// This example shows how to use big.Float to compute the square root of 2 with // a precision of 200 bits, and how to print the result as a decimal number. func Example_sqrt2() { // We'll do computations with 200 bits of precision in the mantissa. const prec = 200 // Compute the square root of 2 using Newton's Method. We start with // an initial estimate for sqrt(2), and then iterate: // x_{n+1} = 1/2 * ( x_n + (2.0 / x_n) ) // Since Newton's Method doubles the number of correct digits at each // iteration, we need at least log_2(prec) steps. steps := int(math.Log2(prec)) // Initialize values we need for the computation. two := new(big.Float).SetPrec(prec).SetInt64(2) half := new(big.Float).SetPrec(prec).SetFloat64(0.5) // Use 1 as the initial estimate. x := new(big.Float).SetPrec(prec).SetInt64(1) // We use t as a temporary variable. There's no need to set its precision // since big.Float values with unset (== 0) precision automatically assume // the largest precision of the arguments when used as the result (receiver) // of a big.Float operation. t := new(big.Float) // Iterate. for i := 0; i <= steps; i++ { t.Quo(two, x) // t = 2.0 / x_n t.Add(x, t) // t = x_n + (2.0 / x_n) x.Mul(half, t) // x_{n+1} = 0.5 * t } // We can use the usual fmt.Printf verbs since big.Float implements fmt.Formatter fmt.Printf("sqrt(2) = %.50f\n", x) // Print the error between 2 and x*x. t.Mul(x, x) // t = x*x fmt.Printf("error = %e\n", t.Sub(two, t)) // Output: // sqrt(2) = 1.41421356237309504880168872420969807856967187537695 // error = 0.000000e+00 }