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f ' (x) = f(exp(x)) ? - tommy1729 - 04/15/2015
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 RE: f ' (x) = f(exp(x)) ? - fivexthethird - 04/17/2015
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) and in general the degree of the polynomial appears to be which obviously grows too fast to converge. RE: f ' (x) = f(exp(x)) ? - tommy1729 - 04/17/2015
(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. Sorry Guys but my 3rd derivative in the op was wrong as clearly shown here. I guess these are named polynomials , not ? Regards Tommy1729 RE: f ' (x) = f(exp(x)) ? - tommy1729 - 04/17/2015
I think this equation has a radius 0 everywhere ? Regards Tommy1729 RE: f ' (x) = f(exp(x)) ? - tommy1729 - 04/17/2015
Therefore Maybe f ' (exp(x)) = f(x) works better. Regards Tommy1729 |