31st Putnam 1970

e^bx cos cx is expanded in a Taylor series ∑ a_n xⁿ. b and c are positive reals. Show that either all a_n are non-zero, or infinitely many a_n are zero.

Solution

e^bx cos cx = Re e^(b+ic)x, so n! a_n = Re( (b + ic)ⁿ ). Let b + ic = k e^iθ, then n! a_n/kⁿ = cos nθ. So a_n = 0 iff nθ = (2m+1)π/2 for some integer m. So if there are any zeros then there are infinitely many.

31st Putnam 1970

© John Scholes
jscholes@kalva.demon.co.uk
25 Aug 2001