# Third Order Taylor Expansion

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.

## 1 Answer

You have to solve $0=a+bs+cs^2+O(s^3)$. You could apply the quadratic solution formula and obtain the original method that Halley used, or you could apply some binomial formula tricks to reduce the equation to a linear equation by multiplying with $(b-cs)$ to get $$0=a(b-cs)+(b^2-c^2s^2)s+O(s^3)=ab+(b^2-ac)s+O(s^3)$$ leading to the Halley method in its hyperbolic form as it is usually known today $$(x^*-x_n)=s\approx -\frac{ab}{b^2-ac}=-\frac{f(x_n)f'(x_n)}{f'(x_n)^2-\frac12f''(x_n)f(x_n)}.$$