StackOverflow2013

Note that there are some explanatory texts on larger screens.

plurals

PO
text
Body
copied!<p><a href="http://books.google.com.au/books?id=h0d8hVA5HyQC&dq=books+Java+Number+Cruncher:+The+Java+Programmer%27s+Guide+to+Numerical+Computing&printsec=frontcover&source=bn&hl=en&ei=MOLjSaPQBdaSkAX4w-nbCw&sa=X&oi=book_result&ct=result&resnum=4" rel="nofollow noreferrer">Java Number Cruncher: The Java Programmer's Guide to Numerical Computing</a> provides a solution using <a href="http://en.wikipedia.org/wiki/Newton%27s_method" rel="nofollow noreferrer">Newton's Method</a>. Source code from the book is available <a href="http://www.apropos-logic.com/nc/download.html" rel="nofollow noreferrer">here</a>. The following has been taken from chapter <em>12.5 Big Decmial Functions</em> (p330 & p331):</p> <pre><code>/** * 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 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; } /** * 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); } </code></pre>

Querying!

Guidance

An individual column

Larger individual text columns get their own page to allow for proper reading.

SQuiL has stopped working due to an internal error.

If you are curious you may find further information in the browser console, which is accessible through the devtools (F12).

Reload