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dethwing
DeTheeThaw

 Posted: Wed Apr 15, 2009 12:28 am    Post subject: 1 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
Daedalian Member

 Posted: Wed Apr 15, 2009 7:54 am    Post subject: 2 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
DeTheeThaw

 Posted: Wed Apr 15, 2009 1:01 pm    Post subject: 3 Fine fine, a non-zero function.
cocoloco
Icarian Member

 Posted: Thu Apr 16, 2009 2:10 pm    Post subject: 4 Suppose f is of the form f(x) = a * x^p. Then f(f(x)) = a*(a*x^p)^p = a^(p+1)*x^pē and its derivative is a^(p+1)*(p+1)*x^(pē-1). Since this polynomial** must have the same degree as f, we get pē-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
DeTheeThaw

 Posted: Thu Apr 16, 2009 3:29 pm    Post subject: 5 Nice, that's the solution I got as well. No idea if any other function would do the job though
Zag
Tired of his old title

Posted: Thu Apr 16, 2009 6:32 pm    Post subject: 6

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

I'm confused why its derivative isn't a^(p+1) * (pē) * x^(pē-1)

a and p are just constants, right?
Zag
Tired of his old title

 Posted: Thu Apr 16, 2009 6:36 pm    Post subject: 7 Oh, I get it. You had plugged into it the result from your next step. Cheater!!
cocoloco
Icarian Member

 Posted: Fri Apr 17, 2009 6:49 am    Post subject: 8 Yeah, looks like I copied the a from the wrong line there. It still works though.
Zag
Tired of his old title

 Posted: Sat Apr 18, 2009 3:22 pm    Post subject: 9 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.
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