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Conference rusure::math

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

1988.0. "Crux Mathematicorum 2056" by RUSURE::EDP (Always mount a scratch monkey.) Fri Aug 11 1995 17:24

    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.
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1988.1CSC32::D_DERAMODan D'Eramo, Customer Support CenterFri Aug 11 1995 19:468
        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
1988.2at least it's all integers...WRKSYS::ROTHGeometry is the real life!Tue Aug 15 1995 06:355
   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
1988.3WRKSYS::ROTHGeometry is the real life!Wed Aug 16 1995 03:0435
   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
1988.4impressedJOBURG::BUCHANANWed Aug 16 1995 09:315
    Re -.1
    
    	That's clever.
    
    Andrew.