Math Challenge - April 2020

I'm sorry but it's hard to read your solution. You need to use the formatting tools as @Not anonymous did. Or if you feel like it you can learn LaTeX.
  • #141
wrobel said:
It is clear there is an interval ##[a,b]\subseteq[0,1]## such that ##f(a)=f(b)=0,\quad |f|\mid_{(a,b)}>0##.
Consider a function ##F(x)=\ln|f(x)|-x##. We have ##F(x)\to-\infty## as ##x\to b-## or ##x\to a+##.
Thus for some ##\tilde x\in(a,b)## we have ##F'(\tilde x)=0##
This solves the problem

This is right, but I do think your first claim deserves a proof. If ##f## is the constant zero function, then the problem is solved. Otherwise, suppose ##f(p)\neq 0## for some ##p##. Then the set ##\{x\in [0,p]: f(x)=0\}## is nonempty, closed and so has a maximal element ##a## (that is not ##p##), and similarly, there is a minimal zero ##b## larger than ##p##. Then ##[a,b]## is the interval you are looking for.

Also, there still some outstanding problems in the March II thread (https://www.physicsforums.com/threads/math-challenge-march-2020-part-ii.985311/) that you might be interested in :)
Not anonymous said:
Is it valid to assume that for any differentiable function ##f:[0,1] \rightarrow \mathbb{R}## that is not the constant-value function but has infinite number of zeros, its zeros will form a countably infinite set?
This is not true, consider ##f(x)=\begin{cases}
0 & x\leq 1/2 \\
(x-1/2)^2 & x\geq 1/2\\
\end{cases}.##

Not anonymous said:
Or that for any such function, there will be some interval ##I ⊂[0,1]## containing 2 or more zeros of ##f## with all zeros falling in that interval forming a finite or a countably infinite set (i.e. set of zeros of ##f## contains a finite or countably infinite subset, with all remaining zeros, if any, lying outside the range defined by this finite or countably infinite subset)?

The issue isn't having countable/uncountably many zeros, but see my above response to @wrobel to see how to find an interval that you're looking for. You should be wary of taking "consecutive elements" of a countable set. For example, there are no consecutive elements in ##\mathbb{Q}##.

Also, this problem has a cute one-line solution: apply Rolle's theorem to the function ##g(x)=e^{-x}f(x)## on the interval ##[0,1].## This is just the exponential of @wrobel's function, and works for the same reason, but has the advantage of being defined everywhere, so there isn't any need to find a sub-interval to work with.
 
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  • #142
@Infrared I am trying to have you have a look at my last solution to question 9.
archaic said:
Saying ##f(f(f(x)))=x## is like saying ##f(u\in[0,1])=x##. As such, ##f:[0,1]\to[0,1]## is bijective.
Since ##f## is a continuous bijection on ##[0,1]##, then it is either strictly decreasing or strictly increasing, the same for its inverse (which has the same direction of variation as ##f##).

Proving ##\forall x\in [0,1],\,f(f(f(x)))=x\implies f(x)=x## is the same as proving ##f(x)\neq x\implies\exists x\in [0,1]\,|\,f(f(f(x)))\neq x##.

Suppose that ##f(x)\neq x##, i.e, there exists a real number in ##[0,1]## such that its image by ##f## is different than itself. In the following, by ##a## I mean this real.
All inequalities will be strict because ##f## is a continuous bijection.
Suppose that ##f(x)## is strictly increasing, and let there be ##a\in[0,1]## such that ##f(a)=b\neq a##, then:
If ##b<a##, then ##f(b)=c<f(a)=b##, and thus ##f(c)<f(a)=b##. It follows that ##f(f(f(a)))<b<a##.

If ##b>a##, then ##f(b)=c>f(a)=b##, and thus ##f(c)>f(a)=b##. It follows that ##f(f(f(a)))>b>a##.

I have proved that if ##f(x)## is strictly increasing, then ##\exists x\in [0,1]\,|\,f(f(f(x)))\neq x##.

Suppose that ##f(x)## is strictly decreasing, and let there be ##a\in[0,1]## such that ##f(a)=b\neq a##, then:
If ##b<a##, then ##f(b)=c>f(a)=b##.

If ##b<c<a##, then ##f(b)=c>f(c)>f(a)=b##. It follows that ##f(f(f(a)))>f(a)##

If ##a<c##, then ##f(a)=b>f(c)##. It follows that ##f(f(f(a)))<f(a)##

If ##b>a##, then ##f(b)=c<f(a)=b##.

If ##b>c>a##, then ##f(b)=c<f(c)<f(a)=b##. It follows that ##f(f(f(a)))<f(a)##

If ##a>c##, then ##f(a)=b<f(c)##. It follows that ##f(f(f(a)))>f(a)##

I have proved that if ##f(x)## is strictly decreasing, then ##\exists x\in [0,1]\,|\,f(f(f(x)))\neq x##.

I have proved that ##f(x)\neq x\implies\exists x\in [0,1]\,|\,f(f(f(x)))\neq x##!
 
  • #143
Adesh said:
What is the table in that figure? What does that cuboid represents in the figure?

@Adesh I think that after my explanations in post #130 things are sufficiently clear. What you see in Fig.1. is the surface of a table. Now, you see the two handles beneath the surface of the table. These are used to pull an internal movable surface - as I said in post #130 you can think of it as a drawer, and obviously, this "drawer" has an upper and a lower surface - it is a rectangular parallelepiped and so there is friction between its left and right sides (as we see it in Fig.1) and the internal part of the respective sides of the table. So, effectively, we have two parallelepipeds: the upper part of the table as we see it in Fig.1 i.e. without table's legs, is the external one and the movable surface is the internal one which is inside the external parallelepiped of the upper part of the table.
 
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  • #144
For #10, SPOILER:I had in mind something like this: as Infrared pointed out, indeed f-f' has the Darboux property (intermediate value property) because it is a derivative, (since f has an antiderivative by the fundamental theorem of calculus).

Thus either the problem is true, or else f-f' is everywhere positive or everywhere negative on (0,1). Suppose f-f' > 0 on (0,1), hence f >0 at every critical point in (0,1), so the global minimum is 0, and occurs only at the endpoints. So let a be a point in (0,1) where f has a global maximum on [0,1] and f(a) > 0. Then by the MVT on [0,a], there is a point b with 0<b<a, where f'(b) = {f(a)-f(0)}/a. Then we get the contradiction f(b) > f'(b) = f(a)/a > f(a).

If f' > f on (0,1), then f<0 at every critical point on (0,1), so choose a in (0,1) as a global minimum with f(a) < 0. We get a similar contradiction at a point b with 0<b<a, and f(b) < f'(b) = {f(a)-f(0)}/a < f(a).
 
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  • #145
Adesh said:
Question 12
Questions 11-15 are for high schoolers only!
 
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  • #146
archaic said:
Saying ##f(f(f(x)))=x## is like saying ##f(u\in[0,1])=x##
This is not properly written. I think what you're trying to say is that is that for any ##x\in [0,1]##, there exists ##u\in [0,1]## such that ##f(u)=x##. This only means that ##f## is surjective, but you need injectivity to conclude monotonicity. Anyway ##f## is a bijection because ##f\circ f## is a (both left and right) inverse function.

archaic said:
Since ##f## is a continuous bijection on ##[0,1]##, then it is either strictly decreasing or strictly increasing
This is true, but I think it's worthwhile to write out a proof. See my post 133 for two arguments, but maybe you want to think about it yourself too.

The casework that follows looks okay, but it could be cleaned up a little bit. For example, the condition ##f\circ f\circ f## is automatically inconsistent with ##f## being decreasing, so there is no need to check all of those cases (and even in this work, I think the sub-cases relating to ##c## are unnecessary). Also, introducing the new variable ##c=f(b)## makes things harder to read, at least for me.
 
  • #147
Infrared said:
This is not properly written. I think what you're trying to say is that is that for any ##x\in [0,1]##, there exists ##u\in [0,1]## such that ##f(u)=x##. This only means that ##f## is surjective, but you need injectivity to conclude monotonicity. Anyway ##f## is a bijection because ##f\circ f## is a (both left and right) inverse function.
I meant that since ##f## is taking ##[0,1]## as an entry point and giving back an ordered ##[0,1]##, or ##x##, then it's bijective as the RHS also is. There can't be a point that has two images because it is a function, and it has pin point the same ##[0,1]## "elements".
 
  • #148
archaic said:
I meant that since ##f## is taking ##[0,1]## as an entry point and giving back an ordered ##[0,1]##, or ##x##
I don't know what this means.

But think about the function ##f:[0,2]\to [0,2]## (re-scaling just to make the numbers easier) given by ##f(x)=2(x-1)^2.## This map is not a bijection because it is not injective, but if we set ##g(x)=\sqrt{\frac{x}{2}}+1##, then we still have ##f(g(x))=x##.
 
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  • #149
Let [itex]U\subseteq X[/itex] be dense and [itex]A:U\to Y[/itex] bounded linear, where [itex]Y[/itex] is a Banach space. Fix [itex]x\in X[/itex]. Due to density we have [itex]u^n\in U, n\in\mathbb N,[/itex] such that [itex]u^n\xrightarrow [n\to\infty]{} x[/itex] .

The sequence [itex]u^n, n\in\mathbb N,[/itex] is Cauchy. We show [itex]A(u^n), n\in\mathbb N,[/itex] is Cauchy in [itex]Y[/itex], hence due to completeness there exists the limit [itex]A(u^n) \xrightarrow [n\to\infty]{} y_x \in Y[/itex].

Let [itex]\varepsilon > 0[/itex]. By boundedness of [itex]A[/itex] we have [itex]M>0[/itex] such that [itex]\|A(u)\|_Y \leq M\|u\|_X[/itex] for every [itex]u\in U[/itex]. Since [itex]u^n,n\in\mathbb N,[/itex] is also Cauchy, pick a suitable [itex]K(\frac{\varepsilon}{M})\in\mathbb N[/itex] by definition. Suppose [itex]m,n>K[/itex], then
[tex]
\|A(u^m) - A(u^n)\| = \|A(u^m-u^n)\| \leq M\|u^m-u^n\| < \varepsilon .
[/tex]
Therefore [itex]A(u^n),n\in\mathbb N,[/itex] is Cauchy.

Define [itex]\tilde{A} : X\to Y[/itex] such that [itex]\tilde{A}(x) = y_x[/itex] (if [itex]x\in U[/itex], then the sequence can be assumed to be constant, hence [itex]\tilde{A}(x) = A(x)[/itex]). We have [itex]\tilde{A}|_U = A[/itex]. Linearity follows from properties of the limit for continuous maps and linearity of [itex]A[/itex]. Additionally, we have by continuity of the norm
[tex]
\|\tilde{A}(x)\| = \left \|\lim _{n\to\infty} A(u^n)\right \| = \lim _{n\to\infty} \|A(u^n)\| \leq \|A\|\lim _{n\to\infty} \|u^n\| = \|A\|\|x\|. \tag{1}
[/tex]
Thus [itex]\tilde{A}[/itex] is bounded (i.e continuous).

Next, we show equality of the operator norms. On the one hand [itex]\|A\| \leq \|\tilde{A}\| [/itex] holds trivially, because by extending, the norm can't become smaller. By definition, the operator norm is given as
[tex]
T:X\to Y\qquad \|T\| := \sup \left \{ \frac{\|T(x)\|}{\|x\|} \colon x \neq 0\right \}
[/tex]
By (1) we have [itex]\|\tilde{A}\| \leq \|A\|[/itex]. Therefore the operator norms are equal.

Note also that if [itex]B:X\to Y[/itex] was another such extension, then
[tex]
\|B(x) - \tilde{A}(x)\| = \lim _{n\to\infty} \|A(u^n) - A(v^n)\| = \lim _{n\to\infty} \|A(u^n-v^n)\| = 0,
[/tex]
because restrictions to [itex]U[/itex] would be equal and [itex]u^n,v^n[/itex] would both converge to [itex]x[/itex]. Thus the extension is unique.
 
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  • #150
Infrared said:
You should be wary of taking "consecutive elements" of a countable set. For example, there are no consecutive elements in ##\mathbb{Q}##.

I agree. Dealing with infinity or infinite sized sets is very tricky, more so for novices. Like everyone else, I have an infinite amount of things to learn, but the infinity I am yet to learn is larger than those of the experts in this forum :smile:

I have a doubt, since you mentioned ##\mathbb{Q}##. Is there any function that can be expressed using only common standard functions and does not need to be written as a sum of infinite series that is continuous and differentiable in ##(0,1)## which has all rational numbers in its range and no other value as its zeros?

Infrared said:
Also, this problem has a cute one-line solution: apply Rolle's theorem to the function ##g(x)=e^{-x}f(x)## on the interval ##[0,1]##. This is just the exponential of @wrobel's function, and works for the same reason, but has the advantage of being defined everywhere, so there isn't any need to find a sub-interval to work with.

Thanks for that example! It is nice to be able to prove something using just one simple theorem. But I am a bit confused about this function's relation to @wrobel's. @wrobel's function uses absolute value of ##f(x)## whereas yours does not, so your example appears to be of similar form but not exactly the same as the exponential of @wrobel's function. Am I missing something?
 
  • #151
Not anonymous said:
Is there any function that can be expressed using only common standard functions and does not need to be written as a sum of infinite series that is continuous and differentiable in ##(0,1)## which has all rational numbers in its range and no other value as its zeros?
The range of any continuous map ##f:(0,1)\to\mathbb{R}## is connected, so if it contains ##\mathbb{Q}##, then it must be all of ##\mathbb{R}.## There are continuous surjections like this, e.g. ##f(x)=\tan(\pi(x-1/2)).## What do you mean by "no other values as its zeros"?

Not anonymous said:
Thanks for that example! It is nice to be able to prove something using just one simple theorem. But I am a bit confused about this function's relation to @wrobel's. @wrobel's function uses absolute value of ##f(x)## whereas yours does not, so your example appears to be of similar form but not exactly the same as the exponential of @wrobel's function. Am I missing something?
Okay, you're right, the exponential of wrobel's function is ##e^{-x}|f(x)|.## But since he is working on an interval where ##f## has no zeros, this is the same up to a possible minus sign. The point is that ##f## and ##e^f## have the same critical points, and negating doesn't change this.
 

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