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(Free42) roundoff for complex SQRT
04-04-2018, 10:24 AM
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RE: (Free42) roundoff for complex SQRT
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:

[Image: Image%203_zps7bpzdhhn.jpg]

[Image: Image%204_zps5axhpxup.jpg]

The second image shows the value of the real and imaginary parts of the square root in memory (2nd line from top). I know nothing about the 42S, but I know early HP calculators used the "Friden" algorithm for square root. Is it possible the 42S did too?
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RE: (Free42) roundoff for complex SQRT - Dan - 04-04-2018 10:24 AM

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