Composes with itself to be its antiderivative

[dethwing]

This was UNL's "Problem of the Fortnight" recently, and thought it was kind of cute. Find a function f: R --> R such that :

(f o f)' = f

Or if you prefer this notation:

(d/dx) f(f(x)) = f(x)

Secondly, I know how to find such a function, but I don't know if its unique with this property. Any ideas?

[lostdummy]

well, one function that should satisfy is: f(x)=0

probably there is another solution, since I doubt above solution would be interesting enough to create problem, but that indicate that such "another" solution at least is not unique ;p

[dethwing]

Fine fine, a non-zero function.

[cocoloco]

Suppose f is of the form f(x) = a * x^p. Then f(f(x)) = a*(a*x^p)^p = a^(p+1)*x^p² and its derivative is a^(p+1)*(p+1)*x^(p²-1).

Since this polynomial** must have the same degree as f, we get p²-1 = p which resolves to the well-known p = (1+sqrt(5))/2.

And to have the same linear factor before x^p we need to solve a = a^(p+1)*(p+1), so a is the p-th root of 1/(p+1).

**Not sure if 'polynomial' is the correct term since it has irrational exponents, but it should work.

[dethwing]

Nice, that's the solution I got as well. No idea if any other function would do the job though

[Zag]

[Zag]

Oh, I get it. You had plugged into it the result from your next step. Cheater!!

[cocoloco]

Yeah, looks like I copied the a from the wrong line there. It still works though.

[Zag]

How about if it has this form:

f(x) = ab^x
f(f(x)) = ab^(ab^x)
f(f(x))' = ab^(ab^x) 2 ln(ab) (ab^x)

So to make f(f(x))' equal to f(x)

ab^(ab^x) = 1/[2 ln(ab)]

So I guess we could solve for x and find the point at which they cross, but they are never going to be equivalent.