本文整理汇总了Java中org.apache.commons.math3.util.FastMath.acosh方法的典型用法代码示例。如果您正苦于以下问题：Java FastMath.acosh方法的具体用法？Java FastMath.acosh怎么用？Java FastMath.acosh使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类org.apache.commons.math3.util.FastMath的用法示例。
在下文中一共展示了FastMath.acosh方法的4个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Java代码示例。
示例1: compute
import org.apache.commons.math3.util.FastMath; //导入方法依赖的package包/类
@Override
protected double compute(double value) {
return Double.isNaN(value) ? Double.NaN : FastMath.acosh(value);
}
示例2: value
import org.apache.commons.math3.util.FastMath; //导入方法依赖的package包/类
/** {@inheritDoc} */
public double value(double x) {
return FastMath.acosh(x);
}
示例3: acosh
import org.apache.commons.math3.util.FastMath; //导入方法依赖的package包/类
/** {@inheritDoc} */
public SparseGradient acosh() {
return new SparseGradient(FastMath.acosh(value), 1.0 / FastMath.sqrt(value * value - 1.0), derivatives);
}
示例4: acosh
import org.apache.commons.math3.util.FastMath; //导入方法依赖的package包/类
/** Compute inverse hyperbolic cosine of a derivative structure.
* @param operand array holding the operand
* @param operandOffset offset of the operand in its array
* @param result array where result must be stored (for
* inverse hyperbolic cosine the result array <em>cannot</em> be the input
* array)
* @param resultOffset offset of the result in its array
*/
public void acosh(final double[] operand, final int operandOffset,
final double[] result, final int resultOffset) {
// create the function value and derivatives
double[] function = new double[1 + order];
final double x = operand[operandOffset];
function[0] = FastMath.acosh(x);
if (order > 0) {
// the nth order derivative of acosh has the form:
// dn(acosh(x)/dxn = P_n(x) / [x^2 - 1]^((2n-1)/2)
// where P_n(x) is a degree n-1 polynomial with same parity as n-1
// P_1(x) = 1, P_2(x) = -x, P_3(x) = 2x^2 + 1 ...
// the general recurrence relation for P_n is:
// P_n(x) = (x^2-1) P_(n-1)'(x) - (2n-3) x P_(n-1)(x)
// as per polynomial parity, we can store coefficients of both P_(n-1) and P_n in the same array
final double[] p = new double[order];
p[0] = 1;
final double x2 = x * x;
final double f = 1.0 / (x2 - 1);
double coeff = FastMath.sqrt(f);
function[1] = coeff * p[0];
for (int n = 2; n <= order; ++n) {
// update and evaluate polynomial P_n(x)
double v = 0;
p[n - 1] = (1 - n) * p[n - 2];
for (int k = n - 1; k >= 0; k -= 2) {
v = v * x2 + p[k];
if (k > 2) {
p[k - 2] = (1 - k) * p[k - 1] + (k - 2 * n) * p[k - 3];
} else if (k == 2) {
p[0] = -p[1];
}
}
if ((n & 0x1) == 0) {
v *= x;
}
coeff *= f;
function[n] = coeff * v;
}
}
// apply function composition
compose(operand, operandOffset, function, result, resultOffset);
}