Calculus Gems: Brief Lives and Memorable Mathematics by George F. Simmons

By George F. Simmons

This article is a spin-off of Appendices A ("A number of extra Topics") and B ("Biographical Notes") of Simmons' winning CALCULUS WITH ANALYTIC GEOMETRY. The textual content is appropriate as a complement for a calculus direction and/or background of arithmetic direction. The textual content can also be acceptable for a liberal arts arithmetic direction for college students with minimum arithmetic history.

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In the same way, we get that the limiting polynomial is of degree 2. 3) B (ω∞ ) · ω∞ = 0 . where B denotes the Hessian. We assume that B (ω∞ ) restricted to (Rω∞ )⊥ is non degenerate and has at least one strictly positive eigenvalue. Then no limiting polynomial can have a local maximum. 30 16 MAKHLOUF DERRIDJ AND BERNARD HELFFER The philosophy is that Tr`eves condition gives immediately the control of the non-degenerate representations in the sense of C. Rockland (see [5] for this notion). But maximal hypoellipticity can fail if the criterion is not satisfied for some degenerate representation.

4 We have 3 1 φ (s0 ( )) ∼ −( )3 4 ϕ(0)4 4 4 and 9 2 ϕ(0)2 . 16 So φ has a unique minimum at s0 ( ) and, depending on the sign of ϕ(0), we have φ (s0 ( )) ∼ −σ < 0 < s0 ( ) < s1 ( ) < σ , or −σ < s1 ( ) < s0 ( ) < 0 < σ . We will show that a suitable configuration is A+ B − B − A+ . [E:] ≥ 4. We have φ (s) = s4 (1 + ϕ(s)) . We will show that a suitable configuration is A+ A+ . 2. Determination of the configurations. 1. As for the homogeneous case of degree 3, we write, between −σ and σ, in increasing order the sequence consisting of the zeroes of φ and φ .

The Tr`eves condition implies ϕ(0) > 0. Hence we get 1 p s , C and 0 is the only zero and this zero is of even order. 9) ϕ (s) ≥ φ (s) = sp−1 (pϕ + sϕ + s −p ψ+ s −p+1 ψ ), 50 36 MAKHLOUF DERRIDJ AND BERNARD HELFFER we obtain that φ has a unique zero at 0. Associated with the sequence −σ < 0 < σ, we get two sectors with configuration A+ A+ . Subcase = 0. So φ = sp ϕ + ψ with ϕ(0) > 0 and ψ(0) = 0. 10) 1 s± ( ) ∼ ±(− ψ(0)) p We have then to analyze φ = sp−1 [pϕ + sϕ ] + ψ , and to discuss in function of the behavior of ψ at 0.

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