- #36
nuuskur
Science Advisor
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Why is my formatting disappearing all the time? This is the second time this happened to #33
nuuskur said:Why is my formatting disappearing all the time? This is the second time this happened to #33
Edited #33 with explanation.Math_QED said:How do you deduce that you have ##\mathcal{P}(\mathbb{N})## elements?
oh lord please have mercy, I don't have the mental capacity required to type mathematical text in wordPi-is-3 said:I don't know a lot about ##\LaTeX## on this site, but writing the entire answer in a word
nuuskur said:Edited #33 with explanation.
oh lord please have mercy, I don't have the mental capacity required to type mathematical text in word
nuuskur said:Let [itex]\Sigma[/itex] be an infinite sigma algebra on a set [itex]X[/itex] and pick pairwise different [itex]A_n\in\Sigma, n\in\mathbb N[/itex]. Put [itex]S_n := \bigcup _{k=1}^nA_k\in\Sigma,n\in\mathbb N[/itex]. The sequence [itex]S_n,n\in\mathbb N[/itex] is increasing. Put
[tex]
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.
[/tex]
The [itex]T_n[/itex] are pairwise disjoint, thus [itex]\Sigma[/itex] contains at least [itex]\mathcal P(\mathbb N)[/itex] many elements, thus [itex]\Sigma[/itex] is uncountable. Indeed, the map
[tex]
\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}
[/tex]
is injective.
Mhm, good catch. Picking pairwise different [itex]\emptyset \subset A_n\subset X[/itex] gives injectivity.Math_QED said: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 said:Mhm, good catch. Picking pairwise different [itex]\emptyset \subset A_n\subset X[/itex] gives injectivity.
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 [itex]\mathcal P(\mathbb N)[/itex] (the [itex]T_n[/itex] correspond to the singletons)Math_QED said: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##.
https://www.physicsforums.com/help/latexhelp/ (short instruction help for PF)Pi-is-3 said: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.
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}}##.nuuskur said:If I get this right the situation is something like this
As time passes the triangle inside gets smaller and smaller until it vanishes. Call the side length [itex]L(t)[/itex]. After [itex]\delta t[/itex] seconds has passed, the new side length would be [itex]L(t+\delta t)[/itex]. Apply the cosine law to the triangle [itex]EAF[/itex] which yields
[tex]
|FE|^2 = |AF|^2 + |AE|^2 - 2\cos \frac{\pi}{3} |AF||AE| = |AF|^2 + |AE|^2 - |AF||AE|.
[/tex]
i.e
[tex]
L(t+\delta t)^2 = (v\delta t)^2 + (L(t) - v\delta t)^2 - (v\delta t)(L(t) - v\delta t).
[/tex]
Rearranging we get
[tex]
L(t+\delta t)^2 - L(t)^2 = (3v\delta t - 3L(t))v\delta t
[/tex]
Divide by [itex]\delta t[/itex] and take the limit as [itex]\delta t \to 0[/itex], thus
[tex]
\frac{d}{dt}L(t)^2 = -3vL(t).
[/tex]
Apply the chain rule and we get an initial value problem:
[tex]
L'(t) = -\frac{3}{2}v, \quad L(0) = L.
[/tex]
Hence [itex]L(t) = L-\frac{3}{2}vt[/itex] and at [itex]t=\frac{2L}{3v}[/itex] the triangle vanishes.
archaic said: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 think it's ok more or less. I don't have time now to check in detail either.nuuskur said:Oops, I forgot about the trajectory. Well, in short, it's ugly. Actually, this problem is much better in polar coordinates..
I think I made a calculation error somewhere, too tired to (double check) [itex]^{10}[/itex] right now.Suffices to find explicitly one trajectory. Denote position of top plane by [itex](r(t), \varphi (t))[/itex], where at [itex]t=0[/itex] we have [itex]r(0) = L[/itex] and [itex]\varphi (0) = 0[/itex] (apply translation or reverse traversion later to line up with some initial parameters if necessary).
The plane that is being followed has position [itex](r(t), \varphi (t) + \alpha)[/itex], where [itex]\alpha = \frac{2\pi}{3}[/itex]. So we get
[tex]
\frac{(r\sin \varphi)_t}{(r\cos \varphi )_t} = \frac{r(\sin (\varphi +\alpha) - \sin\varphi)}{r(\cos (\varphi +\alpha) - \cos\varphi)}
[/tex]
which yields (after a while)
[tex]
r'\sin\alpha + r\varphi '(\cos\alpha -1) = 0,
[/tex]
i.e
[tex]
r' - \sqrt{3}r\varphi ' =0\tag{1}
[/tex]
The flight speed is constant so we have
[tex]
v^2 = (r\sin\varphi)_t^2 + (r\cos\varphi)_t^2 = r'^2 + r^2\varphi '^2 \tag{2}
[/tex]
From (1) and (2) we get [itex]r' = \frac{\sqrt{3}}{2}v[/itex] 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
[tex]
\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
[/tex]
where [itex]C = \frac{\sqrt{3}}{3} \ln 2L[/itex]
nuuskur said:Making use of induction. Suppose [itex]a,b,c[/itex] is a nonzero solution. Due to [itex]x^2\equiv 0,1\pmod{3}[/itex] it must hold that [itex]a^2,b^2[/itex] are multiples of three. Putting [itex]d:= (a,b)[/itex], if [itex]d\mid c[/itex] holds, then [itex]\left (\frac{a}{d},\frac{b}{d},\frac{c}{d}\right )=:(a',b',c')[/itex] would be a solution with [itex](a',b')=1[/itex] which is impossible.
Could we show [itex]d\mid c[/itex]?
Pi-is-3 said: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.
... plus ##f'(x)\neq 0## for all ##x##.archaic said: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.
nuuskur said:Suppose [itex]g\mapsto g^{-1}[/itex] is an endomorphism and take [itex]g,h\in G[/itex]. Then [itex]gh \mapsto (gh)^{-1}[/itex]. Due to compatibility [itex]gh \mapsto g^{-1}h^{-1}[/itex], thus
[tex]
(gh)^{-1} = g^{-1}h^{-1} = (hg)^{-1} \iff gh = hg.
[/tex]
Suppose [itex]G[/itex] is abelian then for every [itex]g,h\in G[/itex]
[tex]
gh \mapsto (gh)^{-1} = h^{-1}g^{-1} = g^{-1} h^{-1}.
[/tex]The map [itex]g\mapsto g^{-1}\sigma (g)[/itex] on [itex]G[/itex] is injective, because if [itex]g^{-1}\sigma (g) = h^{-1}\sigma (h)[/itex], then [itex]hg^{-1} = \sigma (hg^{-1})[/itex]. The only fixpoint is [itex]e[/itex], thus [itex]h=g[/itex]. The map is also surjective, because [itex]G[/itex] is a finite set.
Now, pick [itex]g\in G[/itex] and write [itex]g= h^{-1}\sigma (h)[/itex], then
[tex]
\sigma (g) = \sigma (h^{-1}\sigma (h)) = (\sigma (h))^{-1} (h^{-1})^{-1} = (h^{-1}\sigma (h))^{-1} = g^{-1}.
[/tex]
By 7a [itex]G[/itex] is abelian.
nuuskur said:Making use of induction. Suppose [itex]a,b,c[/itex] is a nonzero solution. Due to [itex]x^2\equiv 0,1\pmod{3}[/itex] it must hold that [itex]a^2,b^2[/itex] are multiples of three. Putting [itex]d:= (a,b)[/itex], if [itex]d\mid c[/itex] holds, then [itex]\left (\frac{a}{d},\frac{b}{d},\frac{c}{d}\right )=:(a',b',c')[/itex] would be a solution with [itex](a',b')=1[/itex] which is impossible.
Could we show [itex]d\mid c[/itex]?
nuuskur said:Let [itex]\Sigma[/itex] be an infinite sigma algebra on a set [itex]X[/itex], choose [itex]\emptyset\subset A_1\subset X[/itex] and pick for every [itex]n\in\mathbb N[/itex]
[tex]
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 ).
[/tex]
[itex]\Sigma[/itex] is infinite, so we can do this. Put [itex]S_n := \bigcup _{k=1}^nA_k\in\Sigma,n\in\mathbb N[/itex]. Observe that all [itex]S_n\subset X[/itex].
The sequence [itex]S_n^c,n\in\mathbb N,[/itex] is strictly decreasing. Put
[tex]
T_n := S_n^c\setminus S_{n+1}^c\in \Sigma, \quad n\in\mathbb N.
[/tex]
The [itex]T_n[/itex] are pairwise disjoint. The map
[tex]
\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}
[/tex]
is injective. Note that if [itex]A\subset\mathbb N[/itex] then [itex]\bigcup _{x\in A}T_x \subseteq S_1^c \subset X[/itex].
This problem is of the devil mhh, the following is a cheapshotMath_QED said:What happens when ...
nuuskur said:This problem is of the devil mhh, the following is a cheapshot
Take a sequence of pairwise disjoint nonempty subsets [itex]S_n\in\Sigma, n\in\mathbb N[/itex]. Then
[tex]
\mathcal P(\mathbb N) \setminus \{\emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} S_x
[/tex]
is injective, because of disjointedness.
nuuskur said:This problem is of the devil mhh, the following is a cheapshot
Take a sequence of pairwise disjoint nonempty subsets [itex]S_n\in\Sigma, n\in\mathbb N[/itex]. Then
[tex]
\mathcal P(\mathbb N) \setminus \{\emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} S_x
[/tex]
is injective, because of disjointedness. Construction of sequence given in #33. We take pairwise distinct [itex]\emptyset \subset A_n \subset X\in\Sigma, n\in\mathbb N[/itex] and put [itex]S_n := \bigcup _{k=1}^n A_k\in\Sigma[/itex]. Then [itex]T_n := S_n^c \setminus S_{n+1}^c\in\Sigma, n\in\mathbb N,[/itex] is a sequence of pairwise disjoint non-empty subsets.
You're right, we need a change of gears ..Math_QED said:Again, it can happen that ##T_i=\emptyset## for some i. See my last counterexample.
nuuskur said:Suppose GG is of order 2m2m, where m>1m>1 is odd.
Consider the map
[tex]
\varphi : G \to \mbox{Sym}(G) : \varphi (g)(x) := gx, x\in G.
[/tex]
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), becausethe 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
[tex]
\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)
[/tex]
[tex]
\varphi (a)(g) = ga \Rightarrow \varphi (a)(ga) = gaa = ge =g.
[/tex]
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
[tex]
G / \mbox{Ker} (\varepsilon\varphi) \cong \{-1,1\}
[/tex]
which implies |Ker(εφ)|=m|Ker(εφ)|=m. Also Ker(εφ)◃GKer(εφ)◃G.
summer's almost over ..
fresh_42 said: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?
I sort of started at the back. My hopes and dreams were that I could compose the signum morphism with some [itex]\varphi : G\to\mbox{Sym}(G)[/itex] such that I would get [itex]\varepsilon \varphi : G\to \{-1,1\}[/itex] surjective and apply the first isomorphism theorem. Of course, getting a preimage for [itex]1[/itex] is trivial, but why woud there be an odd permutation? Then I thought maybe there is a permutation in [itex]\mbox{Sym}(G)[/itex] which contains an odd number of disjoint cycles of length two making it an odd permutation, this is where the order [itex]2m[/itex] gave a clue. A way to get such cycles is to put [itex](g,ga)[/itex] where [itex]a[/itex] is of order two and at that point the definition of [itex]\varphi[/itex] became self-evident.Math_QED said:Correct! Well done! Could you maybe give some insight in why you thought about that action?
nuuskur said:I think P6 generalises via induction to the case [itex]2^rm[/itex].
I sort of started at the back. My hopes and dreams were that I could compose the signum morphism with some [itex]\varphi : G\to\mbox{Sym}(G)[/itex] such that I would get [itex]\varepsilon \varphi : G\to \{-1,1\}[/itex] surjective and apply the first isomorphism theorem. Of course, getting a preimage for [itex]1[/itex] is trivial, but why woud there be an odd permutation? Then I thought maybe there is a permutation in [itex]\mbox{Sym}(G)[/itex] which contains an odd number of disjoint cycles of length two making it an odd permutation, this is where the order [itex]2m[/itex] gave a clue. A way to get such cycles is to put [itex](g,ga)[/itex] where [itex]a[/itex] is of order two and at that point the definition of [itex]\varphi[/itex] became self-evident.
This seems to be correct and is similar to what @nuuskur didNot anonymous said: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.
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
$$
-L / (- \frac {3} {2} v) = \frac {2L} {3v}
$$