Newton-Raphson Method估算函数的根
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
#include
int main(void){
int a,error;
double x0,x1 = 1;
do {
printf("Input a positive integer: ");
scanf("%d",&a);
if (error = (a <=0))
printf("\nERROR: Do it again!\n\n");
}
while (error);
while (x0 != x1) {
x0 = x1; /* save the current value of x1 */
x1 = 0.5 * (x1 + a / x1); /* compute a new value of x1 */
}
printf("%lf\n",x1);
return 0;
}
R implemantation
2010-01-11 用R来实现一下,不单是求开方,估算函数的根。
newton <- function(fun, x0, tol=1e-7, niter=100) {
fun.list = as.list(fun)
var <- names(fun.list[1])
fun.exp = fun.list[[2]]
diff.fun = D(fun.exp, var)
df = list(x=0, diff.fun)
df = as.function(df)
for (i in 1:niter) {
x = x0 - fun(x0)/df(x0)
if (abs(fun(x)) < tol)
return(x)
x0 = x
}
stop("exceeded allowed number of iterations")
}
> f = function(x) x^2 – 5
> newton(f, 4)
[1] 2.236068
> g = function(x) x^3 – 5
> newton(g, 4)
[1] 1.709976