Here you can find the source of cuberoot(BigDecimal b)
public static BigDecimal cuberoot(BigDecimal b)
//package com.java2s; /*//from w ww. j a v a 2s . com * $Id$ * * Copyright 2013 Valentyn Kolesnikov * * Licensed under the Apache License, Version 2.0 (the "License"); * you may not use this file except in compliance with the License. * You may obtain a copy of the License at * * http://www.apache.org/licenses/LICENSE-2.0 * * Unless required by applicable law or agreed to in writing, software * distributed under the License is distributed on an "AS IS" BASIS, * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. * See the License for the specific language governing permissions and * limitations under the License. */ import java.math.BigDecimal; import java.math.MathContext; public class Main { public static long ITER = 1000; public static BigDecimal cuberoot(BigDecimal b) { // Specify a math context with 40 digits of precision. MathContext mc = new MathContext(40); BigDecimal x = new BigDecimal("1", mc); // Search for the cube root via the Newton-Raphson loop. Output each // successive iteration's value. for (int i = 0; i < ITER; i++) { x = x.subtract( x.pow(3, mc).subtract(b, mc).divide(new BigDecimal("3", mc).multiply(x.pow(2, mc), mc), mc), mc); } return x; } public static BigDecimal pow(BigDecimal savedValue, BigDecimal value) { BigDecimal result = null; result = exp(ln(savedValue, 32).multiply(value), 32); return result; } /** * Compute e^x to a given scale. * Break x into its whole and fraction parts and * compute (e^(1 + fraction/whole))^whole using Taylor's formula. * @param x the value of x * @param scale the desired scale of the result * @return the result value */ public static BigDecimal exp(BigDecimal x, int scale) { // e^0 = 1 if (x.signum() == 0) { return BigDecimal.valueOf(1); } // If x is negative, return 1/(e^-x). else if (x.signum() == -1) { return BigDecimal.valueOf(1).divide(exp(x.negate(), scale), scale, BigDecimal.ROUND_HALF_EVEN); } // Compute the whole part of x. BigDecimal xWhole = x.setScale(0, BigDecimal.ROUND_DOWN); // If there isn't a whole part, compute and return e^x. if (xWhole.signum() == 0) { return expTaylor(x, scale); } // Compute the fraction part of x. BigDecimal xFraction = x.subtract(xWhole); // z = 1 + fraction/whole BigDecimal z = BigDecimal.valueOf(1).add(xFraction.divide(xWhole, scale, BigDecimal.ROUND_HALF_EVEN)); // t = e^z BigDecimal t = expTaylor(z, scale); BigDecimal maxLong = BigDecimal.valueOf(Long.MAX_VALUE); BigDecimal result = BigDecimal.valueOf(1); // Compute and return t^whole using intPower(). // If whole > Long.MAX_VALUE, then first compute products // of e^Long.MAX_VALUE. while (xWhole.compareTo(maxLong) >= 0) { result = result.multiply(intPower(t, Long.MAX_VALUE, scale)).setScale(scale, BigDecimal.ROUND_HALF_EVEN); xWhole = xWhole.subtract(maxLong); Thread.yield(); } return result.multiply(intPower(t, xWhole.longValue(), scale)).setScale(scale, BigDecimal.ROUND_HALF_EVEN); } /** * Compute the natural logarithm of x to a given scale, x > 0. */ public static BigDecimal ln(BigDecimal x, int scale) { // Check that x > 0. if (x.signum() <= 0) { throw new IllegalArgumentException("x <= 0"); } // The number of digits to the left of the decimal point. int magnitude = x.toString().length() - x.scale() - 1; if (magnitude < 3) { return lnNewton(x, scale); } // Compute magnitude*ln(x^(1/magnitude)). else { // x^(1/magnitude) BigDecimal root = intRoot(x, magnitude, scale); // ln(x^(1/magnitude)) BigDecimal lnRoot = lnNewton(root, scale); // magnitude*ln(x^(1/magnitude)) return BigDecimal.valueOf(magnitude).multiply(lnRoot).setScale(scale, BigDecimal.ROUND_HALF_EVEN); } } /** * Compute e^x to a given scale by the Taylor series. * @param x the value of x * @param scale the desired scale of the result * @return the result value */ private static BigDecimal expTaylor(BigDecimal x, int scale) { BigDecimal factorial = BigDecimal.valueOf(1); BigDecimal xPower = x; BigDecimal sumPrev; // 1 + x BigDecimal sum = x.add(BigDecimal.valueOf(1)); // Loop until the sums converge // (two successive sums are equal after rounding). int i = 2; do { // x^i xPower = xPower.multiply(x).setScale(scale, BigDecimal.ROUND_HALF_EVEN); // i! factorial = factorial.multiply(BigDecimal.valueOf(i)); // x^i/i! BigDecimal term = xPower.divide(factorial, scale, BigDecimal.ROUND_HALF_EVEN); // sum = sum + x^i/i! sumPrev = sum; sum = sum.add(term); ++i; Thread.yield(); } while (sum.compareTo(sumPrev) != 0); return sum; } /** * Compute x^exponent to a given scale. Uses the same * algorithm as class numbercruncher.mathutils.IntPower. * @param x the value x * @param exponent the exponent value * @param scale the desired scale of the result * @return the result value */ public static BigDecimal intPower(BigDecimal x, long exponent, int scale) { // If the exponent is negative, compute 1/(x^-exponent). if (exponent < 0) { return BigDecimal.valueOf(1).divide(intPower(x, -exponent, scale), scale, BigDecimal.ROUND_HALF_EVEN); } BigDecimal power = BigDecimal.valueOf(1); // Loop to compute value^exponent. while (exponent > 0) { // Is the rightmost bit a 1? if ((exponent & 1) == 1) { power = power.multiply(x).setScale(scale, BigDecimal.ROUND_HALF_EVEN); } // Square x and shift exponent 1 bit to the right. x = x.multiply(x).setScale(scale, BigDecimal.ROUND_HALF_EVEN); exponent >>= 1; Thread.yield(); } return power; } /** * Compute the natural logarithm of x to a given scale, x > 0. * Use Newton's algorithm. */ private static BigDecimal lnNewton(BigDecimal x, int scale) { int sp1 = scale + 1; BigDecimal n = x; BigDecimal term; // Convergence tolerance = 5*(10^-(scale+1)) BigDecimal tolerance = BigDecimal.valueOf(5).movePointLeft(sp1); // Loop until the approximations converge // (two successive approximations are within the tolerance). do { // e^x BigDecimal eToX = exp(x, sp1); // (e^x - n)/e^x term = eToX.subtract(n).divide(eToX, sp1, BigDecimal.ROUND_DOWN); // x - (e^x - n)/e^x x = x.subtract(term); Thread.yield(); } while (term.compareTo(tolerance) > 0); return x.setScale(scale, BigDecimal.ROUND_HALF_EVEN); } /** * Compute the integral root of x to a given scale, x >= 0. * Use Newton's algorithm. * @param x the value of x * @param index the integral root value * @param scale the desired scale of the result * @return the result value */ public static BigDecimal intRoot(BigDecimal x, long index, int scale) { // Check that x >= 0. if (x.signum() < 0) { throw new IllegalArgumentException("x < 0"); } int sp1 = scale + 1; BigDecimal n = x; BigDecimal i = BigDecimal.valueOf(index); BigDecimal im1 = BigDecimal.valueOf(index - 1); BigDecimal tolerance = BigDecimal.valueOf(5).movePointLeft(sp1); BigDecimal xPrev; // The initial approximation is x/index. x = x.divide(i, scale, BigDecimal.ROUND_HALF_EVEN); // Loop until the approximations converge // (two successive approximations are equal after rounding). do { // x^(index-1) BigDecimal xToIm1 = intPower(x, index - 1, sp1); // x^index BigDecimal xToI = x.multiply(xToIm1).setScale(sp1, BigDecimal.ROUND_HALF_EVEN); // n + (index-1)*(x^index) BigDecimal numerator = n.add(im1.multiply(xToI)).setScale(sp1, BigDecimal.ROUND_HALF_EVEN); // (index*(x^(index-1)) BigDecimal denominator = i.multiply(xToIm1).setScale(sp1, BigDecimal.ROUND_HALF_EVEN); // x = (n + (index-1)*(x^index)) / (index*(x^(index-1))) xPrev = x; x = numerator.divide(denominator, sp1, BigDecimal.ROUND_DOWN); Thread.yield(); } while (x.subtract(xPrev).abs().compareTo(tolerance) > 0); return x; } }