334 Answers to Exercises and Solutions to Problems
Remark. These polynomials occur as the denominators of a sequence
of convergents in a continued fraction representation of the infinite series
1 - x + 2x2 - 6x3 + 24x4 - 120x5 +. .. studied by L. Euler around 1760.
See E.J. Barbeau, Euler subdues a very obstreperous series Amer. Math.
Monthly 86 (1979), 356-372.
4.15. First solution. Let the three zeros be u, v, w with u > v 2 w. Let
u - w = r and u - v = s, so that v - w = r - s. Hence
2(a2 -3b) = 2(u+v+w)2-6(uv+vw+wu)
= (u - w)2 + (u - v)Z + (v - w)2
= r2+s2+(r-s)2
= 2(s - r/2)2 + 3r2/2 2 0.Under the condition 0 5 s 5 r, the quadratic (in r and s) is minimized
when s = r/2 and maximized when s = 0 or s = r. Hence(3/2)r2 5 2(a2 - 3b) 5 2r2,so that(u” - 3b)1’2 5 u - w < (2/&)(a2 - 3b)1’2 < 2(a2 - 3b)1’2.(Solution due to David Ash.)
Second solution. By a change of variable x = t + a/3, we can write
t3 + at2 + bt + c = x3 + pz + q where p = b - a2/3. Consider the graph of
y = x3 + px + q. It is centrally symmetric about the point (0, q) and has
parallel tangents at (zl,q) and (x2,(1).The distance between the roots of x3 +px +q = 0 is the distance between
the outer points of intersection of the curve and the line y = 0. Prom the
shape of the curve, it can be seen that this distance lies somewhere between
the values 22 - 21 and 24 -23, where xl, 0, zz are the roots of the equation
x3 + px + q = q and x3, x4 are the roots of the equation x3 + px + q = r,
where r is chosen so that the equation has a double root.
Clearly, c2 - r1 = 2(-p)lj2 (so that, in particular a2 - 3b = -3p >
0). Now, x3 + px + (q - r) has a double zero when it shares a zero with
its derivative 3x2 + p, i.e. when its zeros are either (-~/3)“~, (-p/3)l’“,
-2(-~/3)l’~ or the negative of these. Hence x4 - 2s = 3(-~/3)“~.