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July 13th, 2012, 11:14 AM   #11
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Re: real part of (A+Bi)^(1/3)

You may need to restrict the domain to a one-to-one portion of the cubic.
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July 13th, 2012, 12:17 PM   #12
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Re: real part of (A+Bi)^(1/3)

I am restricting it to a one-to-one region
For example I have...
AX^3+BX^2+CX+D=Y
where
A=3.5855x10^-6
B=9.3864x10^-5
C=-0.17327
D=13.7534

Which is one to one in the region 100<x<300

But when I invert and solve for X I get the above problem in that region
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July 13th, 2012, 12:26 PM   #13
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Re: real part of (A+Bi)^(1/3)

Differentiating, I find the given cubic is one-to-one on the approximate intervals:

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July 13th, 2012, 12:54 PM   #14
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Re: real part of (A+Bi)^(1/3)

Yes you are correct.
Looked again at region from 200<x<300 and this still has a variable K value
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July 13th, 2012, 02:12 PM   #15
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Re: real part of (A+Bi)^(1/3)

My suggestion was really just thinking out loud. I did have doubts that would fix the problem, but it was the first thing I thought of. I'll give this more consideration.
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