# composition of functions

by mnb96
Tags: composition, functions
 P: 603 This might be a silly question: given a function $$g$$ is it possible to find a function $$f$$ such that $$f = f \circ g$$?
P: 322
 Quote by mnb96 This might be a silly question: given a function $$g$$ is it possible to find a function $$f$$ such that $$f = f \circ g$$?
f=const.
 P: 290 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.
P: 813

## composition of functions

If g(x)=x Then f(x) can be anything.

Let
g(n)=2n

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.
P: 322
 Quote by csprof2000 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 the following examples?
a) f(x)=abs(x), g(x)=-x
b) f(x)=cos(x), g(x)=x+2pi
 P: 290 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.
 P: 603 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: $$f = (f \circ g) g'$$ where g is invertible, and g' denotes its derivative If you find it interesting, suggestions are always welcome. Thanks!
 P: 290 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?
 P: 603 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!

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