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1D Newton's search

Suppose we want to find the root of $ f(x)=x^2-3$ (see Figure 13.4). If our current estimate of the answer is $ x_k=2$, we can get a better estimate $ x_{k+1}$ by drawing the line that is tangent to $ f(x)$ at $ (2,f(2))=(2,1)$, and find the point $ x_{k+1}$ where this line crosses the x axis. Since,

$\displaystyle x_{k+1}=x_k- \Delta x, $


$\displaystyle f'(x_k)= \frac{\Delta y}{\Delta
x}= \frac{f(x_k)}{\Delta x}, $

we have that

$\displaystyle f'(x_k) \Delta x=$   F$\displaystyle (x_k)$


$\displaystyle x_{k+1}= x_k - \frac{f(x_k)}{f'(x_k)}$ (13.30)

which gives $ x_{k+1}=2-\frac{1}{4}=1.75$. We apply the same process and iterate on $ k$.

Figure 13.4: A plot of $ \psi (\lambda )$ for $ H$ indefinite.
\centering\epsfig{figure=figures/1DNewton.eps, width=14cm,

Frank Vanden Berghen 2004-04-19