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Infinite tetration and superroot of infinitesimal
Steven in Physorg forum came up with an interesting limit, whihc I extrapolated a little, and will write here just to check if its right, as i used some web complex number calculator to check limits. I wonder if i/x as x-> infinity can be considered as imaginary infinitesimal in the scale where x is.

lim ( 1+- (i/x))^ (pi*(x^2 ))= e^(pi/2) as x-> infinity

if You substitute x=sgrt(y) than it is valid as well, true for both roots of y.

So lim ( 1+-(i/x))^(n*pi*(x^2)) when x-> infinity = e^(n*(pi/2)) for n>=1

Now what happens if n is real? For n=1/2

lim (1+-(i/x))^(pi/2)*(x^2) as x-> infinity e^(pi/4), formula works also for all 1/n

lim (1+-(i/x))^ a*(pi/2)*(x^2) = e^(a*(pi/2)) , a= real, including 0.

But we know that e^(pi/2) = i^ (1/i) or i-th root of i.

So lim (1+-(i/x))^(pi*(x^2)) x-> infinity= i^(1/i) and generally,

lim (1+-i/x)^(a*(pi*(x^2)) x-> infinity = (i^(1/i))^a= (i^-i)^a.

This is probably well known, but still looks nice. I do not know if it really holds for all a I have indicated.

Now if we drop the x-> infinity in a way that we include x in numbers itself, so that i gets "smaller" by infinity and correspondingly pi "longer" by infinity^2, the result would still hold, but in each scale of x ( since we can substitute x with x^2, x^n, x^infinity) the identity will hold, but i and pi will be different, as will e.


Messages In This Thread
RE: Infinite tetration and superroot of infinitesimal - by Ivars - 01/10/2008, 09:19 PM

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