- #1
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Find a function [tex]f\left(z\right)[/tex] such that [tex]f\left(f\left(z\right)\right) = \exp\left(z\right)[/tex]
No, it wasn't obvious. I'm upset and will not come back to this tread. You know how much time I spent on it? What were you thinking? Let's pull the stupid belgian's leg? How do think Count Iblis is feeling when he finds out his question is not taken serious?Yeah, so it doesn't work out right. It was meant as a joke anyway, hope that was obvious.
Cool off... you know, this is a very difficult question, and I'm not sure a solution is known. It seems related to the problem of a continuous extension of the hyperpowers to the real numbers, and that makes me suspect it's difficult.No, it wasn't obvious. I'm upset and will not come back to this tread. You know how much time I spent on it? What were you thinking? Let's pull the stupid belgian's leg? How do think Count Iblis is feeling when he finds out his question is not taken serious?
Oh my god, thanks, I don't know why my brain broke like that! But the princaple still holds I think. I'll retype it all again tomorrow morning when I'm not being an idiot.e^{ln(x)}=x, so e^{ln(x)+2a}=xe^{2a}
I don't think this makes sense, or at least it doesn't make sense to me as an answer for the question the OP posed.Bah, I realised my whole method was fundamentally boring anyway, in more simplified way this is what I was trying to do:
f(x) = e^{a}
f(f(x)) = e^{a}
For each point let x = a
No, I don't think it makes sense either.I don't think this makes sense, or at least it doesn't make sense to me as an answer for the question the OP posed.
I tried a tayloer series before but did it with a full infinite series, I like this method a lot, keep trying it for larger and larger terms, will try it later when I have time to mess about with it. Perhaps only bother trying to approximate it for the first number of terms you have, for example in your last one there, ignore everything x^4 and above.Can it be done using Taylor series?
2 terms
f(x)=a+bx
f(f(x))=a+b(a+bx)=1+x -> a=1/2 b=1
3 terms
f(x)=a+bx+cxx
ff(x)=a+b(a+bx+cxx)+c(a+bx+cxx)^2
=(a+ba+caa)+(bb+2abc)x+(bc+2acc+bbc)xx+(2bcc)xxx+(ccc)xxxx
a+ba+caa=1.
bb+2abc=1
bc+2acc+bbc=1/2!
2bcc=1/3!
ccc=1/4!
Overdefined?- 3 unknowns/4 eqns.