Can a Function be Composed with Itself to Solve a Differential Equation?

[mnb96]

This might be a silly question:

given a function $$g$$ is it possible to find a function $$f$$ such that $$f = f \circ g$$?

 

[yyat]

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.

 

[csprof2000]

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.

 

[John Creighto]

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.

 

[yyat]

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

 

[csprof2000]

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.

 

[mnb96]

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!

 

[csprof2000]

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?

 

[mnb96]

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!