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 |
Proposed by Stanley Rabinowitz, Westford, Massachusetts. Find a polynomial of degree 5 whose roots are the tenth powers of the roots of the polynomial x^5-x-1.
T.R | Title | User | Personal Name | Date | Lines |
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1988.1 | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Fri Aug 11 1995 19:46 | 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 06:35 | 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 03:04 | 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 09:31 | 5 | |
Re -.1 That's clever. Andrew. |