41st Putnam 1980

For which real k do we have cosh x ≤ exp(k x²) for all real x?

Solution

Answer: k ≥ 1/2.

cosh x = 1 + x²/2! + x⁴/4! + ... + x²ⁿ/(2n)! + ... and exp(x²/2) = 1 + x²/2 + x⁴/8 + ... + x²ⁿ/(2ⁿn!). But 2ⁿn! = 2n.(2n - 2).(2n - 4) ... 2 < (2n)! for n > 1, so cosh x < e^x2/2 for all x. Hence the inequality holds for k ≥ 1/2.

cosh x = 1 + x²/2 + o(x), exp(k x²) = 1 + kx² + o(x). So if k < 1/2, then cosh x > exp(k x²) for sufficiently small x. Thus the inequality does not hold for k < 1/2.

41st Putnam 1980

© John Scholes
jscholes@kalva.demon.co.uk
10 Dec 1999
Last corrected/updated 21 Nov 02