| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1988.1 | | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Fri Aug 11 1995 16:46 | 8 |
| Find the roots (call them r0 through r4), raise them to the
tenth power, multiply out (x - r0^10)...(x - r4^10) and the
answer is x^5 - 5x^4 + 8x^3 - 30x^2 - 4x - 1.
But that's the hard way to do it. :-) You're probably
supposed to find the easy way.
Dan
|
| 1988.2 | at least it's all integers... | WRKSYS::ROTH | Geometry is the real life! | Tue Aug 15 1995 03:35 | 5 |
| Not that it's any "easier", but you can raise the companion matrix
for x^5 - x - 1 to the 10'th power and write down its characteristic
polynomial.
- Jim
|
| 1988.3 | | WRKSYS::ROTH | Geometry is the real life! | Wed Aug 16 1995 00:04 | 35 |
| What Stan and Dan are probably thinking of is that since x^5 = x + 1,
the "shifted" polynomial (x - 1)^5 - (x - 1) - 1 will have
roots equal to the fifth powers of x^5 - x - 1.
Since p(x)*p(-x) = q(x^2) has roots that are squares of p(x),
we can expand out [(x - 1)^5 - x][(-x - 1)^5 + x] and the
resulting polynomial in, say, y = x^2 is his desired polynomial:
[(x - 1)^5 - x][(x + 1)^5 - x]
((x - 1)(x + 1))^5 - x((x - 1)^5 + (x + 1)^5) + x^2
(x^2 - 1)^5 - 2x(x^5 + 10 x^3 + 5 x) + x^2
(y - 1)^5 - (2 y^3 + 20 y^2 + 10 y) + y
y^5 - 5 y^4 + 10 y^3 - 10 y^2 + 5 y - 1
- 2 y^3 - 20 y^2 - 10 y
+ y
-----------------------------------------
y^5 - 5 y^4 + 8 y^3 - 30 y^2 - 4 y - 1
Miracle - the same answer as Dan posted, without touching a computer
algebra system :-)
One could also get a poly with k'th powers of roots by taking
a product
k - 1
PROD P(w^j x) = Q(x^k), with w = k'th root of unity
j = 0
And there's also a way by fiddling with symmetric functions to do it.
- Jim
|
| 1988.4 | impressed | JOBURG::BUCHANAN | | Wed Aug 16 1995 06:31 | 5 |
| Re -.1
That's clever.
Andrew.
|