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| 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 |
1671.0. "Super Cross-Ratio" by RUSURE::EDP (Always mount a scratch monkey.) Wed Oct 07 1992 11:56
Stan would like to reduce the following expression to a more aesthetic
form:
- y0*y1*y2 + 9*y0*y1*y3 - 18*y0*y2*y3 + 10*y1*y2*y3 - 18*y0*y1*y4 +
64*y0*y2*y4 - 45*y1*y2*y4 - 36*y0*y3*y4 + 45*y1*y3*y4 - 10*y2*y3*y4 +
10*y0*y1*y5 - 45*y0*y2*y5 + 36*y1*y2*y5 + 45*y0*y3*y5 - 64*y1*y3*y5 +
18*y2*y3*y5 - 10*y0*y4*y5 + 18*y1*y4*y5 - 9*y2*y4*y5 + y3*y 4*y5
Here is how the expression was derived. Let y[n] =
(an^2+bn+c)/(dn^2+en+f). Then if y0, y1, y2, y3, y4, and y5 are
consecutive terms of {y[n]}, the above expression is zero. This is
true if y0 is y[0] or some other term; it doesn't matter where you
start, just that the terms are consecutive.
If y[n] were just (ax^2+bx+c)/(dx+e), the following relationship would
hold:
(y0-2y1+y2)(y3-2y4+y5) 1
---------------------- = --.
(y0-2y2+y4)(y3-2y5+y6) 16
And if y[n] were just (ax+b)/(cx+d), an even simpler expression called
a cross-ratio applies. Can anybody reduce the first expression above
to something similar?
-- edp
| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1671.1 | would MAPLE help? | STAR::ABBASI | life without the DECspell ? | Wed Oct 07 1992 15:55 | 3 |
| have you tried MAPLE?
(asking without any idea what a super corss-ratio is)
|
| 1671.2 | | RUSURE::EDP | Always mount a scratch monkey. | Wed Oct 07 1992 16:46 | 8 |
| Re .1:
Stan has worked with it with Mathematica, but it probably needs to be
expressed as a sum of products of polynomials, rather than just
factored, so it needs a person to massage it.
-- edp
|
| 1671.3 | 3 remarks | HERON::BUCHANAN | The was not found. | Wed Oct 21 1992 10:14 | 25 |
| > Stan would like to reduce the following expression to a more aesthetic
> form.
So would I, but I haven't spotted it yet.
> If y[n] were just (ax^2+bx+c)/(dx+e), the following relationship would
> hold:
>
> (y0-2y1+y2)(y3-2y4+y5) 1
> ---------------------- = --.
> (y0-2y2+y4)(y3-2y5+y6) 16
^^^^^^^^^
Typo here: I think the factor should be (y1-2y3+y5).
> And if y[n] were just (ax+b)/(cx+d), an even simpler expression called
> a cross-ratio applies.
Namely:
(y0-y1)(y2-y3) 1
-------------- = -.
(y0-y2)(y1-y3) 4
Andrew.
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