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 f ' (x) = f(exp(x)) ? tommy1729 Ultimate Fellow Posts: 1,365 Threads: 333 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 « Next Oldest | Next Newest »

 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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