Define $x_0=0$ and $x_{n+1}=x_n-[(\tan x_n-1)/\sec^2x_n]$. What is $\lim_{n\to \infty}x_n$ in this example? Relate this to Newton's method.

Question

Define $x_0=0$ and $x_{n+1}=x_n-[(\tan x_n-1)/\sec^2x_n]$. What is $\lim_{n\to \infty}x_n$ in this example? Relate this to Newton's method.

I know that in this case I am trying to find the zeros of the $f(x)=\tan x-1$ function according to Newton's method but I do not know how to calculate this limit, could someone help me please? Thank you very much.

Answer 1

Here $f(x)=\tan{x}-1$ is zero if $\tan{x}=1$, from where we have $x=\frac{\pi}{4}$. That is limit you wanted.