Putnam 39/A7 solution

2nd Putnam 1939

Problem A7

Do either (1) or (2):

(1) Let C_a be the curve (y - a²)² = x²(a² - x²). Find the curve which touches all C_a for a > 0. Sketch the solution and at least two of the C_a.

(2) Given that (1 - hx)^-1(1 - kx)^-1 = ∑_i≥0 a_i xⁱ, prove that (1 + hkx)(1 - hkx)^-1(1 - h²x)^-1(1 - k²x)^-1 = ∑_i≥0 a_i² xⁱ.

Solution

(1) It is not hard to see that C₁ is a figure 8 lying on its side with the double point at (1,0), vertical tangents at (1,1) and (-1,1) and horizontal tangents at (1/√2, 1/2), (1/√2, 3/2), (-1/√2,1/2), (-1/√2,3/2). C_a is obtained from C₁ by transforming (x,y) to (ax,a²y) (so stretching by a factor a in the x-direction and a factor a² in the y-direction). So we get a sequence of ever-larger horizontal 8s centered ever-further up the y-axis. This suggests that the envelope is something like a parabola y = k x².

At this point it is distinctly helpful to know more classical differential geometry than today's undergraduate. The classical method for finding the envelope for a 1-parameter family of curves (which usually works) is to differentiate wrt the parameter. Thus we eliminate a from:

(y - a²)² - x²(a² - x²) = 0, and 4a(y - a²) + 2a x² = 0

giving y = 3/4 x² (or the y-axis, which is presumably a spurious solution arising from the double points).

Alternatively, we might guess that that the envelope is a curve y = k x² for some k. The intersection of this with C_a is given by:
(k² + 1)x⁴ - (2k + 1) a²x² + a⁴ = 0, or x² = (2k + 1)a²/(2k² + 2) +/- a²(4k - 3)^1/2/(2k² + 2).

For this to be a tangent we need double roots and hence k = 3/4. It is now easy to check that this parabola meets C_a at (2a/√5,3a²/5), (-2a/√5,3a²/5) and to check that the gradients match.

(2) Fairly easy.

This is obviously not a general result (whereby we derive ∑ a_i²xⁱ from ∑ a_ixⁱ), so we need to evaluate the a_i. In fact, it is easily seen that a_i = (hⁱ⁺¹ - kⁱ⁺¹)/(h - k). [For example, multiply the expansions of (1 - hx)^-1 and (1 - kx)^-1 to get a_i = hⁱ + h^i-1k + ... + h k^i-1 + kⁱ = (hⁱ⁺¹ - kⁱ⁺¹)/(h - k).]

Multiplying across to try to show that higher powers of (1 - hkx)(1 - h²x)(1 - k²x) ∑ a_i²xⁱ have zero coefficients is a mistake (doable, but much algebra). It is better to evaluate ∑ a_i²xⁱ directly. After substituting for a_i, we are going to get terms of the form ∑ zⁱ which evaluates immediately to (1 - z)^-1, giving the expression in the question in partial fraction form.

2nd Putnam 1939