Day 7, Mat360: Numerical Analysis

We discovered that other values of \(a\) couldn't solve the problem, which is that \(f'(x) \approx 1\) at values of \(x\) near 1, and that's a problem for fixed point iteration.

• FPI is linear, unless \(f'(\hat{x})=0\) at the fixed point \(\hat{x}\).

  Proposition 5: Let $f$ be a differentiable function with fixed point $\hat{x}$ and let $[a,b]$ be an interval containing $\hat{x}$. If $|f'(x)|\leq M<1$ for all $x\in[a,b]$ and $f([a,b])\subseteq[a,b]$, then for any initial value $x_{0}\in[a,b]$, fixed point iteration, with $x_{k+1}=f(x_{k})$ for all $k\geq0$, gives an approximation of $\hat{x}$ with absolute error no more than $M^{k}|x_{0}-\hat{x}|$.

• FPI is quadratic if \(f'(\hat{x})=0\) at the fixed point \(\hat{x}\).
• Even though FPI may be linear, we can accelerate it with just a little cleverness (Aitken's)!

  \[ p\approx p_{n}-\frac{(p_{n+1}-p_{n})^{2}}{p_{n+2}-2p_{n+1}+p_{n}} \]

  We'll derive the method in two different ways.

• Aitken's delta-squared method uses the starting value and two successive iteratest to generate an improved estimate. Steffensen's strategy was then to use that improved estimate as the next starting value, and then do it again: \[ \begin{array}{c} {x_1=f(x_0)}\cr {x_2=f(x_1)}\cr {\hat{x}\approx a = x_2 - \frac{(x_2-x_1)^2}{x_0-2x_1+x_2}}\cr {x_0=a} \end{array} \] and iterate....