The 63 rd William Lowell Putnam Mathematical Competition Saturday

Solutions to the 64 th William Lowell Putnam Mathematical Competition Saturday , December 6 , 2003

Solutions to the 71 st William Lowell Putnam Mathematical Competition Saturday

A3 If a = b = 0, then the desired result holds trivially, so we assume that at least one of a, b is nonzero. Pick any point (a0, b0) ∈ R, and let L be the line given by the parametric equation L(t) = (a0, b0) + (a, b)t for t ∈ R. By the chain rule and the given equation, we have d dt (h◦L) = h◦L. If we write f = h◦L : R → R, then f (t) = f(t) for all t. It follows that f(t) = Ce for some consta...

Solutions to the 78th William Lowell Putnam Mathematical Competition Saturday, December 2, 2017

Remark: A similar argument shows that any secondorder linear recurrent sequence also satisfies a quadratic second-order recurrence relation. A familiar example is the identity Fn−1Fn+1 − F2 n = (−1)n for Fn the nth Fibonacci number. More examples come from various classes of orthogonal polynomials, including the Chebyshev polynomials mentioned below. Second solution. We establish directly that ...

The William Lowell Putnam Mathematical Competition.

Solutions to the 71st William Lowell Putnam Mathematical Competition Saturday, December 4, 2010

A3 If a = b = 0, then the desired result holds trivially, so we assume that at least one of a, b is nonzero. Pick any point (a0, b0) ∈ R, and let L be the line given by the parametric equation L(t) = (a0, b0) + (a, b)t for t ∈ R. By the chain rule and the given equation, we have d dt (h◦L) = h◦L. If we write f = h◦L : R→ R, then f ′(t) = f(t) for all t. It follows that f(t) = Ce for some consta...