How would one solve the equation f '(-2x) + g '(4x) = - x?

[coverband]

How would one solve the equation f '(-2x) + g '(4x) = - x?

 

[arildno]

Functional equations like these are generally VERY nasty.

If we are to restrict our problem to find those solutions f and g that can both be represented as power series, we might make some headway:
We set, as trial solutions:
f(x)=\sum_{n=0}^{\infty}a_{n}x^{n}, g(x)=\sum_{n=0}^{\infty}b_{n}x^{n},
we get:
f^{'}(x)=\sum_{n=0}^{\infty}(n+1)a_{n+1}x^{n}, g^{'}(x)=\sum_{n=0}^{\infty}(n+1)b_{n+1}x^{n}
whereby we have:
f^{'}(-2x)=\sum_{n=0}^{\infty}(n+1)a_{n+1}(-2)^{n}x^{n}, g^{'}(4x)=\sum_{n=0}^{\infty}(n+1)b_{n+1}4^{n}x^{n}

Inserting this into our equation, and rearranging, we get:
(a_{1}+b_{1})+(-2a_{2}+4b_{2}+1)x+\sum_{n=2}^{\infty}(n+1)((-2)^{n}a_{n}+4^{n}b_{n})x^{n}=0

This is to hold for ALL x, so therefore, the coefficients in front of each power in x need to vanish, so we get:
a_{1}+b_{1}=0
-2a_{2}+4b_{2}+1=0
(-2)^{n}a_{n}+4^{n}b_{n}=0\to{b}_{n}=(-2)^{-n}a_{n}, n\geq{2}

Fixing, say, the power series for f, yields then an explicit scheme for g.

 

[flatmaster]

You're allowed to use the same index n for both?

oh, I supose for any an, there's only one unique bn.

 

[Hurkyl]

This one isn't very hard, though: f is pretty much unconstrained, leaving you with an ordinary differential equation in g.