# f(x) = 0인 x를 찾기
>>> newton = function(f, tol = 1e-7, x0 = 1, N = 100){
>>>     h = 1e-7;
>>>     i = 1
>>>     x1 = x0;
>>>     p = numeric(N)
>>>     while(i <= N) {
>>>         df.dx = (f(x0 + h)-f(x0)) / h
>>>         x1 = (x0 - (f(x0) / df.dx))
>>>         p[i] = x1
>>>         i = i + 1
>>>         if(abs(x1 - x0) < tol) break
>>>         x0 = x1
>>>     }
>>>     return(p[1 : (i - 1)])
>>> }

# 함수 적용
>>> f <- function(x) (x^2 - 2)   # x2-2 함수에 적용
>>> newton(f)
1.500000 1.416667 1.414216 1.414214 1.414214
>>> f <- function(x) (x^3 + 3 * x^2 - 6 * x - 8)   # x3+3x2-6x-8 함수에 적용
>>> newton(f)
4.333333 2.974769 2.270963 2.030331 2.000449 2.000000 2.000000 2.000000