Newton, Fixed Point and Secant Method

Newton Method

This method is based on the Taylor polynomial series. It uses the tangent line to find the next point ($x$) until its $f(x)$ approaches zero.

Given the \[ P(x_0) = \frac{f(x_0)}{0!} + \frac{f'(x_0)(x-x_0)}{1!} + \cdots\]
Equate it 0 and solve for $x$.
\[ \begin{split} \frac{f(x_0)}{0!} + \frac{f'(x_0)(x-x_0)}{1!} + \cdots &= 0\\
f(x_0) &= -f'(x_0)(x-x_0) \\
x-x_0 &= \frac{f(x_0)}{-f'(x_0)}\\
x&=x_0 - \frac{f(x_0)}{f'(x_0)} \end{split} \]

Example: As solved in the numericalmethods.xlsx

Given $x^2-10x+9$ Find a root of the equation using the starting points: 15,-1000 and 5.

We first solve for the derivative of  $x^2-10x+9$ which is $2x-10$. And then we look for the next $x$'s until its $f(x)$ is zero or we followed our stopping rules.

Problem arises if the first derivative of the function is 0 and we must have a good intuition at the starting point.

Secant Method: Assignment

Compute for the zeroes of $x^2-10x+9$ using the Secant Method. Show how you derive at the iterative process.

This method uses finite differences instead of differential calculus. We estimate for the first derivative as "rise over run" and there is a need for three points this time to secure succeeding estimate.

Fixed Point Method

We plug in $x=g(x)$.

Example:

Given $f(x)=x^2-10x+9$, we first find the $g(x)$ by equating $f(x)=0$. And then isolate $x$.

\[ \begin{split} x^2-10x+9 &=0\\
10x &= x^2+9\\
x&= \frac{x^2}{10}+\frac{9}{10} \end{split} \]

We don't know how many iteration it will end but end it when $x=g(x)$ or when they are approximately the same.