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)−a1…an
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 n−1 "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
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
−32−19999360t+1189578592599703552000t2−575209184743927083111118222131200000t3+2825592263310460874099698222880422457709166592000000t4−812974696648768399931666747186847737020565295881781248000000000t5+…
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).
No comments:
Post a Comment