Saturday, 23 February 2013

Roots of a Polynomial Minus It's Constant Term




Suppose we have a sequence of integers a1,,an. Is there any way to determine the roots of the polynomial



P(x)=(x+a1)(x+an)a1an



Clearly P(0)=0, but can anything be said about the other roots? Can they be expressed in some way related to the original integers a1,,an?



Any answer or reference would be appreciated.



Edit: If need be, you may assume ai|ai+1 for all i.




Comment: The new roots need not be integers. I would be satisfied with finding complex roots.


Answer



It doesn't appear to be easier than solving a polynomial of degree n1 "from scratch". For example, if Q(x)=(x+1)(x+2)(x+4)(x+8)(x+16)(x+32), the Galois group of P(z)/z=(Q(z)Q(0))/z=z5+63z4+1302z3+11160z2+41664z+64512

is S5, so this is not solvable by radicals.



On the other hand, it may be interesting to look at the roots of Q(x)t as functions of t: these are analytic except at the points where they collide (the roots of the discriminant of Q(x)t), which can be branch points.
In the above example, the root that is 32 at t=0 has the Maclaurin series
3219999360t+1189578592599703552000t2575209184743927083111118222131200000t3+2825592263310460874099698222880422457709166592000000t4812974696648768399931666747186847737020565295881781248000000000t5+


I believe this has radius of convergence approximately 1.61741×107 (one of the roots of that discriminant), so it should certainly converge quite nicely at t=Q(0).


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