can anyone shed some light into this for me.
I am asked to numerically solve $$x\arctan(x) = 1$$ and to ensure that the error $\epsilon$ is less than $10^{-3}$. Now I usually use Newton's method for this and I have seen the following formula$$n \leq \frac{1}{\log(2)} \cdot \log \left( \frac{\log(\epsilon)}{\log(|p_0 - p|)}\right)$$ where $n$ is the number of iterations. It also says that $p_0$ is our initial guess and $p$ is the actual root. Now how would one go about this if they did not know the exact root? How can we use this formula?
I'm not sure if this is what you were looking for.
Let $f(x) = x \arctan x -1$, note that $f$ is even, $f(0) = -1$ and $f$ is increasing and unbounded on $x \ge 0$. In particular, a unique solution exists on $x \ge 0$.
Since you are only concerned about finding a point $x^*$ such that $|f(x^*)| < {1 \over 10^3}$, there is a particularly simple & fast strategy.
In the following I am assuming that $x \ge 0$ and $x^*$ is the unique solution.
Note that $f'(x) \ge 0$, $f''(x) >0$ (hence $f'$ is strictly increasing) and $f'(0) = 0$. In particular, $f$ is convex.
Since $f$ is convex, if we start with any $x_0 >0$ (so that the iteration is defined) and use Newton's method $x_{n_1} = x_n - {f(x_n) \over f'(x_n)}$, then $x_1 \ge x^*$ and subsequently $x^* \le x_{n+1} \le x_n$.
In particular, as long as we start with $x_0 >0$, the iteration is well defined and $x_n \to x^*$. Since $f'(x^*) >0$, the convergence is quadratic.
So, just iterate until $f(x_n) < {1 \over 10^3}$.
If I pick $x_0 = 1$ then $x_1 \approx 1.670$, $x_2 \approx 1.623$ and $f(x_2) < 2 \times 10^{-12}$.
More detail:
A brief analysis shows that $|x_{n+1}-x^*| \le {1 \over 2} |{f''(\xi_n) \over f'(x_n)}| |x_n-x^*|^2$, where $\xi_n \in [x^*,x_{n+1}]$.
We can show that $f''(x_n) \in (0,1]$ for all $x$ and $f'(x_n) \ge f'(x_0) \ge 1$, so we have $|x_{n+1}-x^*| \le {1 \over 2} |x_n-x^*|^2$.
Since $f(1)<0$ and $f(2)>0$ we see that $|x_0-x^*| \le 1$ and a little work shows that $|x_n-x^*| \le {1 \over 2}^{{1 \over 2} n (n+1)}$. Choosing $n=5$ gives $ {1 \over 2}^{{1 \over 2} n (n+1)} < {1 \over 10^4}$, so five iterations with exact numerics will guarantee that we are close to the solution and since $f'(x) \le 2$ for $x \le 2$ we know that $|f(x_5)| = |f(x_5)-f(x^*)| \le 2|x_5-x^*| \le 2 {1 \over 10^4}$ which is adequate.