# Thread: Extracurricular Elementary Mathematics Marathon

1. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by seanieg89
I do not believe your first solution is correct, neither is the statement that if f(x) is a solution then f(2x/(3x-2)) is one too.

This would be the case if the functional equation was symmetric in these two unknowns, but the factor of 1/2 stops that from happening.

A quick way to verify that this solution doesn't work without plugging into the functional equation directly is by considering the limit L of f(x)/x as x->0. The functional equation implies that L must be 1008 if it exists, whereas the limit in your first proposed solution is -1008.

The second solution is the unique solution and you are done after solving the simultaneous equations. (Because all of the previous lines are quantified over all x, you can actually be sure that this is the unique solution after reaching the line f(x)=blah).

(I obviously cooked up the numbers so that 2016 featured in the unique solution.)
Ah. Knew I should have just done a simple substitution instead of the involution property.

Was that intentional, by the way?

2. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by Paradoxica
Ah. Knew I should have just done a simple substitution instead of the involution property.

Was that intentional, by the way?
Yep, you got the key idea, just a silly error.

Was what intentional?

3. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by seanieg89
Yep, you got the key idea, just a silly error.

Was what intentional?
Involution

4. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by Paradoxica
Involution
Oh yes, entirely. That was why I considered it cute, I quite like simple ideas exploiting symmetries like that.

5. ## Re: Extracurricular Elementary Mathematics Marathon

Find with proof, all continuous functions f:R->R such that:

f(f(f(x)))=x for all real x.

6. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by seanieg89
Find with proof, all continuous functions f:R->R such that:

f(f(f(x)))=x for all real x.
I'm doubt the validity of my solution, and I'm certain there are massive logical gaps, but here's what I have:

$\noindent Inverting the three-fold composition gives us f^{-1}(f^{-1}(f^{-1}(x)))=x \\ Thus, f(x) = f^{-1}(x), since f is arbitrary, and hence f(f(x)) = x. Substituting this back into the original functional equation gives f(x)=x, which obviously satisfies the original functional equation.$

7. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by Paradoxica
I'm doubt the validity of my solution, and I'm certain there are massive logical gaps, but here's what I have:

$\noindent Inverting the three-fold composition gives us f^{-1}(f^{-1}(f^{-1}(x)))=x \\ Thus, f(x) = f^{-1}(x), since f is arbitrary, and hence f(f(x)) = x. Substituting this back into the original functional equation gives f(x)=x, which obviously satisfies the original functional equation.$
Yeah, you have done something fishy in deducing that f=f^{-1}, care to explain your reasoning if you still believe this fact after thinking more?

Note also that you have used continuity nowhere. This is essential, as we have a vast array of solutions to the functional equation if continuity is not required:

Partition the reals into an uncountable union of sets, each with either 1 or 3 elements.
Define f to map elements of singleton sets to themselves and to cycle the three elements in each of the other sets.

Any such function f will satisfy the functional equation, but almost all of them will be highly discontinuous.

8. ## Re: Extracurricular Elementary Mathematics Marathon

$\noindent f^3(x) = x \\ f^3(-x)=-x \\ Since composition preserves any existing parity (favouring even over odd) and cannot create parity, f(x) is odd. If f is a polynomial with degree m, then we have the following fact: O(f^3(x)) = O(x^{m^3}) = O(x) \Leftrightarrow m^3 = 1 \Leftrightarrow m=1 \\ Since f is odd, the polynomial f is f(x) = kx \\ Systematically bashing out the linear function yields f(x)=x \\ Not sure where to go from here...$

9. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by Paradoxica
$\noindent f^3(x) = x \\ f^3(-x)=-x \\ Since composition preserves any existing parity (favouring even over odd) and cannot create parity, f(x) is odd. If f is a polynomial with degree m, then we have the following fact: O(f^3(x)) = O(x^{m^3}) = O(x) \Leftrightarrow m^3 = 1 \Leftrightarrow m=1 \\ Since f is odd, the polynomial f is f(x) = kx \\ Systematically bashing out the linear function yields f(x)=x \\ Not sure where to go from here...$
Composition cannot create parity? Why not? Eg f(x)=1-x is not odd or even, but composed with itself is the identity which is odd.

It is true that the only polynomial function that works is the identity and yes degree considerations give you this quickly as you note.

This isn't too fruitful a way of thinking about what general continuous functions can solve the equation though.

10. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by seanieg89
Composition cannot create parity? Why not? Eg f(x)=1-x is not odd or even, but composed with itself is the identity which is odd.

It is true that the only polynomial function that works is the identity and yes degree considerations give you this quickly as you note.

This isn't too fruitful a way of thinking about what general continuous functions can solve the equation though.
No discontinuities allowed, this includes singularities of rational functions? for example, 1/(1-x) is clearly a solution to the functional equation, but has a simple pole at x=1

11. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by Paradoxica
No discontinuities allowed, this includes singularities of rational functions? for example, 1/(1-x) is clearly a solution to the functional equation, but has a simple pole at x=1
No, singularities are not allowed, because the function is continuous and has domain R. (We cannot have any domain "holes".)

12. ## Re: Extracurricular Elementary Mathematics Marathon

$\noindent Fact 1: f is injective. \\ Proof: Suppose f is not injective, with x and y being distinct reals that satisfy f(x) = f(y). Then f(f(x)) = f(f(y)) and f(f(f(x))) = f(f(f(y))). But from the functional equation, f(f(f(x))) = x and f(f(f(y))) = y. So x=y, which contradicts the non-injectivity of f. Thus, f is injective. \\Fact 2: f is monotone. \\This follows from the requirement that f is entirely continuous in conjunction with Fact 1.\\ Condition 1: f(x) \leq x \\Proof: Suppose for some x, f(x)>x. So f(f(x)) > f(x) > x and f(f(f(x))) > f(f(x)) > f(x) > x. Then the functional equation is no longer true. Thus, f(x) \leq x \\ Condition 2: f(x) \geq x \\Proof: Suppose for some x, f(x)

$\noindent Since both conditions must be simultaneously true, then the only such possibility is f(x) = x. Obviously, this satisfies the functional equation. \blacksquare\\ \\ Extension of this question:\\ Find all rational functions that satisfy the functional equation.$

13. ## Re: Extracurricular Elementary Mathematics Marathon

$\noindent Show there are no solutions to the diophantine equation 2^n + 3^n = k^3 in positive integers (n,k)$

14. ## Re: Extracurricular Elementary Mathematics Marathon

$Find all distinct integers n such that n+3^2 and n^2+3^3 are perfect cubes.$

15. ## Re: Extracurricular Elementary Mathematics Marathon

$\noindent Prove the following inequality: \\\sqrt{a^2 - \sqrt{3}ab+b^2}+\sqrt{b^2-bc+c^2} \geq \sqrt{c^2+a^2}$

16. ## Re: Extracurricular Elementary Mathematics Marathon

$\noindent Solve the following Diophantine Equation in \mathbb{Z} \\\\ a^3 + b^3 + c^3 = a^2 + b^2 + c^2$

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