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 Operator similar to tetration rsgerard Junior Fellow Posts: 11 Threads: 5 Joined: May 2008 05/28/2008, 07:05 PM bo198214, Quote:Thats not infinite tetration. If you expand this, you get $F(0)=1/2$ $F(1)=(1/2)^{1/2}$ $F(2)=\left((1/2)^{1/2}\right)^{1/2}=(1/2)^{1/4}$ $F(3)=\left((1/2)^{1/4}\right)^{1/2}=(1/2)^{1/8}$ $F(n)=(1/2)^{2^{-n}}$ $\lim_{n\to\infty} F(n)=1$ I would guess what you mean is (at least this resembles infinite tetration): $F(0)=1$ $F(1)=1/2$ $F(2)=(1/2)^(1/2)$ $F(3)=(1/2)^{(1/2)^{1/2}}$ $F(n+1)=(1/2)^{F(n)}$ yes, no? You are completely correct, the recurrence relation I suggested does not evaluate the exponent from "highest to lowest". I agree that the rewritten expression accurately describes a tetration. Quote:Ok, taking your original definition a bit more generally: $F(n+1) = (c*F(n))^\alpha$ $F(1)=c^{\alpha} F(0)^{\alpha}$ $F(2)=\left(c c^{\alpha} F(0)^{\alpha}\right)^{\alpha}=c^{\alpha+\alpha^2} F(0)^{\alpha^2}$ $F(3)=\left(c c^{\alpha+\alpha^2} F(0)^{\alpha^2}\right)^\alpha= c^{\alpha+\alpha^2+\alpha^3} F(0)^{\alpha^3}$ $F(n)=c^{\alpha+\alpha^2+\dots+\alpha^n} F(0)^{\alpha^n}$ Now we know that for $0<\alpha<1$ $\lim_{n\to\infty} 1+\alpha+\alpha^2+\dots+\alpha^n = 1/(1-\alpha)$ hence for $F(0)>0$ $\lim_{n\to\infty} F(n) = c^{\alpha/(1-\alpha)}$ in our case $\alpha=1/2$ and $c=-1$ we get $\lim_{n\to\infty} F(n) = (-1)^{(1/2)/(1/2)} = -1$ however the derivation is somewhat sloppy as the exponential laws are generally applicable for complex numbers (as painfully observed by Gottfried ). So the above should be considered only valid for $c>0$. Indeed one sees that the the sequence $F(n)$ has no limit, but oscillates between two values, i.e. has two limit points. The task for the reader is to determine these both points First, thanks a lot for expanding the math out here. This is very helpful to see. When I attempt to evaluate this recurrence, I see the result oscillate between 2 values when the principle root is used and -1 otherwise. So, I get the following solutions: $\lim_{n\to\infty} F(n) = -1, .5 \pm \sqrt{3}/2 i$ I believe these are the roots for $x^3=-1$ My two specific questions are: 1. Is this above correct? 2. Has something similar ever been shown in a more general case: For example: $F(n)=\sqrt[3]{-1 * F(n-1)}$ $\lim_{n\to\infty} F(n)$ seems to be the roots for $x^4=-1$ ??? Sorry for being a bit off topic to tetrations but this operation performed an infinite number of times relates to taking the 1/(n+1) exponent. Please let me know if this is on the right track, or have I derailed somewhere Thanks. -Ryan « Next Oldest | Next Newest »

 Messages In This Thread Operator similar to tetration - by rsgerard - 05/27/2008, 03:57 AM RE: Operator similar to tetration - by bo198214 - 05/27/2008, 08:09 AM RE: Operator similar to tetration - by rsgerard - 05/28/2008, 07:05 PM RE: Operator similar to tetration - by bo198214 - 05/29/2008, 03:41 PM RE: Operator similar to tetration - by rsgerard - 05/29/2008, 10:34 PM RE: Operator similar to tetration - by bo198214 - 05/30/2008, 06:11 AM

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