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 Operator similar to tetration bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/27/2008, 08:09 AM (This post was last modified: 05/29/2008, 01:57 PM by bo198214.) rsgerard Wrote:Daniel Geisler referred me to this forum, and I'm surprised how active a forum is on a topic like tetration Quote:I have a question about an operator that is similar to a tetration that I'm trying to locate. I'm a bit new to this topic so forgive the poor terminology. An infinite tetriation with exponent (1/2) could be written as a recurrence relation, I believe: F(n) = square_root( F(n-1) ) F(0) =.5 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? Quote:I'm trying to locate the mathematical operator that describes something very similar, but just multiplying a negative through each iteration (see example). F(n) = square_root( -1 * F(n-1) ) 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 not 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 « 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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