(Free42) roundoff for complex SQRT
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05-11-2018, 09:06 AM
Post: #20
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RE: (Free42) roundoff for complex SQRT
(04-04-2018 03:03 PM)Claudio L. Wrote:(04-04-2018 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 Actually on my system it's slower (2 - 3 times), probably because I'm using a shift and add algorithm for division, so division is much slower than square root. |
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