| 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 |
P(x) is a polynomial in x. Then P(a + b*sqrt(s)) can be expressed as:
P(a + b*sqrt(s)) = Q(a,b,s) + sqrt(s)*R(a,b,s)
where Q and R are polynomials of their parameters (e.g, the sqrt(s) has
been 'separated' from the rest of the stuff). The problem is to express
Q(a,b,s) and R(a,b,s) in terms of P.
[ FWIW, this is a 'real' problem. I don't have a solution, but I suspect ]
[ Q is something like: Q(a,b,s) = P(a + b*sqrt(s)) - P(a - b*sqrt(s)). ]
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1876.1 | AUSSIE::GARSON | achtentachtig kacheltjes | Thu May 26 1994 07:51 | 18 | |
re .0
>The problem is to express Q(a,b,s) and R(a,b,s) in terms of P.
>[FWIW, this is a 'real' problem. I don't have a solution, but I suspect]
>[Q is something like: Q(a,b,s) = P(a + b*sqrt(s)) - P(a - b*sqrt(s)).]
It seems that
Q(a,b,s) = � [ P(a+b*sqrt(s)) + P(a-b*sqrt(s)) ]
and
R(a,b,s) = �[ P(a+b*sqrt(s)) - P(a-b*sqrt(s)) ]
but I don't see that this buys anything.
{I suppose that you intend that a, b and s are rational and that sqrt(s)
is irrational.}
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| 1876.2 | RTL::GILBERT | Thu May 26 1994 21:47 | 12 | ||
Right; thanks!
If P(x) is a polynomial, then
P(a+b*sqrt(s)) = Q(a,b,s) + sqrt(s)*R(a,b,s)
where
Q(a,b,s) = � [ P(a+b*sqrt(s)) + P(a-b*sqrt(s)) ]
and
R(a,b,s) = � [ P(a+b*sqrt(s)) - P(a-b*sqrt(s)) ]/sqrt(s)
are polynomials in a, b, and s.
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| 1876.3 | AUSSIE::GARSON | achtentachtig kacheltjes | Fri May 27 1994 04:29 | 4 | |
re .2
Blush. Sorry about the factor of sqrt(s) in R(a,b,s).
Just checking whether you were awake. (-:
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