Challenge Math Challenge - August 2019

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I have not solved question 12, there has been a mistake.
 

fresh_42

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I have not solved question 12, there has been a mistake.
Why? We have continuity, ##f(0)=f(1)##, no differentiability at ##x=\frac{1}{2}## and where it's differentiable, it has ##f'(x_0)=\pm 1 \neq 0##.
 
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Why? We have continuity, ##f(0)=f(1)##, no differentiability at ##x=\frac{1}{2}## and where it's differentiable, it has ##f'(x_0)=\pm 1 \neq 0##.
But shouldn't it be differentiable on ##(0,1)##?
 

fresh_42

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But shouldn't it be differentiable on ##(0,1)##?
If it is differentiable on ##(0,1)##, continuous on ##[1,0]## and ##f(0)=f(1)## then Rolle's theorem guarantees the existence of a point ##x_0\in (0,1)## such that ##f'(x_0)=0##. The theorem is important, since it is used to prove the mean value theorem. (Basically the same with an inclined statement, not only horizontal: ##f'(x_0)=\dfrac{f(a)-f(b)}{a-b}##.)
 
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If it is differentiable on ##(0,1)##, continuous on ##[1,0]## and ##f(0)=f(1)## then Rolle's theorem guarantees the existence of a point ##x_0\in (0,1)## such that ##f'(x_0)=0##. The theorem is important, since it is used to prove the mean value theorem. (Basically the same with an inclined statement, not only horizontal: ##f'(x_0)=\dfrac{f(a)-f(b)}{a-b}##.)
I do know that, but isn't the question about finding a function that satisfies the assumptions but does not comply to the conclusion? The function I gave fails to meet the differentiability assumption.
 

fresh_42

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I do know that, but isn't the question about finding a function that satisfies the assumptions but does not comply to the conclusion? The function I gave fails to meet the differentiability assumption.
If a function satisfies all assumptions, then the theorem holds because it is proven. So the conclusion can only fail if an assumption (or all) doesn't hold.

The logic is ##A \wedge B \wedge C \Longrightarrow D## where
A - f is continuous on a closed interval
B - f is differentiable on the interior
C - f has the same value at the starting and the end point
D - there is a point in between with horizontal tangent
 
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If a function satisfies all assumptions, then the theorem holds because it is proven. So the conclusion can only fail if an assumption (or all) doesn't hold.

The logic is ##A \wedge B \wedge C \Longrightarrow D## where
A - f is continuous on a closed interval
B - f is differentiable on the interior
C - f has the same value at the starting and the end point
D - there is a point in between with horizontal tangent
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.
 
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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].
 
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Math_QED

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Let [itex]\Sigma[/itex] be an infinite sigma algebra and pick freely [itex]A_n\in\Sigma, n\in\mathbb N[/itex]. Put [itex]S_n := \bigcup _{k=1}^n A_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.
How do you deduce that you have ##\mathcal{P}(\mathbb{N})## elements?
 
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Here you should specify that ##l,m,n## are non-zero positive integers. This seems important for what follows.

In what follows, you distinguish between two cases. Can you explain why it is enough to consider WLOG only these two cases? (I see why it is sufficient, but other readers may wonder why this is enough).

Also explain how it follows that for example ##x## is even.
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.
 
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Why is my formatting disappearing all the time? This is the second time this happened to #33 :mad:
 
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Why is my formatting disappearing all the time? This is the second time this happened to #33 :mad:
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.
 
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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 :oops:
 
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Edited #33 with explanation.

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

Math_QED

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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.
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.
 
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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 [itex]\emptyset \subset A_n\subset X[/itex] gives injectivity.
 

Math_QED

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Mhm, good catch. Picking pairwise different [itex]\emptyset \subset A_n\subset X[/itex] 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##.
 
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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 [itex]\mathcal P(\mathbb N)[/itex] (the [itex]T_n[/itex] correspond to the singletons)

I Changed the picking scheme in #33. You can't pick complements now to make some pathological cases.
 
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fresh_42

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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{}{})
 
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If I get this right the situation is something like this
triangle.png
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.
 

fresh_42

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Yes, and the flight path (parametrized by time)?
 
129
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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.
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)##
 
Last edited:
40
10
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.
 
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Oops, I forgot about the trajectory. Well, in short, it's ugly. Actually, this problem is much better in polar coordinates..
Suffices to find explicitly one trajectory. Denote position of 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]
I think I made a calculation error somewhere, too tired to (double check) [itex]^{10}[/itex] right now.
 
Last edited:

fresh_42

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Oops, I forgot about the trajectory. Well, in short, it's ugly. Actually, this problem is much better in polar coordinates..
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]
I think I made a calculation error somewhere, too tired to (double check) [itex]^{10}[/itex] 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.
 

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