Checking to see if is correct

Question

Suppose Newton’s method is applied to the function f(x) = 1/x. If the initial guess is x0 = 1, find x50.

Solution F(x) =1/x x0=1 where n=1,2,3,... F'(x) =-1/(x^2) Xn+1=xn-(f(xn))/(f'(xn)) And I had x50=1125899907*10^6

Answer 1

I get that $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{\frac{1}{x_n}}{-\frac{1}{x_n^2}} = x_n + x_n = 2x_n,$$ thus $x_n = 2^n$, given $x_0 = 1$. According to Google, we have $$x_{50} = 2^{50} \approx 1.1258999 \cdot 10^{15},$$ which appears roughly the same as what you've got.