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Infinite tetration and superroot of infinitesimal
#7
jaydfox Wrote:The absolute value of (1-i) is sqrt(2), and the angle from the origin is -pi/4, so the natural logarithm is ln(sqrt(2))-pi*I/4, or 0.5*ln(2)-(pi/4)*I if you prefer it that way.

Yes I know I am outstepping convetions, probably this forum is not the right place to do it, but:

Argand diagram is not applicable to my idea. The fact that mod(i) = 1 is not obvious from definition of i as sqrt(-1).

The fact that (sgrt^(-1))^2 = 1 does not mean module of a negative number squared is 1. At least when applied to physics since negative numbers as such are artificial construct:If someone ows me money, he has negative money in balance sheet. But in fact he does not have negative money, just more or less positive. So negative numebr as such involves delay, some time period over which it can be represented back into positives, and this period may be such that the scale of negative numbers is totally different from positives- e.g. interest or inflation may reduce the scale of negative money because of time.


I think that the diagram would be right, but i being infintiesimal and -i being infinity of the same grade, and -1 also having different modulus than 1.

mod(1-i) = infinity so ln (1-i) = infinity as well (1/i)
mod(1+i)= 1 so ln (1+i) = i ( not 0).
mod(1)=1
mod(-1) = mod(i^2) = second order infinitesimal - smaller than i
mod(1)=mod(i^4) = 4th order infinitesimal even smaller than i

In this case the diagram would have totally different scales on i, -1, -i and 1 axis.
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Messages In This Thread
RE: Infinite tetration and superroot of infinitesimal - by Ivars - 12/19/2007, 08:55 PM

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