I am having difficulty understanding the Taylor expansion and solving for $f(x) = 0$
Using evaluations of $f(x_n)$, $f'(x_n)$, and $f''(x_n)$. The following remarks may be helpful in constructing the algorithm"
$\quad \quad \bullet$ Use the Taylor expansion with three terms plus a remainder term.
$\quad \quad \bullet$ Show that in the course of derivation a quadratic equation arises, and therefor two distainct schemes can be derived.
Here is the question exactly.
a. Another method for solving $f(x) = 0$ is to consider the third-order approximation of $f$ around the point $x_n$:
b. Show that the order of convergence (under the appropriate conditions ) is cubic.
From my professors notes he says,
for some point $\xi$ between $x^*$ and $x_n$. If $x_n$ is already fairly close to $x^*$, then $(x^* - x_n)^2$ will be very small, so we have
$$0 \approx f(x_k) + f'(x_k)(x^* - x_k).$$
But that was when he only had the Taylor Series Theorem at second order I believe, so I am supposed to solve it for third order.
Directly from his notes and using his logic about $(x^*-x_n)^2$ being too small so I figured that if that cancels out when squared then perhaps it does when it is cubed.
Here is my attempt at solving it applying my profs notes.
$$ f(x^*) = f(x_n) + f'(x_n)(x^* - x_n) + \frac{f''(x_n)}{2}(x^* - x_n)^2 + \frac{f'''(\xi)}{6}(x^* - x_n)^3 $$
$$0 \approx f(x_n) + f'(x_n)(x^* - x_n) + \frac{f''(x_n)}{2}(x^* - x_n)^2.$$
Right here is where the quadratic arises
Solving for $x^*$, we obtain
$$ 2 [ -f(x_n) - f'(x_n)(x^* - x_n) ] \approx f''(x_n)(x^* - x_n)^2$$
$$ 2 \frac{-f(x_n) - f'(x_n)(x^* - x_n)}{f''(x_n)} \approx (x^* - x_n)^2$$
Which is where I get stuck so I don't know how to approac the problem further. Any help would be appreciated.