Newton-Raphson Method

曲线f(x)有根c,取曲线上一点$(x_1,f(x_1))$, 过此点的切线交x轴$x_2$,过曲线上$(x_2,f(x_2))$的切线交x轴$x_3$,如此反复得到一个序列 $x_1,x_2,\cdot \cdot \cdot,x_n$ 逼近c值.

过$(x_n,f(x_n))$的切线方程为 $y-f(x_n) = f’(x_n)\,(x-xn)$,假设此方程与x轴的交点为$x{n+1}$, 即有: $0 - f(x_n) = f’(x_n)(x_n+1 - xn)$, 即$x{n+1} = x_n - \frac{f(x_n)}{f’(x_n)}$ <Eq. 1>.

下面利用此法来求一个数的开方。 $f(x) = x^2 - a$ 有根$\sqrt{a}$, 由$f’(x_n) = 2xn$, 代入式<Eq. 1>可得$x{n+1} = (x_n + a/x_n)/2$; 当i -> INF 时, $x_i$ -> $\sqrt{a}$;

C implementation

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Guangchuang Yu

a senior-in-age-but-not-senior-in-knowledge bioinformatician

Postdoc researcher

Hong Kong