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Answers to Exercises; Chapter (^6 343)
= t6 - (Cmn)t5 + (Cm2np + 3mnpq)t4

• (Cm3npq + 2Cm2n2pq + Cm2n2p2)t3

• (Cm3n2p2q + 3m2n2p2q2)t2 - (Cm3n3p2q2)t


• m3n3p3q3
  = t6 - s2t5 + (SlS3 - S4)t4 - (sfs, + s; - 2S&)P


• (SlS3 - s4)s4t2 - s& + s;
  = t6 - 2t5 - 2t4 + 4t3 + 2t2 - 2t - 1.

The given quartic has 1 as a double zero; indeed,

t4 - 3t3 + 2t2 + t - 1 = (t - l)“(? -t - 1).

The sextic will have two double zeros, and we find that

t6 - 2t5 - 2t4 + 4t3 + 2t2 - 2t - 1 = (t - l)(t + l)(P -t - l)?

1.4. (1/5)(5t3 + 6t2 - 2t + 1).

1.5. Ifs, is the rth symmetric function of the zeros ti of p(t) and u, is the
rth symmetric function of tf’, then

ur = s,-,/s, = (-l)ra,-,/a,,

whence the manic polynomial with zeros ti’ is

(00)-l fJ-1)2rn,-rt’.

r=O

1.6. a,t” + ka,,-lP1 + e-e+ k”-‘alt + k”ao.

1.7. (a) The polynomial must have the form a,P + a,-2tn-2 +... so that
its derivative is na,t”-l + (n - 2)a,-2tn-3 +.... Since the next to leading
coefficient of the derivative vanishes, the result follows.

w: + * * .w;=(wl+.**+w,)2 - 2CWiWj = -2(n - 2)a,-2/n

= (n - 2)[(%1 + **a+ Zn)2 - 2CZiZj]/n.

It is conjectured by I.J. Schoenberg in Amer. Math. Monthly 93 (1986), 8-
11 that in fact Clwi12 5 (n-2)(Clzi12/n). The reader may wish to establish
this when degp = n = 3. See also problem E3115, Amer. Math. Monthly
92 (1985), 666; 94 (1987), 689.

1.8. (a) The sum of the squares of the zeros is (a,-l/a,)2 - (2a,+z/a,).
If the zeros are real, then the sum of their squares is positive and the
first inequality follows. The second inequality follows from the fact that, if