Examples of org.apache.commons.math3.fraction.BigFraction.compareTo()

Package org.apache.commons.math3.fraction

Class org.apache.commons.math3.fraction.BigFraction

Examples of org.apache.commons.math3.fraction.BigFraction.compareTo()

org.apache.commons.math3.fraction.BigFraction.compareTo()

Compares this object to another based on size.
@param object the object to compare to, must not be null. @return -1 if this is less than {@code object}, +1 if this is greater than {@code object}, 0 if they are equal. @see java.lang.Comparable#compareTo(java.lang.Object)

        /*
         * [1] states: "For 1/2 < h < 1 the bottom left element of the matrix
         * should be (1 - 2*h^m + (2h - 1)^m )/m!" Since 0 <= h < 1, then if h >
         * 1/2 is sufficient to check:
         */
        if (h.compareTo(BigFraction.ONE_HALF) == 1) {
            Hdata[m - 1][0] = Hdata[m - 1][0].add(h.multiply(2).subtract(1).pow(m));
        }


        /*
         * Aside from the first column and last row, the (i, j)-th element is

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        /*
         * [1] states: "For 1/2 < h < 1 the bottom left element of the matrix should be (1 - 2*h^m +
         * (2h - 1)^m )/m!" Since 0 <= h < 1, then if h > 1/2 is sufficient to check:
         */
        if (h.compareTo(BigFraction.ONE_HALF) == 1) {
            Hdata[m - 1][0] = Hdata[m - 1][0].add(h.multiply(2).subtract(1).pow(m));
        }


        /*
         * Aside from the first column and last row, the (i, j)-th element is 1/(i - j + 1)! if i -

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        /*
         * [1] states: "For 1/2 < h < 1 the bottom left element of the matrix
         * should be (1 - 2*h^m + (2h - 1)^m )/m!" Since 0 <= h < 1, then if h >
         * 1/2 is sufficient to check:
         */
        if (h.compareTo(BigFraction.ONE_HALF) == 1) {
            Hdata[m - 1][0] = Hdata[m - 1][0].add(h.multiply(2).subtract(1).pow(m));
        }


        /*
         * Aside from the first column and last row, the (i, j)-th element is

View Full Code Here


        /*
         * [1] states: "For 1/2 < h < 1 the bottom left element of the matrix should be (1 - 2*h^m +
         * (2h - 1)^m )/m!" Since 0 <= h < 1, then if h > 1/2 is sufficient to check:
         */
        if (h.compareTo(BigFraction.ONE_HALF) == 1) {
            Hdata[m - 1][0] = Hdata[m - 1][0].add(h.multiply(2).subtract(1).pow(m));
        }


        /*
         * Aside from the first column and last row, the (i, j)-th element is 1/(i - j + 1)! if i -

View Full Code Here

        /*
         * [1] states: "For 1/2 < h < 1 the bottom left element of the matrix
         * should be (1 - 2*h^m + (2h - 1)^m )/m!" Since 0 <= h < 1, then if h >
         * 1/2 is sufficient to check:
         */
        if (h.compareTo(BigFraction.ONE_HALF) == 1) {
            Hdata[m - 1][0] = Hdata[m - 1][0].add(h.multiply(2).subtract(1).pow(m));
        }


        /*
         * Aside from the first column and last row, the (i, j)-th element is

View Full Code Here

        /*
         * [1] states: "For 1/2 < h < 1 the bottom left element of the matrix
         * should be (1 - 2*h^m + (2h - 1)^m )/m!" Since 0 <= h < 1, then if h >
         * 1/2 is sufficient to check:
         */
        if (h.compareTo(BigFraction.ONE_HALF) == 1) {
            Hdata[m - 1][0] = Hdata[m - 1][0].add(h.multiply(2).subtract(1).pow(m));
        }


        /*
         * Aside from the first column and last row, the (i, j)-th element is

View Full Code Here

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