/**
* Returns {@code true} if there is no double value strictly between the arguments or the reltaive
* difference between them is smaller or equal to the given tolerance.
*
* @param x First value.
* @param y Second value.
* @param eps Amount of allowed relative error.
* @return {@code true} if the values are two adjacent floating point numbers or they are within
* range of each other.
* @since 3.1
*/
public static boolean equalsWithRelativeTolerance(double x, double y, double eps) {
if (equals(x, y, 1)) {
return true;
}
final double absoluteMax = FastMath.max(FastMath.abs(x), FastMath.abs(y));
final double relativeDifference = FastMath.abs((x - y) / absoluteMax);
return relativeDifference <= eps;
}
/**
* Evaluates the continued fraction at the value x.
*
* <p>The implementation of this method is based on equations 14-17 of:
*
* <ul>
* <li>Eric W. Weisstein. "Continued Fraction." From MathWorld--A Wolfram Web Resource. <a
* target="_blank" href="http://mathworld.wolfram.com/ContinuedFraction.html">
* http://mathworld.wolfram.com/ContinuedFraction.html</a>
* </ul>
*
* The recurrence relationship defined in those equations can result in very large intermediate
* results which can result in numerical overflow. As a means to combat these overflow conditions,
* the intermediate results are scaled whenever they threaten to become numerically unstable.
*
* @param x the evaluation point.
* @param epsilon maximum error allowed.
* @param maxIterations maximum number of convergents
* @return the value of the continued fraction evaluated at x.
* @throws ConvergenceException if the algorithm fails to converge.
*/
public double evaluate(double x, double epsilon, int maxIterations) {
double p0 = 1.0;
double p1 = getA(0, x);
double q0 = 0.0;
double q1 = 1.0;
double c = p1 / q1;
int n = 0;
double relativeError = Double.MAX_VALUE;
while (n < maxIterations && relativeError > epsilon) {
++n;
double a = getA(n, x);
double b = getB(n, x);
double p2 = a * p1 + b * p0;
double q2 = a * q1 + b * q0;
boolean infinite = false;
if (Double.isInfinite(p2) || Double.isInfinite(q2)) {
/*
* Need to scale. Try successive powers of the larger of a or b
* up to 5th power. Throw ConvergenceException if one or both
* of p2, q2 still overflow.
*/
double scaleFactor = 1d;
double lastScaleFactor = 1d;
final int maxPower = 5;
final double scale = FastMath.max(a, b);
if (scale <= 0) { // Can't scale
throw new ConvergenceException(
LocalizedFormats.CONTINUED_FRACTION_INFINITY_DIVERGENCE, x);
}
infinite = true;
for (int i = 0; i < maxPower; i++) {
lastScaleFactor = scaleFactor;
scaleFactor *= scale;
if (a != 0.0 && a > b) {
p2 = p1 / lastScaleFactor + (b / scaleFactor * p0);
q2 = q1 / lastScaleFactor + (b / scaleFactor * q0);
} else if (b != 0) {
p2 = (a / scaleFactor * p1) + p0 / lastScaleFactor;
q2 = (a / scaleFactor * q1) + q0 / lastScaleFactor;
}
infinite = Double.isInfinite(p2) || Double.isInfinite(q2);
if (!infinite) {
break;
}
}
}
if (infinite) {
// Scaling failed
throw new ConvergenceException(LocalizedFormats.CONTINUED_FRACTION_INFINITY_DIVERGENCE, x);
}
double r = p2 / q2;
if (Double.isNaN(r)) {
throw new ConvergenceException(LocalizedFormats.CONTINUED_FRACTION_NAN_DIVERGENCE, x);
}
relativeError = FastMath.abs(r / c - 1.0);
// prepare for next iteration
c = p2 / q2;
p0 = p1;
p1 = p2;
q0 = q1;
q1 = q2;
}
if (n >= maxIterations) {
throw new MaxCountExceededException(
LocalizedFormats.NON_CONVERGENT_CONTINUED_FRACTION, maxIterations, x);
}
return c;
}