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

  1. Jul 22, 2009 #1
    How would one solve the equation f '(-2x) + g '(4x) = - x?
     
  2. jcsd
  3. Jul 22, 2009 #2

    arildno

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    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:
    [tex]f(x)=\sum_{n=0}^{\infty}a_{n}x^{n}, g(x)=\sum_{n=0}^{\infty}b_{n}x^{n}[/tex],
    we get:
    [tex]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}[/tex]
    whereby we have:
    [tex]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}[/tex]

    Inserting this into our equation, and rearranging, we get:
    [tex](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[/tex]

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

    Fixing, say, the power series for f, yields then an explicit scheme for g.
     
  4. Jul 22, 2009 #3
    You're allowed to use the same index n for both?

    oh, I supose for any an, there's only one unique bn.
     
  5. Jul 22, 2009 #4

    Hurkyl

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