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(Free42) roundoff for complex SQRT
04-03-2018, 01:28 PM
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(Free42) roundoff for complex SQRT
Nitpicking, perhaps, but if the original 42S gets it right - then so should Free42 ;-)
Try SQRT(a+ib) with a=0 and b/2 a perfect square eg. b=8
Free42 shows 2+i2, which SHOW reveals to be really 2+i(2-1E-33).
Almost any b you throw at it has a rounding error like that, in the real or complex part.
The 12-digit 42S gets the exact result - it must have special code for a=0, and return SQRT(ABS(b)/2)*(1+i*SIGN(b)) in that case - even when b/2 would underflow.

Cheers, Werner
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(Free42) roundoff for complex SQRT - Werner - 04-03-2018 01:28 PM

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