Java BigDecimal Log log10(BigDecimal b)

Here you can find the source of log10(BigDecimal b)

Description

log

License

Apache License

Declaration

public static BigDecimal log10(BigDecimal b) 

Method Source Code


//package com.java2s;
/*/*from w w w  .j  a v  a2  s .  c o  m*/
 * $Id$
 *
 * 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;
import java.math.RoundingMode;

public class Main {
    private static final int SCALE = 18;

    public static BigDecimal log10(BigDecimal b) {
        final int NUM_OF_DIGITS = SCALE + 2;
        // need to add one to get the right number of dp
        //  and then add one again to get the next number
        //  so I can round it correctly.

        MathContext mc = new MathContext(NUM_OF_DIGITS, RoundingMode.HALF_EVEN);
        //special conditions:
        // log(-x) -> exception
        // log(1) == 0 exactly;
        // log of a number lessthan one = -log(1/x)
        if (b.signum() <= 0) {
            throw new ArithmeticException("log of a negative number! (or zero)");
        } else if (b.compareTo(BigDecimal.ONE) == 0) {
            return BigDecimal.ZERO;
        } else if (b.compareTo(BigDecimal.ONE) < 0) {
            return (log10((BigDecimal.ONE).divide(b, mc))).negate();
        }

        StringBuilder sb = new StringBuilder();
        //number of digits on the left of the decimal point
        int leftDigits = b.precision() - b.scale();

        //so, the first digits of the log10 are:
        sb.append(leftDigits - 1).append(".");

        //this is the algorithm outlined in the webpage
        int n = 0;
        while (n < NUM_OF_DIGITS) {
            b = (b.movePointLeft(leftDigits - 1)).pow(10, mc);
            leftDigits = b.precision() - b.scale();
            sb.append(leftDigits - 1);
            n++;
        }

        BigDecimal ans = new BigDecimal(sb.toString());

        //Round the number to the correct number of decimal places.
        ans = ans.round(new MathContext(ans.precision() - ans.scale() + SCALE, RoundingMode.HALF_EVEN));
        return ans;
    }

    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;
    }
}

Related

  1. log(int base_int, BigDecimal x)
  2. log10AbsCeil(BigDecimal x)
  3. log2AbsCeil(BigDecimal x)