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Java LocalizedFormats.CANNOT_COMPUTE_NTH_ROOT_FOR_NEGATIVE_N属性代码示例

本文整理汇总了Java中org.apache.commons.math3.exception.util.LocalizedFormats.CANNOT_COMPUTE_NTH_ROOT_FOR_NEGATIVE_N属性的典型用法代码示例。如果您正苦于以下问题:Java LocalizedFormats.CANNOT_COMPUTE_NTH_ROOT_FOR_NEGATIVE_N属性的具体用法?Java LocalizedFormats.CANNOT_COMPUTE_NTH_ROOT_FOR_NEGATIVE_N怎么用?Java LocalizedFormats.CANNOT_COMPUTE_NTH_ROOT_FOR_NEGATIVE_N使用的例子?那么, 这里精选的属性代码示例或许可以为您提供帮助。您也可以进一步了解该属性所在org.apache.commons.math3.exception.util.LocalizedFormats的用法示例。


在下文中一共展示了LocalizedFormats.CANNOT_COMPUTE_NTH_ROOT_FOR_NEGATIVE_N属性的1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: nthRoot

/**
 * Computes the n-th roots of this complex number.
 * The nth roots are defined by the formula:
 * <pre>
 *  <code>
 *   z<sub>k</sub> = abs<sup>1/n</sup> (cos(phi + 2&pi;k/n) + i (sin(phi + 2&pi;k/n))
 *  </code>
 * </pre>
 * for <i>{@code k=0, 1, ..., n-1}</i>, where {@code abs} and {@code phi}
 * are respectively the {@link #abs() modulus} and
 * {@link #getArgument() argument} of this complex number.
 * <p>
 * If one or both parts of this complex number is NaN, a list with just
 * one element, {@link #NaN} is returned.
 * if neither part is NaN, but at least one part is infinite, the result
 * is a one-element list containing {@link #INF}.
 *
 * @param n Degree of root.
 * @return a List of all {@code n}-th roots of {@code this}.
 * @throws NotPositiveException if {@code n <= 0}.
 * @since 2.0
 */
public List<Complex> nthRoot(int n) throws NotPositiveException {

    if (n <= 0) {
        throw new NotPositiveException(LocalizedFormats.CANNOT_COMPUTE_NTH_ROOT_FOR_NEGATIVE_N,
                                       n);
    }

    final List<Complex> result = new ArrayList<Complex>();

    if (isNaN) {
        result.add(NaN);
        return result;
    }
    if (isInfinite()) {
        result.add(INF);
        return result;
    }

    // nth root of abs -- faster / more accurate to use a solver here?
    final double nthRootOfAbs = FastMath.pow(abs(), 1.0 / n);

    // Compute nth roots of complex number with k = 0, 1, ... n-1
    final double nthPhi = getArgument() / n;
    final double slice = 2 * FastMath.PI / n;
    double innerPart = nthPhi;
    for (int k = 0; k < n ; k++) {
        // inner part
        final double realPart = nthRootOfAbs *  FastMath.cos(innerPart);
        final double imaginaryPart = nthRootOfAbs *  FastMath.sin(innerPart);
        result.add(createComplex(realPart, imaginaryPart));
        innerPart += slice;
    }

    return result;
}
 
开发者ID:biocompibens,项目名称:SME,代码行数:57,代码来源:Complex.java


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