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Infinite tetration and superroot of infinitesimal
Ivars Wrote:
andydude Wrote:I think trying to understand that statement intuitively is doomed, as recursive interpretation may lead to madness...

Andrew Robbins
So infinitely working beginning with end ( from right) in infinite number of dimensions on e^(pi/2) is equal to -imaginary unit.

I think you mean: rotating the length through infinite number of dimensions gives a hyper-volume equal to -imaginary unit. I only knew what you were talking about from context. If I were not in this forum, and I saw your words, they would not describe what you are trying to describe. Try and be more careful about how to say things.

Ivars Wrote:So log(log(log(...... ) is tetration? if we take e^pi/2=i^-i, and log with base i, we get (from right) (only main value) :

So, if we ever get to the end, we might have either
or .
(sorry, I had to reorganize your math to understand it)

True and true. I think your issue here is that you are calculating with base i, and as such, the only way to calculate the infinitely iterated exponential with base i is to use exponentials, not logarithms. The base-b logarithms can be infinitely iterated for all b such that where a is a fixed point . Exponentials on the other hand converge everywhere logs don't. So the base-b exponentials can be infinitely iterated for all b such that . Now the fixed point of base i is which means Since this value (sometimes called the Lyapunov number), means that exponentials will converge, but logarithms will not, because it is less than 1. So what you have found is that when you try and use logarithms for base i, then it doesn't converge, it oscillates between values, as expected.

Ivars Wrote:Is this the same as tetration of (e^pi/2)? How do we calculate it in this case?
You would do it just as you would with base i, only we are looking at base this time. According to the infinitely iterated exponential (with the usual branch structure) we get . This means that -i is the fixed point of the base- exponential function. To determine whether we should be using exponentials or logarithms to find this fixed point, we do the test: so we should be using logarithms instead of exponentials, because this number (the Lyapunov number) is greater than 1. So with this, we can say that: for all n. Since this is true for all n, the version when is true.

Andrew Robbins

Messages In This Thread
RE: Infinite tetration and superroot of infinitesimal - by andydude - 01/09/2008, 12:50 AM

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