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Source. A. W. Goodman, On the convexity of the level curves of a polynomial, Proc. Amer. Math. Soc. 17 (1966), no. 2, 358--361, DOI 10.1090/S0002-9939-1966-0188408-3, identified on the source card: section 2, "The first counterexample", p. 359, with the setting on p. 358.

Read depth. Claims checked: the example was read clause by clause on the page image, and P′P', the critical points and the values (3) and (4) were recomputed here. The topological claims (the double points and the count of three components) are taken as printed. Nothing here is independently reviewed.

Statement

With the paper's notation (p. 358), E(c)={z:∣P(z)∣<c}E(c)=\{z:|P(z)|<c\} is open and its boundary Γ(c)\Gamma(c) is the lemniscate ∣P(z)∣=c|P(z)|=c.

Example (p. 359). Let P(z)=(z2+1)(z−2)2P(z)=(z^2+1)(z-2)^2 (2). Then P′(z)=2(z−2)(2z2−2z+1)P'(z)=2(z-2)(2z^2-2z+1), with zeros z1∗,z2∗=(1±i)/2z_1^*,z_2^*=(1\pm i)/2 and z3∗=2z_3^*=2. Take c=∣P(z1∗)∣=∣P(z2∗)∣=55/4c=|P(z_1^*)|=|P(z_2^*)|=5\sqrt5/4 (3). At this cc the curve Γ(c)\Gamma(c) has double points at z1∗z_1^* and z2∗z_2^*, and E(c)E(c) has three components, as many as PP has distinct roots (±i\pm i and 22). The component E3(c)E_3(c) containing 22 is not convex: z1∗z_1^* and z2∗z_2^* lie on its boundary, but their midpoint x∗=1/2x^*=1/2 has ∣P(1/2)∣=45/16|P(1/2)|=45/16 (4), and 45/16>55/445/16>5\sqrt5/4, so x∗x^* lies outside the closure of E3(c)E_3(c).

The root 22 is double, so m=3m=3 is less than the degree 44; the paper's Theorem (p. 361) gives a quartic with four simple roots.

Proof pointer

P. 359. Everything is direct computation: the factorization of P′P', the values ∣P((1±i)/2)∣=55/4|P((1\pm i)/2)|=5\sqrt5/4 and ∣P(1/2)∣=45/16|P(1/2)|=45/16, and the comparison 45/16=2.8125>2.795…=55/445/16=2.8125>2.795\ldots=5\sqrt5/4. If E3(c)E_3(c) were convex its closure would contain the segment from z1∗z_1^* to z2∗z_2^*, and so x∗x^*.

Dependencies

None; elementary computation.

Bears on

  • Problem 1047: the example answers Grunsky's question no for the open set E(c)E(c) at the critical level c=55/4c=5\sqrt5/4. At that level the closed set {z:∣P(z)∣≤c}\{z:|P(z)|\le c\} of the problem contains the two double points, at which branches of the lemniscate cross, so its components are not those of E(c)E(c) and the example as printed does not answer the problem as posed.