java bigdecimal 开方_Java中BigDecimal的平方根

末蓝、 2022-11-03 10:56 310阅读 0赞

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繁星coding

我已经用过了，效果很好。 这是该算法在更高层次上如何工作的示例。编辑：我很好奇，看看这是多么准确如下定义。这是来自官方来源的sqrt(2)：(first 200 digits) 1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702492483605585073721264412149709993583141322266592750559275579995050115278206057147这里使用的是我在下面概述的SQRT\_DIG等于150的方法：(first 200 digits) 1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702492483605585073721264412149709993583141322266592750559275579995050115278206086685第一个偏差出现在195位精度之后。如果您需要如此高的精度，则需要您自担风险。更改SQRT\_DIG为1000可获得1570位精度。private static final BigDecimal SQRT\_DIG = new BigDecimal(150);private static final BigDecimal SQRT\_PRE = new BigDecimal(10).pow(SQRT\_DIG.intValue());/\*\* \* Private utility method used to compute the square root of a BigDecimal. \*  \* @author Luciano Culacciatti  \* @url http://www.codeproject.com/Tips/257031/Implementing-SqrtRoot-in-BigDecimal \*/private static BigDecimal sqrtNewtonRaphson  (BigDecimal c, BigDecimal xn, BigDecimal precision)\{    BigDecimal fx = xn.pow(2).add(c.negate());    BigDecimal fpx = xn.multiply(new BigDecimal(2));    BigDecimal xn1 = fx.divide(fpx,2\*SQRT\_DIG.intValue(),RoundingMode.HALF\_DOWN);    xn1 = xn.add(xn1.negate());    BigDecimal currentSquare = xn1.pow(2);    BigDecimal currentPrecision = currentSquare.subtract(c);    currentPrecision = currentPrecision.abs();    if (currentPrecision.compareTo(precision) <= -1)\{        return xn1;    \}    return sqrtNewtonRaphson(c, xn1, precision);\}/\*\* \* Uses Newton Raphson to compute the square root of a BigDecimal. \*  \* @author Luciano Culacciatti  \* @url http://www.codeproject.com/Tips/257031/Implementing-SqrtRoot-in-BigDecimal \*/public static BigDecimal bigSqrt(BigDecimal c)\{    return sqrtNewtonRaphson(c,new BigDecimal(1),new BigDecimal(1).divide(SQRT\_PRE));\}请务必查看barwnikk的答案。它更简洁，看起来精度更高或更高。

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