Operator similar to tetration
#2
rsgerard Wrote:Daniel Geisler referred me to this forum, and I'm surprised how active a forum is on a topic like tetrationSmile
Smile

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 Wink ). 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 Smile
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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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