本文整理匯總了Java中org.apache.commons.math3.util.FastMath.atanh方法的典型用法代碼示例。如果您正苦於以下問題：Java FastMath.atanh方法的具體用法？Java FastMath.atanh怎麽用？Java FastMath.atanh使用的例子？那麽, 這裏精選的方法代碼示例或許可以為您提供幫助。您也可以進一步了解該方法所在類org.apache.commons.math3.util.FastMath的用法示例。
在下文中一共展示了FastMath.atanh方法的4個代碼示例，這些例子默認根據受歡迎程度排序。您可以為喜歡或者感覺有用的代碼點讚，您的評價將有助於係統推薦出更棒的Java代碼示例。
示例1: compute
import org.apache.commons.math3.util.FastMath; //導入方法依賴的package包/類
@Override
protected double compute(double value) {
return Double.isNaN(value) ? Double.NaN : FastMath.atanh(value);
}
示例2: value
import org.apache.commons.math3.util.FastMath; //導入方法依賴的package包/類
/** {@inheritDoc} */
public double value(double x) {
return FastMath.atanh(x);
}
示例3: atanh
import org.apache.commons.math3.util.FastMath; //導入方法依賴的package包/類
/** {@inheritDoc} */
public SparseGradient atanh() {
return new SparseGradient(FastMath.atanh(value), 1.0 / (1.0 - value * value), derivatives);
}
示例4: atanh
import org.apache.commons.math3.util.FastMath; //導入方法依賴的package包/類
/** Compute inverse hyperbolic tangent 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 tangent the result array <em>cannot</em> be the input
* array)
* @param resultOffset offset of the result in its array
*/
public void atanh(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.atanh(x);
if (order > 0) {
// the nth order derivative of atanh has the form:
// dn(atanh(x)/dxn = Q_n(x) / (1 - x^2)^n
// where Q_n(x) is a degree n-1 polynomial with same parity as n-1
// Q_1(x) = 1, Q_2(x) = 2x, Q_3(x) = 6x^2 + 2 ...
// the general recurrence relation for Q_n is:
// Q_n(x) = (1-x^2) Q_(n-1)'(x) + 2(n-1) x Q_(n-1)(x)
// as per polynomial parity, we can store coefficients of both Q_(n-1) and Q_n in the same array
final double[] q = new double[order];
q[0] = 1;
final double x2 = x * x;
final double f = 1.0 / (1 - x2);
double coeff = f;
function[1] = coeff * q[0];
for (int n = 2; n <= order; ++n) {
// update and evaluate polynomial Q_n(x)
double v = 0;
q[n - 1] = n * q[n - 2];
for (int k = n - 1; k >= 0; k -= 2) {
v = v * x2 + q[k];
if (k > 2) {
q[k - 2] = (k - 1) * q[k - 1] + (2 * n - k + 1) * q[k - 3];
} else if (k == 2) {
q[0] = q[1];
}
}
if ((n & 0x1) == 0) {
v *= x;
}
coeff *= f;
function[n] = coeff * v;
}
}
// apply function composition
compose(operand, operandOffset, function, result, resultOffset);
}