I was looking at efficiency of various algorithms involving number theory and I kept on getting variations of one particular functional equation: $f(x) = \sqrt{x}c + 2f(\sqrt{x})$. Clearly, $c$ is $-f(1)$ but that's really all I've got so far. How can you solve this?
Solution to functional equation giving $f(x)$ in terms of $f(x^{1/2})$
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functional-equations
radicals
recursion
1 Answers
0
Don't know if this is of use to you but you can write it as a power series like this :
$f(x)=-\frac{c}{2}x -\frac{c}{4}x^2-\frac{c}{8}x^4-\frac{c}{16}x^8...$
or written as a sum:
$f(x)=-c\sum_{i=0}^\infty {\frac{x^{2^{i}}}{2^{i+1}}}$
$\implies -2f(\sqrt{x})=c\sum_{i=0}^\infty {\frac{x^{2^{i-1}}}{2^{i}}} \implies $
$f(x)-2f(\sqrt{x})=-c\sum_{i=0}^\infty {\frac{x^{2^{i}}}{2^{i+1}}}+c\sum_{i=0}^\infty {\frac{x^{2^{i-1}}}{2^{i}}}= c\sum_{i=0}^\infty {\frac{x^{2^{i-1}}}{2^{i}}} -c\sum_{i=0}^\infty {\frac{x^{2^{i}}}{2^{i+1}}}=c\sqrt{x}+ c\sum_{i=1}^\infty {\frac{x^{2^{i-1}}}{2^{i}}} -c\sum_{i=0}^\infty {\frac{x^{2^{i}}}{2^{i+1}}}
=c\sqrt{x}+ c\sum_{i=0}^\infty {\frac{x^{2^{i}}}{2^{i+1}}} -c\sum_{i=0}^\infty {\frac{x^{2^{i}}}{2^{i+1}}} = c\sqrt{x} \enspace \square $
The sum converges for $\mid x \mid \le 1 $ . Maybe there are other solutions, also I don't know if a closed form expression for this exists. Depending on what you want to do it might give some extra insight.
PS: Found similar (but not exactly the same ) question here : What is known about doubly exponential series?
This might give you some idea of what a closed form solution might look like.