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Java example source code file (DfpMath.java)

This example Java source code file (DfpMath.java) is included in the alvinalexander.com "Java Source Code Warehouse" project. The intent of this project is to help you "Learn Java by Example" TM.

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Java - Java tags/keywords

dfp, dfpfield, dfpmath, pow_trap, string, this, use

The DfpMath.java Java example source code

/*
 * Licensed to the Apache Software Foundation (ASF) under one or more
 * contributor license agreements.  See the NOTICE file distributed with
 * this work for additional information regarding copyright ownership.
 * The ASF licenses this file to You 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.
 */

package org.apache.commons.math3.dfp;

/** Mathematical routines for use with {@link Dfp}.
 * The constants are defined in {@link DfpField}
 * @since 2.2
 */
public class DfpMath {

    /** Name for traps triggered by pow. */
    private static final String POW_TRAP = "pow";

    /**
     * Private Constructor.
     */
    private DfpMath() {
    }

    /** Breaks a string representation up into two dfp's.
     * <p>The two dfp are such that the sum of them is equivalent
     * to the input string, but has higher precision than using a
     * single dfp. This is useful for improving accuracy of
     * exponentiation and critical multiplies.
     * @param field field to which the Dfp must belong
     * @param a string representation to split
     * @return an array of two {@link Dfp} which sum is a
     */
    protected static Dfp[] split(final DfpField field, final String a) {
        Dfp result[] = new Dfp[2];
        char[] buf;
        boolean leading = true;
        int sp = 0;
        int sig = 0;

        buf = new char[a.length()];

        for (int i = 0; i < buf.length; i++) {
            buf[i] = a.charAt(i);

            if (buf[i] >= '1' && buf[i] <= '9') {
                leading = false;
            }

            if (buf[i] == '.') {
                sig += (400 - sig) % 4;
                leading = false;
            }

            if (sig == (field.getRadixDigits() / 2) * 4) {
                sp = i;
                break;
            }

            if (buf[i] >= '0' && buf[i] <= '9' && !leading) {
                sig ++;
            }
        }

        result[0] = field.newDfp(new String(buf, 0, sp));

        for (int i = 0; i < buf.length; i++) {
            buf[i] = a.charAt(i);
            if (buf[i] >= '0' && buf[i] <= '9' && i < sp) {
                buf[i] = '0';
            }
        }

        result[1] = field.newDfp(new String(buf));

        return result;
    }

    /** Splits a {@link Dfp} into 2 {@link Dfp}'s such that their sum is equal to the input {@link Dfp}.
     * @param a number to split
     * @return two elements array containing the split number
     */
    protected static Dfp[] split(final Dfp a) {
        final Dfp[] result = new Dfp[2];
        final Dfp shift = a.multiply(a.power10K(a.getRadixDigits() / 2));
        result[0] = a.add(shift).subtract(shift);
        result[1] = a.subtract(result[0]);
        return result;
    }

    /** Multiply two numbers that are split in to two pieces that are
     *  meant to be added together.
     *  Use binomial multiplication so ab = a0 b0 + a0 b1 + a1 b0 + a1 b1
     *  Store the first term in result0, the rest in result1
     *  @param a first factor of the multiplication, in split form
     *  @param b second factor of the multiplication, in split form
     *  @return a × b, in split form
     */
    protected static Dfp[] splitMult(final Dfp[] a, final Dfp[] b) {
        final Dfp[] result = new Dfp[2];

        result[1] = a[0].getZero();
        result[0] = a[0].multiply(b[0]);

        /* If result[0] is infinite or zero, don't compute result[1].
         * Attempting to do so may produce NaNs.
         */

        if (result[0].classify() == Dfp.INFINITE || result[0].equals(result[1])) {
            return result;
        }

        result[1] = a[0].multiply(b[1]).add(a[1].multiply(b[0])).add(a[1].multiply(b[1]));

        return result;
    }

    /** Divide two numbers that are split in to two pieces that are meant to be added together.
     * Inverse of split multiply above:
     *  (a+b) / (c+d) = (a/c) + ( (bc-ad)/(c**2+cd) )
     *  @param a dividend, in split form
     *  @param b divisor, in split form
     *  @return a / b, in split form
     */
    protected static Dfp[] splitDiv(final Dfp[] a, final Dfp[] b) {
        final Dfp[] result;

        result = new Dfp[2];

        result[0] = a[0].divide(b[0]);
        result[1] = a[1].multiply(b[0]).subtract(a[0].multiply(b[1]));
        result[1] = result[1].divide(b[0].multiply(b[0]).add(b[0].multiply(b[1])));

        return result;
    }

    /** Raise a split base to the a power.
     * @param base number to raise
     * @param a power
     * @return base<sup>a
     */
    protected static Dfp splitPow(final Dfp[] base, int a) {
        boolean invert = false;

        Dfp[] r = new Dfp[2];

        Dfp[] result = new Dfp[2];
        result[0] = base[0].getOne();
        result[1] = base[0].getZero();

        if (a == 0) {
            // Special case a = 0
            return result[0].add(result[1]);
        }

        if (a < 0) {
            // If a is less than zero
            invert = true;
            a = -a;
        }

        // Exponentiate by successive squaring
        do {
            r[0] = new Dfp(base[0]);
            r[1] = new Dfp(base[1]);
            int trial = 1;

            int prevtrial;
            while (true) {
                prevtrial = trial;
                trial *= 2;
                if (trial > a) {
                    break;
                }
                r = splitMult(r, r);
            }

            trial = prevtrial;

            a -= trial;
            result = splitMult(result, r);

        } while (a >= 1);

        result[0] = result[0].add(result[1]);

        if (invert) {
            result[0] = base[0].getOne().divide(result[0]);
        }

        return result[0];

    }

    /** Raises base to the power a by successive squaring.
     * @param base number to raise
     * @param a power
     * @return base<sup>a
     */
    public static Dfp pow(Dfp base, int a)
    {
        boolean invert = false;

        Dfp result = base.getOne();

        if (a == 0) {
            // Special case
            return result;
        }

        if (a < 0) {
            invert = true;
            a = -a;
        }

        // Exponentiate by successive squaring
        do {
            Dfp r = new Dfp(base);
            Dfp prevr;
            int trial = 1;
            int prevtrial;

            do {
                prevr = new Dfp(r);
                prevtrial = trial;
                r = r.multiply(r);
                trial *= 2;
            } while (a>trial);

            r = prevr;
            trial = prevtrial;

            a -= trial;
            result = result.multiply(r);

        } while (a >= 1);

        if (invert) {
            result = base.getOne().divide(result);
        }

        return base.newInstance(result);

    }

    /** Computes e to the given power.
     * a is broken into two parts, such that a = n+m  where n is an integer.
     * We use pow() to compute e<sup>n and a Taylor series to compute
     * e<sup>m.  We return e*n × em
     * @param a power at which e should be raised
     * @return e<sup>a
     */
    public static Dfp exp(final Dfp a) {

        final Dfp inta = a.rint();
        final Dfp fraca = a.subtract(inta);

        final int ia = inta.intValue();
        if (ia > 2147483646) {
            // return +Infinity
            return a.newInstance((byte)1, Dfp.INFINITE);
        }

        if (ia < -2147483646) {
            // return 0;
            return a.newInstance();
        }

        final Dfp einta = splitPow(a.getField().getESplit(), ia);
        final Dfp efraca = expInternal(fraca);

        return einta.multiply(efraca);
    }

    /** Computes e to the given power.
     * Where -1 < a < 1.  Use the classic Taylor series.  1 + x**2/2! + x**3/3! + x**4/4!  ...
     * @param a power at which e should be raised
     * @return e<sup>a
     */
    protected static Dfp expInternal(final Dfp a) {
        Dfp y = a.getOne();
        Dfp x = a.getOne();
        Dfp fact = a.getOne();
        Dfp py = new Dfp(y);

        for (int i = 1; i < 90; i++) {
            x = x.multiply(a);
            fact = fact.divide(i);
            y = y.add(x.multiply(fact));
            if (y.equals(py)) {
                break;
            }
            py = new Dfp(y);
        }

        return y;
    }

    /** Returns the natural logarithm of a.
     * a is first split into three parts such that  a = (10000^h)(2^j)k.
     * ln(a) is computed by ln(a) = ln(5)*h + ln(2)*(h+j) + ln(k)
     * k is in the range 2/3 < k <4/3 and is passed on to a series expansion.
     * @param a number from which logarithm is requested
     * @return log(a)
     */
    public static Dfp log(Dfp a) {
        int lr;
        Dfp x;
        int ix;
        int p2 = 0;

        // Check the arguments somewhat here
        if (a.equals(a.getZero()) || a.lessThan(a.getZero()) || a.isNaN()) {
            // negative, zero or NaN
            a.getField().setIEEEFlagsBits(DfpField.FLAG_INVALID);
            return a.dotrap(DfpField.FLAG_INVALID, "ln", a, a.newInstance((byte)1, Dfp.QNAN));
        }

        if (a.classify() == Dfp.INFINITE) {
            return a;
        }

        x = new Dfp(a);
        lr = x.log10K();

        x = x.divide(pow(a.newInstance(10000), lr));  /* This puts x in the range 0-10000 */
        ix = x.floor().intValue();

        while (ix > 2) {
            ix >>= 1;
            p2++;
        }


        Dfp[] spx = split(x);
        Dfp[] spy = new Dfp[2];
        spy[0] = pow(a.getTwo(), p2);          // use spy[0] temporarily as a divisor
        spx[0] = spx[0].divide(spy[0]);
        spx[1] = spx[1].divide(spy[0]);

        spy[0] = a.newInstance("1.33333");    // Use spy[0] for comparison
        while (spx[0].add(spx[1]).greaterThan(spy[0])) {
            spx[0] = spx[0].divide(2);
            spx[1] = spx[1].divide(2);
            p2++;
        }

        // X is now in the range of 2/3 < x < 4/3
        Dfp[] spz = logInternal(spx);

        spx[0] = a.newInstance(new StringBuilder().append(p2+4*lr).toString());
        spx[1] = a.getZero();
        spy = splitMult(a.getField().getLn2Split(), spx);

        spz[0] = spz[0].add(spy[0]);
        spz[1] = spz[1].add(spy[1]);

        spx[0] = a.newInstance(new StringBuilder().append(4*lr).toString());
        spx[1] = a.getZero();
        spy = splitMult(a.getField().getLn5Split(), spx);

        spz[0] = spz[0].add(spy[0]);
        spz[1] = spz[1].add(spy[1]);

        return a.newInstance(spz[0].add(spz[1]));

    }

    /** Computes the natural log of a number between 0 and 2.
     *  Let f(x) = ln(x),
     *
     *  We know that f'(x) = 1/x, thus from Taylor's theorum we have:
     *
     *           -----          n+1         n
     *  f(x) =   \           (-1)    (x - 1)
     *           /          ----------------    for 1 <= n <= infinity
     *           -----             n
     *
     *  or
     *                       2        3       4
     *                   (x-1)   (x-1)    (x-1)
     *  ln(x) =  (x-1) - ----- + ------ - ------ + ...
     *                     2       3        4
     *
     *  alternatively,
     *
     *                  2    3   4
     *                 x    x   x
     *  ln(x+1) =  x - -  + - - - + ...
     *                 2    3   4
     *
     *  This series can be used to compute ln(x), but it converges too slowly.
     *
     *  If we substitute -x for x above, we get
     *
     *                   2    3    4
     *                  x    x    x
     *  ln(1-x) =  -x - -  - -  - - + ...
     *                  2    3    4
     *
     *  Note that all terms are now negative.  Because the even powered ones
     *  absorbed the sign.  Now, subtract the series above from the previous
     *  one to get ln(x+1) - ln(1-x).  Note the even terms cancel out leaving
     *  only the odd ones
     *
     *                             3     5      7
     *                           2x    2x     2x
     *  ln(x+1) - ln(x-1) = 2x + --- + --- + ---- + ...
     *                            3     5      7
     *
     *  By the property of logarithms that ln(a) - ln(b) = ln (a/b) we have:
     *
     *                                3        5        7
     *      x+1           /          x        x        x          \
     *  ln ----- =   2 *  |  x  +   ----  +  ----  +  ---- + ...  |
     *      x-1           \          3        5        7          /
     *
     *  But now we want to find ln(a), so we need to find the value of x
     *  such that a = (x+1)/(x-1).   This is easily solved to find that
     *  x = (a-1)/(a+1).
     * @param a number from which logarithm is requested, in split form
     * @return log(a)
     */
    protected static Dfp[] logInternal(final Dfp a[]) {

        /* Now we want to compute x = (a-1)/(a+1) but this is prone to
         * loss of precision.  So instead, compute x = (a/4 - 1/4) / (a/4 + 1/4)
         */
        Dfp t = a[0].divide(4).add(a[1].divide(4));
        Dfp x = t.add(a[0].newInstance("-0.25")).divide(t.add(a[0].newInstance("0.25")));

        Dfp y = new Dfp(x);
        Dfp num = new Dfp(x);
        Dfp py = new Dfp(y);
        int den = 1;
        for (int i = 0; i < 10000; i++) {
            num = num.multiply(x);
            num = num.multiply(x);
            den += 2;
            t = num.divide(den);
            y = y.add(t);
            if (y.equals(py)) {
                break;
            }
            py = new Dfp(y);
        }

        y = y.multiply(a[0].getTwo());

        return split(y);

    }

    /** Computes x to the y power.<p>
     *
     *  Uses the following method:<p>
     *
     *  <ol>
     *  <li> Set u = rint(y), v = y-u
     *  <li> Compute a = v * ln(x)
     *  <li> Compute b = rint( a/ln(2) )
     *  <li> Compute c = a - b*ln(2)
     *  <li> xy = xu  *   2b * ec
     *  </ol>
     *  if |y| > 1e8, then we compute by exp(y*ln(x))   <p>
     *
     *  <b>Special Cases

* <ul> * <li> if y is 0.0 or -0.0 then result is 1.0 * <li> if y is 1.0 then result is x * <li> if y is NaN then result is NaN * <li> if x is NaN and y is not zero then result is NaN * <li> if |x| > 1.0 and y is +Infinity then result is +Infinity * <li> if |x| < 1.0 and y is -Infinity then result is +Infinity * <li> if |x| > 1.0 and y is -Infinity then result is +0 * <li> if |x| < 1.0 and y is +Infinity then result is +0 * <li> if |x| = 1.0 and y is +/-Infinity then result is NaN * <li> if x = +0 and y > 0 then result is +0 * <li> if x = +Inf and y < 0 then result is +0 * <li> if x = +0 and y < 0 then result is +Inf * <li> if x = +Inf and y > 0 then result is +Inf * <li> if x = -0 and y > 0, finite, not odd integer then result is +0 * <li> if x = -0 and y < 0, finite, and odd integer then result is -Inf * <li> if x = -Inf and y > 0, finite, and odd integer then result is -Inf * <li> if x = -0 and y < 0, not finite odd integer then result is +Inf * <li> if x = -Inf and y > 0, not finite odd integer then result is +Inf * <li> if x < 0 and y > 0, finite, and odd integer then result is -(|x|y) * <li> if x < 0 and y > 0, finite, and not integer then result is NaN * </ul> * @param x base to be raised * @param y power to which base should be raised * @return x<sup>y */ public static Dfp pow(Dfp x, final Dfp y) { // make sure we don't mix number with different precision if (x.getField().getRadixDigits() != y.getField().getRadixDigits()) { x.getField().setIEEEFlagsBits(DfpField.FLAG_INVALID); final Dfp result = x.newInstance(x.getZero()); result.nans = Dfp.QNAN; return x.dotrap(DfpField.FLAG_INVALID, POW_TRAP, x, result); } final Dfp zero = x.getZero(); final Dfp one = x.getOne(); final Dfp two = x.getTwo(); boolean invert = false; int ui; /* Check for special cases */ if (y.equals(zero)) { return x.newInstance(one); } if (y.equals(one)) { if (x.isNaN()) { // Test for NaNs x.getField().setIEEEFlagsBits(DfpField.FLAG_INVALID); return x.dotrap(DfpField.FLAG_INVALID, POW_TRAP, x, x); } return x; } if (x.isNaN() || y.isNaN()) { // Test for NaNs x.getField().setIEEEFlagsBits(DfpField.FLAG_INVALID); return x.dotrap(DfpField.FLAG_INVALID, POW_TRAP, x, x.newInstance((byte)1, Dfp.QNAN)); } // X == 0 if (x.equals(zero)) { if (Dfp.copysign(one, x).greaterThan(zero)) { // X == +0 if (y.greaterThan(zero)) { return x.newInstance(zero); } else { return x.newInstance(x.newInstance((byte)1, Dfp.INFINITE)); } } else { // X == -0 if (y.classify() == Dfp.FINITE && y.rint().equals(y) && !y.remainder(two).equals(zero)) { // If y is odd integer if (y.greaterThan(zero)) { return x.newInstance(zero.negate()); } else { return x.newInstance(x.newInstance((byte)-1, Dfp.INFINITE)); } } else { // Y is not odd integer if (y.greaterThan(zero)) { return x.newInstance(zero); } else { return x.newInstance(x.newInstance((byte)1, Dfp.INFINITE)); } } } } if (x.lessThan(zero)) { // Make x positive, but keep track of it x = x.negate(); invert = true; } if (x.greaterThan(one) && y.classify() == Dfp.INFINITE) { if (y.greaterThan(zero)) { return y; } else { return x.newInstance(zero); } } if (x.lessThan(one) && y.classify() == Dfp.INFINITE) { if (y.greaterThan(zero)) { return x.newInstance(zero); } else { return x.newInstance(Dfp.copysign(y, one)); } } if (x.equals(one) && y.classify() == Dfp.INFINITE) { x.getField().setIEEEFlagsBits(DfpField.FLAG_INVALID); return x.dotrap(DfpField.FLAG_INVALID, POW_TRAP, x, x.newInstance((byte)1, Dfp.QNAN)); } if (x.classify() == Dfp.INFINITE) { // x = +/- inf if (invert) { // negative infinity if (y.classify() == Dfp.FINITE && y.rint().equals(y) && !y.remainder(two).equals(zero)) { // If y is odd integer if (y.greaterThan(zero)) { return x.newInstance(x.newInstance((byte)-1, Dfp.INFINITE)); } else { return x.newInstance(zero.negate()); } } else { // Y is not odd integer if (y.greaterThan(zero)) { return x.newInstance(x.newInstance((byte)1, Dfp.INFINITE)); } else { return x.newInstance(zero); } } } else { // positive infinity if (y.greaterThan(zero)) { return x; } else { return x.newInstance(zero); } } } if (invert && !y.rint().equals(y)) { x.getField().setIEEEFlagsBits(DfpField.FLAG_INVALID); return x.dotrap(DfpField.FLAG_INVALID, POW_TRAP, x, x.newInstance((byte)1, Dfp.QNAN)); } // End special cases Dfp r; if (y.lessThan(x.newInstance(100000000)) && y.greaterThan(x.newInstance(-100000000))) { final Dfp u = y.rint(); ui = u.intValue(); final Dfp v = y.subtract(u); if (v.unequal(zero)) { final Dfp a = v.multiply(log(x)); final Dfp b = a.divide(x.getField().getLn2()).rint(); final Dfp c = a.subtract(b.multiply(x.getField().getLn2())); r = splitPow(split(x), ui); r = r.multiply(pow(two, b.intValue())); r = r.multiply(exp(c)); } else { r = splitPow(split(x), ui); } } else { // very large exponent. |y| > 1e8 r = exp(log(x).multiply(y)); } if (invert && y.rint().equals(y) && !y.remainder(two).equals(zero)) { // if y is odd integer r = r.negate(); } return x.newInstance(r); } /** Computes sin(a) Used when 0 < a < pi/4. * Uses the classic Taylor series. x - x**3/3! + x**5/5! ... * @param a number from which sine is desired, in split form * @return sin(a) */ protected static Dfp sinInternal(Dfp a[]) { Dfp c = a[0].add(a[1]); Dfp y = c; c = c.multiply(c); Dfp x = y; Dfp fact = a[0].getOne(); Dfp py = new Dfp(y); for (int i = 3; i < 90; i += 2) { x = x.multiply(c); x = x.negate(); fact = fact.divide((i-1)*i); // 1 over fact y = y.add(x.multiply(fact)); if (y.equals(py)) { break; } py = new Dfp(y); } return y; } /** Computes cos(a) Used when 0 < a < pi/4. * Uses the classic Taylor series for cosine. 1 - x**2/2! + x**4/4! ... * @param a number from which cosine is desired, in split form * @return cos(a) */ protected static Dfp cosInternal(Dfp a[]) { final Dfp one = a[0].getOne(); Dfp x = one; Dfp y = one; Dfp c = a[0].add(a[1]); c = c.multiply(c); Dfp fact = one; Dfp py = new Dfp(y); for (int i = 2; i < 90; i += 2) { x = x.multiply(c); x = x.negate(); fact = fact.divide((i - 1) * i); // 1 over fact y = y.add(x.multiply(fact)); if (y.equals(py)) { break; } py = new Dfp(y); } return y; } /** computes the sine of the argument. * @param a number from which sine is desired * @return sin(a) */ public static Dfp sin(final Dfp a) { final Dfp pi = a.getField().getPi(); final Dfp zero = a.getField().getZero(); boolean neg = false; /* First reduce the argument to the range of +/- PI */ Dfp x = a.remainder(pi.multiply(2)); /* if x < 0 then apply identity sin(-x) = -sin(x) */ /* This puts x in the range 0 < x < PI */ if (x.lessThan(zero)) { x = x.negate(); neg = true; } /* Since sine(x) = sine(pi - x) we can reduce the range to * 0 < x < pi/2 */ if (x.greaterThan(pi.divide(2))) { x = pi.subtract(x); } Dfp y; if (x.lessThan(pi.divide(4))) { y = sinInternal(split(x)); } else { final Dfp c[] = new Dfp[2]; final Dfp[] piSplit = a.getField().getPiSplit(); c[0] = piSplit[0].divide(2).subtract(x); c[1] = piSplit[1].divide(2); y = cosInternal(c); } if (neg) { y = y.negate(); } return a.newInstance(y); } /** computes the cosine of the argument. * @param a number from which cosine is desired * @return cos(a) */ public static Dfp cos(Dfp a) { final Dfp pi = a.getField().getPi(); final Dfp zero = a.getField().getZero(); boolean neg = false; /* First reduce the argument to the range of +/- PI */ Dfp x = a.remainder(pi.multiply(2)); /* if x < 0 then apply identity cos(-x) = cos(x) */ /* This puts x in the range 0 < x < PI */ if (x.lessThan(zero)) { x = x.negate(); } /* Since cos(x) = -cos(pi - x) we can reduce the range to * 0 < x < pi/2 */ if (x.greaterThan(pi.divide(2))) { x = pi.subtract(x); neg = true; } Dfp y; if (x.lessThan(pi.divide(4))) { Dfp c[] = new Dfp[2]; c[0] = x; c[1] = zero; y = cosInternal(c); } else { final Dfp c[] = new Dfp[2]; final Dfp[] piSplit = a.getField().getPiSplit(); c[0] = piSplit[0].divide(2).subtract(x); c[1] = piSplit[1].divide(2); y = sinInternal(c); } if (neg) { y = y.negate(); } return a.newInstance(y); } /** computes the tangent of the argument. * @param a number from which tangent is desired * @return tan(a) */ public static Dfp tan(final Dfp a) { return sin(a).divide(cos(a)); } /** computes the arc-tangent of the argument. * @param a number from which arc-tangent is desired * @return atan(a) */ protected static Dfp atanInternal(final Dfp a) { Dfp y = new Dfp(a); Dfp x = new Dfp(y); Dfp py = new Dfp(y); for (int i = 3; i < 90; i += 2) { x = x.multiply(a); x = x.multiply(a); x = x.negate(); y = y.add(x.divide(i)); if (y.equals(py)) { break; } py = new Dfp(y); } return y; } /** computes the arc tangent of the argument * * Uses the typical taylor series * * but may reduce arguments using the following identity * tan(x+y) = (tan(x) + tan(y)) / (1 - tan(x)*tan(y)) * * since tan(PI/8) = sqrt(2)-1, * * atan(x) = atan( (x - sqrt(2) + 1) / (1+x*sqrt(2) - x) + PI/8.0 * @param a number from which arc-tangent is desired * @return atan(a) */ public static Dfp atan(final Dfp a) { final Dfp zero = a.getField().getZero(); final Dfp one = a.getField().getOne(); final Dfp[] sqr2Split = a.getField().getSqr2Split(); final Dfp[] piSplit = a.getField().getPiSplit(); boolean recp = false; boolean neg = false; boolean sub = false; final Dfp ty = sqr2Split[0].subtract(one).add(sqr2Split[1]); Dfp x = new Dfp(a); if (x.lessThan(zero)) { neg = true; x = x.negate(); } if (x.greaterThan(one)) { recp = true; x = one.divide(x); } if (x.greaterThan(ty)) { Dfp sty[] = new Dfp[2]; sub = true; sty[0] = sqr2Split[0].subtract(one); sty[1] = sqr2Split[1]; Dfp[] xs = split(x); Dfp[] ds = splitMult(xs, sty); ds[0] = ds[0].add(one); xs[0] = xs[0].subtract(sty[0]); xs[1] = xs[1].subtract(sty[1]); xs = splitDiv(xs, ds); x = xs[0].add(xs[1]); //x = x.subtract(ty).divide(dfp.one.add(x.multiply(ty))); } Dfp y = atanInternal(x); if (sub) { y = y.add(piSplit[0].divide(8)).add(piSplit[1].divide(8)); } if (recp) { y = piSplit[0].divide(2).subtract(y).add(piSplit[1].divide(2)); } if (neg) { y = y.negate(); } return a.newInstance(y); } /** computes the arc-sine of the argument. * @param a number from which arc-sine is desired * @return asin(a) */ public static Dfp asin(final Dfp a) { return atan(a.divide(a.getOne().subtract(a.multiply(a)).sqrt())); } /** computes the arc-cosine of the argument. * @param a number from which arc-cosine is desired * @return acos(a) */ public static Dfp acos(Dfp a) { Dfp result; boolean negative = false; if (a.lessThan(a.getZero())) { negative = true; } a = Dfp.copysign(a, a.getOne()); // absolute value result = atan(a.getOne().subtract(a.multiply(a)).sqrt().divide(a)); if (negative) { result = a.getField().getPi().subtract(result); } return a.newInstance(result); } }

Other Java examples (source code examples)

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my book on functional programming

 

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