(Free42) roundoff for complex SQRT

04042018, 03:03 PM
Post: #9




RE: (Free42) roundoff for complex SQRT
(04042018 10:24 AM)Dan Wrote: I used the "Friden" algorithm for square root on the AriCalculator, and sqrt(z) = sqrt(.5(a+sqrt(a^2+b^2)) )+ sqrt(0.5(a+sqrt(a^2+b^2))). This gives exact results for 8i and other z I tried for which Re(z) = 0 and Im(z)/2 is a perfect square This is what I used on newRPL too, but I'm probably going to change it to what Werner showed above for the 48, it saves one square root, so it's roughly 30% faster. 

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