g\in G$ and write $g= h^{-1}\sigma (h)$, then
$$\sigma (g) = \sigma (h^{-1}\sigma (h)) = (\sigma (h))^{-1} (h^{-1})^{-1} = (h^{-1}\sigma (h))^{-1} = g^{-1}.$$
By 7a $G$ is abelian.

 

[member 587159]

Spoiler: Problem 7a

Suppose $g\mapsto g^{-1}$ is an endomorphism and take $g,h\in G$. Then $gh \mapsto (gh)^{-1}$. Due to compatibility $gh \mapsto g^{-1}h^{-1}$, thus
$$(gh)^{-1} = g^{-1}h^{-1} = (hg)^{-1} \iff gh = hg.$$
Suppose $G$ is abelian then for every $g,h\in G$
$$gh \mapsto (gh)^{-1} = h^{-1}g^{-1} = g^{-1} h^{-1}.$$

Spoiler: Problem 7b

The map $g\mapsto g^{-1}\sigma (g)$ on $G$ is injective, because if $g^{-1}\sigma (g) = h^{-1}\sigma (h)$, then $hg^{-1} = \sigma (hg^{-1})$. The only fixpoint is $e$, thus $h=g$. The map is also surjective, because $G$ is a finite set.

Now, pick $g\in G$ and write $g= h^{-1}\sigma (h)$, then
$$\sigma (g) = \sigma (h^{-1}\sigma (h)) = (\sigma (h))^{-1} (h^{-1})^{-1} = (h^{-1}\sigma (h))^{-1} = g^{-1}.$$
By 7a $G$ is abelian.

Correct! I will give feedback on the measure theory question soon! Bit busy now!

 

[nuuskur]

Spoiler: Problem 8 - an idea

Making use of induction. Suppose $a,b,c$ is a nonzero solution. Due to $x^2\equiv 0,1\pmod{3}$ it must hold that $a^2,b^2$ are multiples of three. Putting $d:= (a,b)$, if $d\mid c$ holds, then $\left (\frac{a}{d},\frac{b}{d},\frac{c}{d}\right )=:(a',b',c')$ would be a solution with $(a',b')=1$ which is impossible.

Could we show $d\mid c$?

Spoiler: Problem 8 - divisibility

Write
$$a = 3 ^{k_1} \ldots p_n^{k_n} \quad b = 3^{l_1}\ldots p_n^{l_n} \quad c = 3^{r_1} \ldots p_n ^{r_n}$$
where the powers are non-negative. Then $(a,b) = 3 ^{m_1} \ldots p_n ^{m_n}$, where $m_j = \min \{k_j,l_j\}$. The goal is to show $m_j\leq r_j$. The initial equality can be written as
$$3^{2m_1} \ldots p_n^{2m_n} \left ( 3^{2k_1 - 2m_1} \ldots p_n^{2k_n-2m_n} + 3^{2l_1 - 2m_1} \ldots p_n^{2l_n-2m_n}\right ) = 3^{2r_1+1}\ldots p_n^{2r_n}$$
Suppose, for a contradiction, some $m_j>r_j$. The case $j\neq 1$ would run into $2m_j + k = 2r_j$, where $k\geq 0$, which is impossible. So it must be that $j=1$. But then $2m_1 + k = 2r_1 + 1$ for some $k\geq 0$. By assumption $m_1 \geq r_1+1$ so
$$2m_1 + k = 2r_1 + 1 \Rightarrow 2r_1 + 1 - k \geq 2r_1 + 2 \Rightarrow k\leq -1,$$
a contradiction. Thus $(a,b) \mid c$.

I think this is overkill, not sure how to optimise.

 

[member 587159]

Spoiler: Problem 4

Let $\Sigma$ be an infinite sigma algebra on a set $X$, choose $\emptyset\subset A_1\subset X$ and pick for every $n\in\mathbb N$
$$A_{n+1} \in \Sigma \setminus \left ( \left\lbrace\emptyset, X\right\rbrace \cup \left\lbrace\bigcup_{k=1}^mA_k\mid m\leq n\right\rbrace\cup\left\lbrace\bigcap _{k=1}^m A_k^c\mid m\leq n\right\rbrace\right ).$$
$\Sigma$ is infinite, so we can do this. Put $S_n := \bigcup _{k=1}^nA_k\in\Sigma,n\in\mathbb N$. Observe that all $S_n\subset X$.

The sequence $S_n^c,n\in\mathbb N,$ is strictly decreasing. Put
$$T_n := S_n^c\setminus S_{n+1}^c\in \Sigma, \quad n\in\mathbb N.$$
The $T_n$ are pairwise disjoint. The map
$$\mathcal P(\mathbb N) \to \Sigma, \quad A\mapsto \begin{cases} \emptyset, &A=\emptyset \\ \bigcup_{x\in A}T_x, &\emptyset\subset A\subset\mathbb N \\ X, &A=\mathbb N \end{cases}$$
is injective. Note that if $A\subset\mathbb N$ then $\bigcup _{x\in A}T_x \subseteq S_1^c \subset X$.

What happens when we take $X =\mathbb{N}, A_1=\{0,1\},A_2=\{0\}$?

Then $S_1=S_2 = A_1$ and $T_1=\emptyset$ so again injectivity is violated.

 

[nuuskur]

What happens when ...

This problem is of the devil mhh, the following is a cheapshot

Spoiler: Problem 4

Take a sequence of pairwise disjoint nonempty subsets $S_n\in\Sigma, n\in\mathbb N$. Then
$$\mathcal P(\mathbb N) \setminus \{\emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} S_x$$
is injective, because of disjointedness. Construction of sequence given in #33. We take pairwise distinct $\emptyset \subset A_n \subset X\in\Sigma, n\in\mathbb N$ and put $S_n := \bigcup _{k=1}^n A_k\in\Sigma$. Then $T_n := S_n^c \setminus S_{n+1}^c\in\Sigma, n\in\mathbb N,$ is a sequence of pairwise disjoint non-empty subsets.

 

[member 587159]

This problem is of the devil mhh, the following is a cheapshot

Spoiler: Problem 4

Take a sequence of pairwise disjoint nonempty subsets $S_n\in\Sigma, n\in\mathbb N$. Then
$$\mathcal P(\mathbb N) \setminus \{\emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} S_x$$
is injective, because of disjointedness.

Now we are going somewhere. Or course, the question is why such a sequence exists. How can we construct one?

 

[member 587159]

This problem is of the devil mhh, the following is a cheapshot

Spoiler: Problem 4

Take a sequence of pairwise disjoint nonempty subsets $S_n\in\Sigma, n\in\mathbb N$. Then
$$\mathcal P(\mathbb N) \setminus \{\emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} S_x$$
is injective, because of disjointedness. Construction of sequence given in #33. We take pairwise distinct $\emptyset \subset A_n \subset X\in\Sigma, n\in\mathbb N$ and put $S_n := \bigcup _{k=1}^n A_k\in\Sigma$. Then $T_n := S_n^c \setminus S_{n+1}^c\in\Sigma, n\in\mathbb N,$ is a sequence of pairwise disjoint non-empty subsets.

Again, it can happen that $T_i=\emptyset$ for some i. See my last counterexample.

 

[nuuskur]

Again, it can happen that $T_i=\emptyset$ for some i. See my last counterexample.

You're right, we need a change of gears ..

Spoiler: Problem 4 - strictly nested sequence

Call a subset $S\in \Sigma$ to have property $(P)$ iff the sub sigma algebra $\Sigma _S := \{S\cap A \mid A\in\Sigma\}$ is infinite. Note that
$$\Sigma = \sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ).\tag{E}$$
On the one hand we have by definition
$$\Sigma _S \cup \Sigma _{S^c} \subseteq \Sigma \Rightarrow\sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ) \subseteq \sigma (\Sigma) = \Sigma.$$
Conversely, take $A\in \Sigma$, then
$$A = A\cap (S\cup S^c) = A\cap S \cup A\cap S^c \in \sigma \left ( \Sigma _S \cup \Sigma _{S^c}\right ).$$
Suffices to show the following: if $S$ has property $(P)$, then there exists a partition $S = T\dot{\cup}T'$ such that $T,T'\in\Sigma _S$ are non-empty and at least one of them has property $(P)$. Since $X$ has property $(P)$ by assumption, we can repeatedly apply this fact and obtain a strictly decreasing sequence of non-empty proper subsets.

Proof of fact. Suppose $S\in\Sigma$ has property $(P)$. Then $\Sigma _S$ is an infinite sub sigma algebra. Pick $T \in \Sigma _S \setminus \{\emptyset, S\}$ Write $S= T\cup (S\cap T^c)$. By (E) at least one of the respective sub sigma algebras must be infinite.

 

[nuuskur]

For the sake of convenience/not cluttering denote

$$\frac{d^n}{dz^n} f(z) =: f_n(z) =: f_n \quad\mbox{and}\quad f^n(z) := (f(z)) ^n =: f^n.$$

Spoiler: Problem 10

Some preliminaries. By the chain and product rules

$$\begin{align*} (gf)_1 &= g_1(f)f_1 \\ (gf)_2 &= g_2(f)f_1^2 + g_1(f)f_2 \\ (gf)_3 &= g_3(f)f_1^2 + 3g_2(f)f_1f_2 + g_1(f)f_3 \end{align*}$$

Checking ahead in wiki - the Schwarzian in general would have to be:

$$S_{gf} = (S_g \circ f) \cdot f_1^2 + S_f$$

Now it's pretty straightforward:

$$\begin{align*} S_{gf} &= \frac{(gf)_3}{(gf)_1} - \frac{3}{2} \left (\frac{(gf)_2}{(gf)_1}\right )^2 \\ &= \frac{g_3(f)f_1^3 + 3g_2(f)f_1f_2 + g_1(f)f_3}{g_1(f)f_1} \\ &-\frac{3}{2} \cdot \frac{g_2^2(f)f_1^4 + 2g_1(f)g_2(f)f_1^2f_2 + g_1^2(f)f_2^2}{g_1^2(f)f_1^2} \\ \end{align*}$$
Post-reduction:
$$\begin{align*} \frac{g_3(f)}{g_1(f)}f_1^2 + \frac{f_3}{f_1} - \frac{3}{2} \left (\frac{g_2^2(f)}{g_1^2(f)}f_1^2 + \frac{f_2^2}{f_1^2}\right ) \end{align*}$$
Re-arrange:
$$\begin{align*} S_{gf} &= \left (\frac{g_3(f)}{g_1(f)} - \frac{3}{2} \left ( \frac{g_2(f)}{g_1(f)}\right )^2\right ) f_1^2 + \left (\frac{f_3}{f_1} - \frac{3}{2} \left (\frac{f_2}{f_1}\right )^2\right ) \\ &= (S_g \circ f) f_1^2 + S_f \end{align*}$$
The Schwarzian is said to have negative derivative if $f'(z) \neq 0$ implies $S_f (z) <0$, thus if both Schwarzians are negative, then by definition the sum is a negative Schwarzian.

 

[nuuskur]

Spoiler: Problem 6

Suppose $G$ is of order $2m$, where $m>1$ is odd.

Consider the map
$$\varphi : G \to \mbox{Sym}(G) : \varphi (g)(x) := gx, x\in G.$$

Spoiler: well-defined, morphism

the map $\varphi (g)$ is injective, because $gx = gy$ implies $x=y$ and for a fixed $h\in G$ we have $\varphi (g) (g^{-1}h) = gg^{-1}h = h$. Take $g,h\in G$, then
$$\varphi (gh)(x) = (gh)x = g(hx) = g \varphi (h)(x) = \varphi (g) (\varphi (h)(x)) = (\varphi (g)\varphi (h)) (x)\quad (x\in G)$$

By Sylow's first theorem there exists $a \in G$ of order two. Note that $(g,ga), g\in G,$ are cycles in the permutation $\varphi (a)$, because
$$\varphi (a)(g) = ga \Rightarrow \varphi (a)(ga) = gaa = ge =g.$$
Due to injectivity, the cycles are disjoint, thus there are precisely $m$ such cycles, which implies $\varphi (a)$ is an odd permutation. Consider the signum morphism $\varepsilon : \mbox{Sym}(G) \to \{-1,1\}$ where $\varepsilon (\sigma ) = 1$ iff $\sigma$ is an even permutation. We have $\varepsilon \varphi : G\to \{-1,1\}$ a surjective morphism, because $e\mapsto 1$ ($\varphi (e)$ has no inversions). By first isomorphism theorem
$$G / \mbox{Ker} (\varepsilon\varphi) \cong \{-1,1\}$$
which implies $|\mbox{Ker} (\varepsilon\varphi)| = m$. Also $\mbox{Ker} (\varepsilon\varphi) \triangleleft G$.

summer's almost over ..

 

[member 587159]

Spoiler: Problem 6

Suppose GG is of order 2m2m, where m>1m>1 is odd.

Consider the map
$$\varphi : G \to \mbox{Sym}(G) : \varphi (g)(x) := gx, x\in G.$$

Spoiler: well-defined, morphism

the map φ(g)φ(g) is injective, because gx=gygx=gy implies x=yx=y and for a fixed h∈Gh∈G we have φ(g)(g−1h)=gg−1h=hφ(g)(g−1h)=gg−1h=h. Take g,h∈Gg,h∈G, then
$$\varphi (gh)(x) = (gh)x = g(hx) = g \varphi (h)(x) = \varphi (g) (\varphi (h)(x)) = (\varphi (g)\varphi (h)) (x)\quad (x\in G)$$

By Sylow's first theorem there exists a∈Ga∈G of order two. Note that (g,ga),g∈G,(g,ga),g∈G, are cycles in the permutation φ(a)φ(a), because
$$\varphi (a)(g) = ga \Rightarrow \varphi (a)(ga) = gaa = ge =g.$$
Due to injectivity, the cycles are disjoint, thus there are precisely mm such cycles, which implies φ(a)φ(a) is an odd permutation. Consider the signum morphism ε:Sym(G)→{−1,1}ε:Sym(G)→{−1,1} where ε(σ)=1ε(σ)=1 iff σσ is an even permutation. We have εφ:G→{−1,1}εφ:G→{−1,1} a surjective morphism, because e↦1e↦1 (φ(e)φ(e) has no inversions). By first isomorphism theorem
$$G / \mbox{Ker} (\varepsilon\varphi) \cong \{-1,1\}$$
which implies |Ker(εφ)|=m|Ker(εφ)|=m. Also Ker(εφ)◃GKer(εφ)◃G.

summer's almost over ..

Correct! Well done! Could you maybe give some insight in why you thought about that action?

 

[Not anonymous]

Summary: 1. - 2. posed and moderated by @QuantumQuest
3. - 8. posed and moderated by @Math_QED
9. - 10. posed and moderated by @fresh_42

keywords: calculus, abstract algebra, measure theory, mechanics, dynamical systems

Questions
9. Three identical airplanes start at the same time at the vertices of an equilateral triangle with side length $L$. Let's say the origin of our coordinate system is the center of the triangle. The planes fly at a constant speed $v$ above ground in the direction of the clockwise next airplane. How long will it take for the planes to reach the same point, and which are the flight paths?

Disclaimer: This spoiler may not be a spoiler at all, as I am no sure if the answer is correct :) Too long since I last used calculus to solve anything other than the most basic and even longer since I last worked on solving a physics problem :)

If the solution is incorrect, please explain which assumption or step is incorrect.

Spoiler: Q9

I am unable to draw here the diagram I sketched on paper to derive the solution, but the fundamental observation is that if all the 3 planes fly at the same speed, then at any point in time till they meet, the positions of the 3 planes will continue to form an equilateral triangle and this triangle keeps shrinking in size as time progresses.

Suppose the current distance between 2 planes is $a$ and $\delta$ is the small distance traveled by each of them, along the direction of edges of the current triangle, in a very small instant of time. Let $b$ denote the distance between the 2 planes after this instant. Using trigonometry, I found that the following equation determines the value of $b$:

$b^2 = a^2 - 3 \delta (a - \delta)$

If $v$ is the constant velocity and $\Delta t$ is the small time instant, then $\delta = v \Delta t$, yielding $b^2 = a^2 - 3 v \Delta t (a - v\Delta t)$

If $D^2$ is used as the variable denoting the squared length of side of equilateral triangle, then we see that $\Delta D^2 = -3 v \Delta t (D - v\Delta t)$.

$$
\lim_{\Delta t \rightarrow 0} {\frac {\Delta D^2} {\Delta t}} = -3 v D \Rightarrow
\frac {dD^2} {dt} = -3 v D \Rightarrow 2 D \frac {dD} {dt} = - 3 v D \Rightarrow \\
\frac {dD} {dt} = - \frac {3} {2} v
$$

So the cumulative change in the value of $D$ over a period of time $T$ would be
$$
\int_0^T - \frac {3} {2} v \, dt = - \frac {3} {2} v T
$$

When the planes meet, the value of D would be 0 whereas its initial value is $L$, so the cumulative change in $D$ has to be $-L$. Substituting this in the previous equation gives the value of $T$ to be

$$
-L / (- \frac {3} {2} v) = \frac {2L} {3v}
$$

 

[nuuskur]

I think P6 generalises via induction to the case $2^rm$.

Correct! Well done! Could you maybe give some insight in why you thought about that action?

I sort of started at the back. My hopes and dreams were that I could compose the signum morphism with some $\varphi : G\to\mbox{Sym}(G)$ such that I would get $\varepsilon \varphi : G\to \{-1,1\}$ surjective and apply the first isomorphism theorem. Of course, getting a preimage for $1$ is trivial, but why woud there be an odd permutation? Then I thought maybe there is a permutation in $\mbox{Sym}(G)$ which contains an odd number of disjoint cycles of length two making it an odd permutation, this is where the order $2m$ gave a clue. A way to get such cycles is to put $(g,ga)$ where $a$ is of order two and at that point the definition of $\varphi$ became self-evident.

 

[member 587159]

I think P6 generalises via induction to the case $2^rm$.

I sort of started at the back. My hopes and dreams were that I could compose the signum morphism with some $\varphi : G\to\mbox{Sym}(G)$ such that I would get $\varepsilon \varphi : G\to \{-1,1\}$ surjective and apply the first isomorphism theorem. Of course, getting a preimage for $1$ is trivial, but why woud there be an odd permutation? Then I thought maybe there is a permutation in $\mbox{Sym}(G)$ which contains an odd number of disjoint cycles of length two making it an odd permutation, this is where the order $2m$ gave a clue. A way to get such cycles is to put $(g,ga)$ where $a$ is of order two and at that point the definition of $\varphi$ became self-evident.

Yes, using actions to find kernels is something that works when all other things seem to fail. You are right btw that this exercise generalises (via induction)

The generalisation is:

If $G$ is a group of order $2^nm$ with $m$ odd and $G$ has a cyclic Sylow 2-subgroup, then $G$ has a normal subgroup of order $m$.

 

[fresh_42]

Disclaimer: This spoiler may not be a spoiler at all, as I am no sure if the answer is correct :) Too long since I last used calculus to solve anything other than the most basic and even longer since I last worked on solving a physics problem :)

If the solution is incorrect, please explain which assumption or step is incorrect.

Spoiler: Q9

I am unable to draw here the diagram I sketched on paper to derive the solution, but the fundamental observation is that if all the 3 planes fly at the same speed, then at any point in time till they meet, the positions of the 3 planes will continue to form an equilateral triangle and this triangle keeps shrinking in size as time progresses.

Suppose the current distance between 2 planes is $a$ and $\delta$ is the small distance traveled by each of them, along the direction of edges of the current triangle, in a very small instant of time. Let $b$ denote the distance between the 2 planes after this instant. Using trigonometry, I found that the following equation determines the value of $b$:

$b^2 = a^2 - 3 \delta (a - \delta)$

If $v$ is the constant velocity and $\Delta t$ is the small time instant, then $\delta = v \Delta t$, yielding $b^2 = a^2 - 3 v \Delta t (a - v\Delta t)$

If $D^2$ is used as the variable denoting the squared length of side of equilateral triangle, then we see that $\Delta D^2 = -3 v \Delta t (D - v\Delta t)$.

$$
\lim_{\Delta t \rightarrow 0} {\frac {\Delta D^2} {\Delta t}} = -3 v D \Rightarrow
\frac {dD^2} {dt} = -3 v D \Rightarrow 2 D \frac {dD} {dt} = - 3 v D \Rightarrow \\
\frac {dD} {dt} = - \frac {3} {2} v
$$

So the cumulative change in the value of $D$ over a period of time $T$ would be
$$
\int_0^T - \frac {3} {2} v \, dt = - \frac {3} {2} v T
$$

When the planes meet, the value of D would be 0 whereas its initial value is $L$, so the cumulative change in $D$ has to be $-L$. Substituting this in the previous equation gives the value of $T$ to be

$$
-L / (- \frac {3} {2} v) = \frac {2L} {3v}
$$

This seems to be correct and is similar to what @nuuskur did
https://www.physicsforums.com/threads/math-challenge-august-2019.975478/page-2#post-6214607

My solution is slightly different, but of course not basically.

The side length of the triangle at $t=0$ is $L(0)=L.$ For the position $\vec{r}(t)$ of the first airplane we have $|\vec{r}(0)|=r(0)=\dfrac{2}{3}L\cos \dfrac{\pi}{6}=\dfrac{L}{\sqrt{3}}.$ The distance between the airplanes are the same at any point in time, because of the symmetry, i.e. the airplanes will always mark the vertices of an equilateral triangle with its center at the origin. Thus the angle between the velocity $\vec{v}(t)$ and the position $\vec{r}(t)$ is always
$$
\sphericalangle (\vec{v}(t),\vec{r}(t))=\psi(t)=\psi(0)=\psi =\pi - \dfrac{\pi}{6}
$$
Thus we have
\begin{align*}
\vec{v}(t)&=\dot{\vec{r}}(t)\\
\vec{r}(t))\vec{v}(t)&=\vec{r}(t)\dot{\vec{r}}(t)\\
r\cdot v \cdot \cos \psi &=\dfrac{1}{2}\dfrac{d}{dt}(\vec{r}(t)\vec{r}(t)\\
r\cdot v \cdot \cos \psi &=\dfrac{1}{2}\dfrac{dr^2}{dt}\\
r\cdot v \cdot \cos \psi &=r\dfrac{dr}{dt}\\
\dfrac{dr}{dt}&= -v \dfrac{\sqrt{3}}{2}\\
r(t)&= \dfrac{L}{\sqrt{3}}-v\dfrac{\sqrt{3}}{2}t
\end{align*}
Hence $r(t_f)=0$ implies $t_f=\dfrac{2L}{3v}.$