| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Which functions f(x) have a polynomial 'Newton's iteration'?
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1810.1 | AUSSIE::GARSON | Hotel Garson: No Vacancies | Tue Oct 26 1993 18:20 | 7 | |
Two examples which spring to mind are
f(x) = 1/(Ax+B)
f(x) = exp(Ax+B)
but of course Newton's method does not converge for either of these
(and they have no zero).
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| 1810.2 | 3D::ROTH | Geometry is the real life! | Tue Oct 26 1993 22:03 | 23 | |
Since the iteration function is x - f/f' it suggests looking
at the logarithmic derivative of f.
For example, let
d log(F) F' 1 A1 A2 A3
-------- = --- = ----- = ------ + ------ + ------
dx F P(x) x - a1 x - a2 x - a3
express the logarithmic derivative of an F in terms of
the reciprocal of a cubic polynomial expanded as in partial
fractions.
We require the A's and a's give a numerator of 1.
Then integrating
log(F) = log( (x - a1)^A1 (x - a2)^A2 (x - a3)^A3 ) + C
P could have repeated roots and so on, but this seems like
an idea that would work.
- Jim
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| 1810.3 | nibbles | AUSSIE::GARSON | Hotel Garson: No Vacancies | Wed Oct 27 1993 01:00 | 9 |
additional to .1 (me)
The first function given should have been generalised to
f(x) = (Ax+B)^n where n is an integer.
For f(x) a quadratic it can be shown that f must be of the form
A(x+B)�.
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