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The convergence rate

We can now analyze which of the particular choices of $\gamma$ is more appropriate as far as the convergence rate is concerned.

If we consider the general form of the square root iteration

\begin{displaymath}x_{n+1}=\frac{s+x_{n}\gamma}{x_n+\gamma}\end{displaymath}

we can estimate the convergence rate by the difference between the actual estimation at step $(n+1)$ and the analytical value $\sqrt{s}$. For the general case, we obtain

\begin{displaymath}x_{n+1}-\sqrt{s} =
\frac{s+\gamma x_n -x_n \sqrt{s}-\gamma \sqrt{s}}{x_n+\gamma}\end{displaymath}

or
\begin{displaymath}
\fbox{$ \displaystyle
x_{n+1}-\sqrt{s} =
\frac{(x_n-\sqrt{s})(\gamma-\sqrt{s})}{x_n+\gamma}
$} \end{displaymath} (7)

sqroot
Figure 2.
Convergence plots for different recursive algorithms, shown in Table 1.
sqroot
[pdf] [png] [matlab]

The possible selections for $\gamma$ from Table 1 clearly show that the recursions described in the preceding subsection generally have a linear convergence rate (that is, the error at step $n+1$ is proportional to the error at step $n$), but can converge quadratically for an appropriate selection of the parameter $\gamma$, as shown in Table 2. Furthermore, the convergence is faster when $\gamma$ is closer to $\sqrt{s}$.


Table 2: Convergence rate
  $\gamma$ Convergence
Muir $1$ linear
Secant $x_{n-1}$ quasi-quadratic
Newton $x_n$ quadratic

We therefore conclude that Newton's iteration has the potential to achieve the fastest convergence rate. Ideally, however, we could use a fixed $\gamma$ which is a good approximation to the square root. The convergence would then be slightly faster than for the Newton-Raphson method, as shown in Figure 2.


next up previous [pdf]

Next: Spectral factorization Up: The square root of Previous: Root-finding recursions

2013-03-03