# Determining if a differential equation has unique solution

Our teacher asked a seemingly simple question:

What is the value of $$f(2)~$$ if $$~f( f(x) ) = 16x-15~$$?

I found a solution assuming $$f(x)$$ in the form of $$ax+b$$, but how do I show that $$f(x)$$ must be in that form?

I thought of taking derivative of both sides and reached

$$f'(f(x)) = \frac{16}{ f'(x) }$$

How can I show that this function has a unique solution or it doesn't have a unique solution?

Apply $$f$$ from the inside and outside, set $$x=f(w)$$ then $$f(f(f(w)))=16f(w)-15$$ and also $$f(f(f(x)))=f(16x-15)$$ so that $$f(16x-15)=16f(x)-15$$, from which one concludes directly $$f(1)=1$$. Set $$g(x)=f(1+x)-1$$, then $$16g(x)=f(1+16x)-1=g(16x)$$, $$g(0)=0$$ or equivalently $$g(x)=16\,g\biggl(\frac{x}{16}\biggr)=16^n\,g\biggl(\frac{x}{16^n}\biggr)~~\forall n\in\Bbb N$$ Assuming differentiability of $$f$$ and thus of $$g$$ one concludes that $$g(x)=ax$$, $$f(x)=ax+1-a$$, validating your first guess as the only solution, $$f(f(x))=a(ax+1-a)+1-a=a^2x+1-a^2\implies a=\pm4, ~~f(2)=a+1\in\{-3,5\}.$$
• Sorry for bumping up, how you getting $g(x)=ax$? Commented Nov 22, 2018 at 18:43
• Differentiability in $x=0$ means $g(x)=ax+o(x)$ for some $a$ and $x\approx 0$, and by the recursion equation for any $x$ and $n$ large enough $$g(x)=16^n(a16^{-n}x+o(16^{-n}x))=ax+16^no(16^{-n}x),$$ where the last term converges to $0$ for $n\to\infty$. Commented Nov 22, 2018 at 20:58