Compute the reciprocal of in. Use the following algorithm. in = c + d. want to find x + y such that x+y = 1/(c+d) and x is much larger than y and x has several zero bits on the right. Set b = 1/(2^22), a = 1 - b. Thus (a+b) = 1. Use following identity to compute (a+b)/(c+d) (a+b)/(c+d) =
(final double in[], final double result[])
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