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f ' (x) = f(exp(x)) ?
#1
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
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Messages In This Thread
f ' (x) = f(exp(x)) ? - by tommy1729 - 04/15/2015, 12:18 PM
RE: f ' (x) = f(exp(x)) ? - by fivexthethird - 04/17/2015, 02:15 AM
RE: f ' (x) = f(exp(x)) ? - by tommy1729 - 04/17/2015, 08:32 AM
RE: f ' (x) = f(exp(x)) ? - by tommy1729 - 04/17/2015, 09:32 PM
RE: f ' (x) = f(exp(x)) ? - by tommy1729 - 04/17/2015, 09:40 PM



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