f ' (x) = f(exp(x)) ? tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/15/2015, 12:18 PM 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 fivexthethird Junior Fellow Posts: 9 Threads: 3 Joined: Nov 2013 04/17/2015, 02:15 AM (This post was last modified: 04/17/2015, 01:13 PM by fivexthethird.) 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. tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/17/2015, 08:32 AM (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 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/17/2015, 09:32 PM I think this equation has a radius 0 everywhere ? Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/17/2015, 09:40 PM Therefore Maybe f ' (exp(x)) = f(x) works better. Regards Tommy1729 « Next Oldest | Next Newest »