# F(f(x)) = exp(x)

1. Aug 31, 2007

### Blouge

If f(f(x)) = exp(x), then what is f(x)?

I don't think that f(x) can be an entire function. It would be nice to have f(x) be monotonically increasing for positive x. I tried to attack this problem by using a method in the book "Bypasses", and it works, but it involves numerical calculation of a power series and reversion of the series, and applying the two power series together, and seems very inelegant and I'm not even sure if it will converge.

2. Aug 31, 2007

### uiulic

if f(x) has inverse, then you can get it in high school, I think.

3. Aug 31, 2007

### Hurkyl

Staff Emeritus
Isn't it enough to simply pick some a < 0, choose f to be a bijection from [-inf, a) to [a, 0), and then extend f by the functional equation?

Last edited: Aug 31, 2007
4. Aug 31, 2007

### Blouge

>if f(x) has inverse, then you can get it in high school, I think.

I'm trying to find out a more direct way of expressing the unknown function f(x) -- not focusing on a particular value of x or known function f.

>choose f to be a bijection

Yes, the whole problem is how to "choose" f that satisfies the functional equation.

Here is one idea on how to progress,

f(f(a)) = exp(a) = a, at certain constants a; one such number being ln ln ln ln ... ln -1.

Therefore

(f^2n)(a) = a for n = 0, 1, 2...

and it's reasonable to insist that

(f^z)(a) = a for any complex number z

Differentiating the function equation gives

f(f(x)) = exp(x) = d/dx exp(x) = f'(x) f'(f(x))
f(f(a)) = a f'(a) f'(f(a))

So that f(a) = a, f'(a) = a^0.5

I can see developing this into a power series expanded about a. Not sure what the radius of convergence is, or if it gives "nice" positive numbers for positive arguments.

Last edited: Aug 31, 2007
5. Aug 31, 2007

### Hurkyl

Staff Emeritus
I claim it's not a problem: any choice works. Even better, if you choose the bijection of [-inf, a) with [a, 0) to be monotone, then the resulting f is monotone.

6. Aug 31, 2007

### Blouge

>I claim it's not a problem: any choice works.

Um, f(x) = x + 3 is not a good choice: x + 3 + 3 = exp(x)?

I'm trying to find a choice that satisfies the equation.

7. Aug 31, 2007

### Hurkyl

Staff Emeritus
What a did you choose? And f(x) = x+3 is not a bijection from [-inf, a) to [a, 0) for any choice of a.

8. Aug 31, 2007

### Blouge

>And f(x) = x+3 is not a bijection from [-inf, a) to [a, 0) for any choice of a.

How about a = -1 and f(x) = -1 - 1/x? I don't see where that gets you.

Going back to the power series idea (where the constant a i used is ln ln ln ... -1 )

d^2/(dx)^2 f(f(x)) = f''(x) f'(f(x)) + f'(x)^2 f''(f(x))
a = f''(a) f'(a) + f'(a)^2 f''(a) = f''(a)[ f'(a) + f'(a)^2] = f''(a)[ a^0.5 + a]
f''(a) = 1/(1+a^-0.5)

So

f(x) = a + (x - a) a^-0.5 + (x - a)^2 / [(1 + a^-0.5) * 2!] + ...

Last edited: Aug 31, 2007
9. Aug 31, 2007

### Hurkyl

Staff Emeritus
Okay, so that's how f is defined on [-inf, -1). Now use the functional equation to figure out what f should be on [-1, 0)...

10. Aug 31, 2007

### Blouge

>Okay, so that's the piece of f defined on [-inf, -1). Now use the functional equation to figure out what f should be on [-1, 0)...

I guess, for y on [-1,0) you can say that f(y) = exp(-1/(1+y)):

f(f(x)) = exp(x)
f(y) = exp(f^-1(y))
f(y) = exp(-1/(1+y))

11. Aug 31, 2007

### Hurkyl

Staff Emeritus
I think that's right. And then you can figure out what the value of f should be on [0, e^a)...

Basically, I'm trying to define f(x) piecewise. I know that f(f(-inf)) = 0, so I choose some value
f(-inf) = a
and then I immediately know that i have to choose
f(a) = 0, f(0) = e^a, f(e^a) = 1, et cetera. Eyeballing this sequence, it appears plausible I can repeat this procedure if I start with any arbitrary increasing bijection [-inf, a) --> [a, 0), and maybe even with any bijection.

12. Aug 31, 2007

### Blouge

I see now. Thanks Hurkyl!

For [-inf,a), f(x) = f0(x) = a - aa/x
For [a,0), f(x) = f1(x) = exp(f0^-1(x)) = exp(aa / (a - x))
For [0,exp(a)), f(x) = f2(x) = exp(f1^-1(x)) = exp( a - aa / ln x )
For [exp(a), exp(exp(a))), f(x) = f3(x) = exp(f2^-1(x)) = ...
etc...

The only problem is that there's a lot of arbitrariness and its piecewise, and I wonder if it can be a "nice" (differentiable) function. And also satisfy f(c) = c for constants c where exp(c) = c.

On the other hand, my power series probably doesn't converge very well, and there are many constants c to choose to expand about. I wonder if it could help suggest a nice bijection.

It seems like a rather unnatural problem overall.

Last edited: Aug 31, 2007
13. Aug 31, 2007

### Hurkyl

Staff Emeritus
Well, if you have additional constraints on f, you can attempt to translate those into conditions on your original choices. You could easily get f to be infintiely differentiable (even analytic!) on the interior of each interval, and it's easy enough to make it infintiely differentiable at a. Does that imply it's everywhere differentiable?

exp doesn't have a fixed point, but if it did, I think the method I described would never reach c: it would define a function [-inf,c) to [-inf,c). You would have to make another choice to get an entire function.

14. Sep 4, 2007

### Blouge

>exp doesn't have a fixed point

There are complex, not real, fixed points. If you take -1 (or some other starting point) and apply the ln function repeatedly, it converges to some complex number c for which exp(c) = c. You can find infinitely many other fixed points by applying (2Npi + ln) instead of ln.