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src/math/big/sqrt.go
// Solving 1/x² - z = 0 avoids Quo calls and is faster, especially // for high precisions. z.sqrtInverse(z) // re-attach halved exponent return z.SetMantExp(z, b/2) } // Compute √x (to z.prec precision) by solving // // 1/t² - x = 0 // // for t (using Newton's method), and then inverting. func (z *Float) sqrtInverse(x *Float) { // let // f(t) = 1/t² - x // then // g(t) = f(t)/f'(t) = -½t(1 - xt²)
Registered: Wed Jun 12 16:32:35 UTC 2024 - Last Modified: Mon Apr 11 16:34:30 UTC 2022 - 2.8K bytes - Viewed (0)