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Full Version: f ' (x) = f(exp(x)) ?
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I was thinking about f ' (x) = f(exp(x))

It reminds me of James recent paper and the Julia equation.
Continuum sums may also be involved and infinite matrices ( carleman equations ) as well.

For instance D^2 f(x) = f '' (x) = f ' (exp(x)) exp(x)
= f(exp(exp(x))) exp(x).

D^3 f(x) = ( f(exp^[2](x)) + f(exp^[3](x)) ) exp(x).

Clearly for D^n f(0) we get a simple expression with using the pascal triangle / binomium identities.

Assuming f(+oo) = Constant We might use tricks such as James Nixon's construction.

Does the generalized binomium analogue hold for the fractional derivatives of f(0) ??
Or does it hold for one solution , assuming there was choice ?
Can we use Ramanujan's master theorem ?

Can we compute tetration from assuming this generalized binomium analogue for the fractional derivative ?

After some confusion and troubles , I also came to consider

f ' (x) = f(exp(x/2)).
And then consider the analogues from above.

( this to have convergeance problems solved )

Many more ideas come to me , but I will see what you guys think.

At least I believe :

D^t f(0) is of the form " Binomial type " * g(t) where g(t) is 1-periodic.

Probably some theorem from fractional calculus for products can solve this part.

As a sidenote I wonder what f ' (x) = f(exp(x)) says about integral f(x) ?

Also I want to note that f ' (x) = f(exp(x)) " probably" cannot hold everywhere for an entire f(x) because exp is chaotic ... probably ...

regards

tommy1729
I'm pretty sure that this function can't be analytic at the fix points of exp.
If it was, then if c is a fix point we could directly evaluate the derivatives of f at c, which are actually polynomials of c multiplied by f(c)
$f $$c$$ = \alpha$
$f' $$c$$ = f$$e^c$$ = f$$c$$ = \alpha$
$f'' $$c$$ = e^c f'$$e^c$$ = c \alpha$
$f^{[3]} $$c$$ = c^3 \alpha + c \alpha$
$f^{[4]} $$c$$ = c^6 \alpha + c^4 \alpha + 3*c^2 \alpha + c \alpha$
$f^{[5]} $$c$$ = c^{10} \alpha + c^8 \alpha + 9*c^6 \alpha + c^5 \alpha + 6c^4 \alpha + 7*c^3 \alpha + c \alpha$

and in general the degree of the polynomial appears to be $\frac{n(n+1)}{2}$ which obviously grows too fast to converge.
(04/17/2015, 02:15 AM)fivexthethird Wrote: [ -> ]I'm pretty sure that this function can't be analytic at the fix points of exp.
If it was, then if c is a fix point we could directly evaluate the derivatives of f at c, which are actually polynomials of c multiplied by f(c)
$f $$c$$ = \alpha$
$f' $$c$$ = f$$e^c$$ = f$$c$$ = \alpha$
$f'' $$c$$ = e^c f'$$e^c )$$ = c \alpha$
$f^{[3]} $$c$$ = c^3 \alpha + c \alpha$
$f^{[4]} $$c$$ = c^6 \alpha + c^4 \alpha + 3*c^2 \alpha + c \alpha$
$f^{[5]} $$c$$ = c^10 \alpha + c^8 \alpha + 9*c^6 \alpha + c^5 \alpha + + 6c^4 \alpha + 7*c^3 \alpha + c \alpha$

and in general the degree of the polynomial appears to be $\frac{n(n+1)}{2}$ which obviously grows too fast to converge.

Sorry Guys but my 3rd derivative in the op was wrong as clearly shown here.

I guess these are named polynomials , not ?

Regards

Tommy1729
I think this equation has a radius 0 everywhere ?

Regards

Tommy1729
Therefore Maybe f ' (exp(x)) = f(x) works better.

Regards

Tommy1729