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Composition of functions

  1. Feb 27, 2009 #1
    This might be a silly question:

    given a function [tex]g[/tex] is it possible to find a function [tex]f[/tex] such that [tex]f = f \circ g[/tex]?
  2. jcsd
  3. Feb 27, 2009 #2
  4. Feb 27, 2009 #3
    That is the trivial solution. Perhaps there are others?

    f(x) = f(g(x)) if...

    f(x) = c, c constant.
    g(x) = x, f any function.

    If these two conditions don't hold, though...

    Assume f has an inverse function. For instance, if f(x) = 2x, then (inv f)(x) = x/2. Then

    f = f o g
    f(x) = f(g(x))
    x = g(x).

    So if f has an inverse, g must equal x if f = f o g.

    So you'd only be looking for functions which don't have an inverse.

    But what we have is a little stronger than that, no? Since my argument makes no reference to intervals, it must be true on any interval. So g = x if you want a function f which is invertible over any interval.

    The only function which is not invertible over any interval is - you guessed it - constant functions.

    So, in summary:

    if g(x) = x, then any function f(x) will do.

    Otherwise, f(x) = c , c constant, is the only solution.

    What about functions of more variables? Or generalized operations like differentiation? No idea.
  5. Feb 27, 2009 #4
    If g(x)=x Then f(x) can be anything.


    Then f(n) can map even numbers to one constant and odd numbers to a different constant. (At least it works if n is discrete). Not sure if it works in the continuous case but I think it might.
    Last edited: Feb 27, 2009
  6. Feb 27, 2009 #5
    What about the following examples?
    a) f(x)=abs(x), g(x)=-x
    b) f(x)=cos(x), g(x)=x+2pi
  7. Feb 27, 2009 #6
    Good point.

    I guess then that my argument only works for functions which don't have an inverse where Domain(inv f) = Range(f). This, naturally, precludes functions such as abs(x) and cos(x)...

    So I guess more though will have to be put into functions which are not bijections.
  8. Feb 28, 2009 #7
    Thanks a lot. You all made very good observations that helped me a lot.
    BTW, it seems that if the function f admits an inverse there are not many choices, while if the function f is not invertible, many solutions exist but the problem is non-trivial, and it is difficult to say what kind of functions f and g would have to be, in order to satisfy f(x)=f(g(x)).

    At the moment I am trying to solve the following (similar) problem:

    [tex]f = (f \circ g) g'[/tex]
    where g is invertible, and g' denotes its derivative

    If you find it interesting, suggestions are always welcome.
  9. Feb 28, 2009 #8
    Well, similar suggestions - cases - are possible.

    Assume f(x) = c, c constant. Then c = cg', g' = 1, and g = x + k, k constant.

    Assume f(x) = x. Then x = gg', gdg = xdx, and (g^2)/2 = (x^2)/2 + k, k constant.

    Assume f(x) = x^n. Then x^n = g^n g', g^n dg = x^n dx, [1/(n+1)]g^(n+1) = [1/(n+1)]x^(n+1) + k, k constant

    Wow, that's an alright result. So for x to any power at all, it's possible. Is is true for any polynomial? Yes, it seems like it should be. So... for any polynomial, I believe you can use the above formula to reduce it.

    f(x) = cos x, cos x = (cos g)g', (cos x)dx = (cos g)dg, sin(x) = sin(g) + k.

    It seems like, unless I'm mistaken, this is the same thing every time:

    f(x) = f(g(x))g'(x) is the same as solving the differential equation f(x)dx = f(g(x))dg.

    So, as long as f is integrable, the problem is actually quite easy. Maybe I'm wrong. Thoughts?
  10. Mar 1, 2009 #9
    csproof2000: I think you just made it!
    You gave the solution to the problem, and for some reason I didn't immediately spot that what I was trying to do is actually solving a differential equation.
    Thanks a lot you all...you made very helpful observations!