Example 1.  Consider the function  , which has a root at  .
1 (b).  Use Halley's formula to find the root.      Use the starting value

Solution 1 (b).

Now we will investigate Halley's iteration for finding square roots.

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Form the Halley iteration function  h(x).

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We start the iteration with    and carry 100 digits in the computations, by telling Mathematica the precision of by issuing the command p[0] = N[2,100].  Next, a short program is written to compute the first five terms in the iteration:

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Since the root is known to be exactly     we can have Mathematica list the error    at each step in the iteration:

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Looking at the error, we see that the number of accurate digits is tripling at each step in the computations, hence convergence is proceeding cubically.
We can conclude that Halley's method is faster than Newton's method.

Verify the convergence rate.  At the simple root    we can explore the ratio .

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Therefore, the Halley iteration is converging cubically.

(c) John H. Mathews 2004