Re: Unstable Simultaneous Linear Equations!

On Apr 3, 5:06 pm, The Phantom <phan...@xxxxxxx> wrote:
On Thu, 3 Apr 2008 07:10:08 -0700 (PDT), monir <mon...@xxxxxxxxxxxx> wrote:

<SNIP>





Rodger;

Okay.
Given: (exact 7 significant digits)
..........x.........y........d=(dy/dx)
P1:: -0.0266255  0.9981449   -0.1308565
P2:: -0.0248588  0.9847863  132.5335679

what poly coeff A, B, C, D do you get for:
y = A + B*x +C*x^2 + D*x^3

Upon receipt, I'll provide you with an Excel plot showing the
comparison and how to construct the correct elliptic interpolation
curve.

Regards.
Monir

I calculated the coefficients in Mathematica with 50 digit arithmetic.  Other
high precision arithmetic packages could be used, such as Matlab, Maple, Scilab,
etc.  Here is a link to an image showing the whole thing, including a plot of
the cubic passing through the endpoints plus the elliptic function you gave in
another post:

http://img2.freeimagehosting.net/image.php?8ada8a8e27.gif

The coefficients I got are:

A = 830.9589672389913
B = 95 850.51921961399
C = 3 687 666.010250186
D = 47 265 155.70059824

I think Excel won't be able to get this result even using double precision
arithmetic, and probably won't even be able to accurately evaluate the cubic and
the derivatives with double precision.

The results and calculations shown in the Mathematica result are quite accurate,
and can be relied on to be correct, for the starting values:

Given: (exact 7 significant digits)
..........x.........y........d=(dy/dx)
P1:: -0.0266255  0.9981449   -0.1308565
P2:: -0.0248588  0.9847863  132.5335679

This all shows that extreme care and high precision arithmetic is required to
get a good result for the cubic because the condition number for the system is
very high.  The cubic is not a good function for the interpolation, but it can
be accurately calculated with high precision arithmetic.- Hide quoted text -

- Show quoted text -

Rodger;

Thank you for the info. and for your effort.

1) I've applied your latest poly. coeffs. in Excel for comparison
purposes (file HERMITE-JCH-Dave-Rodger-Monir-Interpol-6.xls, just
emailed to you).

2) It might surprise you that your cubic poly calculations using DP
arithmetic agree extremely well with the Hermite method.
The difficulty, however and as demonstrated earlier with general cubic
poly interpolation, is the unavoidable presence of extremum and
inflection points within the x-range-of-interest, which is clearly
undesirable!

3) On the other hand, the elliptic interpolation curve appears to
satisfy all the conditions and requirements. It's critical, however,
that one applies the correct scenario in constructing the applicable
curve segment of the ellipse. The 4 possible scenarios were described
earlier (my post April 1, 12:20 am local).
The applicable scenario for the provided example is scen2

4) As a footnote, Dave Dodson has insisted that in the elliptic
interpolation procedure a "cusp" exists where the ellipse joins the
(analytical) curve at point P2.
Dave's observation suggests that although the elliptic curve and the
analytical curve share the same coordinates (x2,y2) and have the same
slope (dy/dx) at P2, the elliptic curve nevertheless has a cusp at
that point!
I haven't personally observed or been able to locate such abnormality
at or near P2 in any of the 10s test cases.
Have you observed a cusp at P2 ??

Any comments ? Thank you.
Monir
.

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