If you are in the mood for a good math Soap opera, take a gander at the story of Tartaglia
Having tumbled into this (my notes are unclear. The immediately previous entry refers to Gaussians, and before that are several pages of Nash problems. I think I got to Tartaglia via Eratosthenes, but who can be sure.
11 pages of trash on cubics, mostly because I can't seem to keep track of negative signs. Once I got that working, and did a bit more reading, I started trying to work out which root would fall out of this mess (you are taking the cube root of a quadatic solution. The +- from the quadratic formula washes out, as it happens. So you just have to pick which cube root you want - oh, did we not mention that most of the time you are looking at complex solutions to the quadratic?
Getting some godawful mess, and converting it to cosine. 2 root 7 cosine of 1/3 of Acos ( root something else ). Bleah. Excel claims the solution is exact, and is in fact one of the roots I used to construct the original equation. Which suggests to me that I can find some identity to solve it exactly.
A brief burst of energy unveils the cos(3x) trig identity that I've never had a use for, but I soon discover that converting that to a cos(x/3) identity is solving a cubic [yes, to those of you snickering, I'm well on my way to duplicating de moivre the hard way]. And the minus signs continue to jump away, and weird things happen - that don't work when I plug them back in - until it finally occurs to me that this would all go a lot smoother if the quadratic had imaginary roots... Oh it does (another minus sign bites the dust).
Great, now I'm at de Moivre; realizing I've deadended myself, I step back where I should have to begin with: e^ix (I skipped a few physics courses I ought to have taken in college; this comes around to haunt me from time to time). Rotate the angle by 2pi/3, and the other loaded roots fall out.
OK, so with one real root, that's what the cubic gives us. When we have three real roots, the quadratic has imaginary solutions, so we're taking the cube root of a complex. Fine, but I still don't know which is which.
I think this means answering "which root will have the smallest theta", but I'm not sure.
If you happen to see a missing minus sign lying about, send it this way.
May 12, 2003 12:53 AM
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Seems like you are trying to calculate an exponential root using Excel?
Try y=10^((log(n)/x))
(x=power)