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(15C) Halley's Method
02-26-2022, 03:48 PM (This post was last modified: 02-26-2022 04:42 PM by Albert Chan.)
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RE: (15C) Halley's Method
(02-26-2022 11:09 AM)Thomas Klemm Wrote:  We start with Halley's Irrational Formula:

\(C_f(x)=x-\frac{2f(x)}{f{'}(x)+\sqrt{[f{'}(x)]^2-2f(x)f{''}(x)}}\)

However, we reduce the fraction with \(f{'}(x)\) to get:

\(C_f(x)=x-\frac{\frac{2f(x)}{f{'}(x)}}{1 + \sqrt{1 - \frac{2f(x)f{''}(x)}{[f{'}(x)]^2}}}\)

This also avoid cancellation if Re(f'(x)) < 0, since Re(√z) ≥ 0

Comment: two Cf expressions are not the same.
First form denominator + is really ± (2 roots of quadratic)
Second form pick the smaller correction, exactly what we wanted.
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Messages In This Thread
(15C) Halley's Method - Thomas Klemm - 02-26-2022, 11:09 AM
RE: (15C) Halley's Method - Albert Chan - 02-26-2022 03:48 PM
RE: (15C) Halley's Method - Albert Chan - 06-11-2022, 04:12 PM
RE: (15C) Halley's Method - Albert Chan - 06-14-2022, 06:22 PM
RE: (15C) Halley's Method - Gil - 06-14-2022, 01:20 AM
RE: (15C) Halley's Method - Albert Chan - 06-14-2022, 02:59 AM
RE: (15C) Halley's Method - Albert Chan - 06-18-2022, 03:37 PM
RE: (15C) Halley's Method - Gil - 06-14-2022, 07:33 AM



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