2nd Putnam 1939

Problem B7

Do either (1) or (2):

(1)   Let ai = ∑n=0 x3n+i/(3n+i)!   Prove that a03 + a13 + a23 - 3 a0a1a2 = 1.

(2)   Let O be the origin, λ a positive real number, C be the conic   ax2 + by2 + cx + dy + e = 0, and Cλ the conic   ax2 + by2 + λcx + λdy + λ2e = 0. Given a point P and a non-zero real number k, define the transformation D(P,k) as follows. Take coordinates (x',y') with P as the origin. Then D(P,k) takes (x',y') to (kx',ky'). Show that D(O,λ) and D(A,-λ) both take C into Cλ, where A is the point (-cλ/(a(1 + λ)), -dλ/(b(1 + λ)) ). Comment on the case λ = 1.



(1)   Moderately hard, unless differentiating is a reflex, in which case it is easy.

The series are all absolutely convergent for all x, so we can carry out whatever operations we want. But it is not at all obvious what to do. Multiplying out the power series to get a complicated sum for the coefficient of xn is offputting. Much scope for algebraic error, and no guarantee that the eventual simplification will be obvious. So we look for some trick. The ai are closely related to the exponential series, eg a0 + a1 + a2 = ex, and the prevalence of 3 in the question may eventually suggest looking at ω, the cube root of 1. Indeed, a0 + ωa1 + ω2a2 = eωx. Since ω2 is also a root, we also have a0 + ω2a1 + ωa2 = eω2x. Remembering that 1 + ω + ω2 = 0 might prompt us to multiply these three expressions together, but it helps to remember the product of the three left hand sides: (a0 + a1 + a2)(a0 + ωa1 + ω2a2)(a0 + ω2a1 + ωa2) = a03 + a13 + a23 - 3a0a1a2. Otherwise, you are faced with multiplying this out the hard way: after collecting terms, you get (a03 + a13 + a23), then 6 expressions of the type a02(1 + ω + ω2), which are all zero, and 3a0a1a2(ω + ω2), which is -3a0a1a2.

A more general approach, which is more likely to work, is to differentiate the expression   a03 + a13 + a23 - 3 a0a1a2. Provided you notice that a0' = a2 etc, this gives the result almost immediately (the derivative is zero, so the expression must be constant, but its value for x = 0 is 1).

(2)   Trivial. [D(O,λ) takes (x,y) to (λx,λy). So if (x,y) satisfies the equation for C, just check that (λx,λy) satisfies the equation for Cλ. Similarly, D(A,-λ) takes (x,y) to (-λx - λc/a, -λy - λd/b). Again, just check by substituting this into the equation for Cλ and using the fact that (x,y) satisfies the equation for C. If λ = 1, then Cλ = C, D(O,λ) is the identity transformation, and D(A,-λ) the central symmetry.]

Comment. I find the contrast between these two parts odd, given that they are alternatives. It is obvious how to attack (2) and the algebra is not at all heavy, so you are almost guaranteed to complete it within 5 minutes or so. (1) may be easy once you have seen how to attack it, but it is far from obvious how to attack it, so you could easily waste half-an-hour getting nowhere. Maybe the question is designed to test your ability to pick the easy option.



2nd Putnam 1939

© John Scholes
4 Sep 1999