a1, a2, ... , aN are real and aN is non-zero. f(x) = a1 sin 2πx + a2 sin 4πx + a3 sin 6πx + ... + aN sin 2Nπx. Show that the number of zeros of f(i)(x) = 0 in the interval [0, 1) is a non-decreasing function of i and tends to 2N (as i tends to infinity).
Solution
By Rolle's theorem there is a root of f ' between two adjacent roots of f. Also the function is periodic with period 1, so there number of zeros is non-decreasing.
If we write z = e2πix, then the nth derivative is 1/zN times a polynomial degree 2N in z, so it has at most 2N complex roots. Hence there are at most 2N real roots for x in the interval [0, 1).
Finally, we need to show that for sufficiently high derivatives the term aN sin 2Nπx dominates and hence gives exactly 2N roots. We use Rolle's theorem again. At the values x = 0, 1/2N, 2/2N, ... , 2N/2N, cos 2Nπx is alternately ±1, so if we choose k sufficiently large, then the odd derivative f(k)(x) is alternately positive and negative at these values (because (2N)k |aN| > 2k |a1| + 4k |a2| + ... + (2N-1)k |aN-1| and hence f(k)(x) has at least 2N zeros in the interval (0, 1).
© John Scholes
jscholes@kalva.demon.co.uk
1 Jan 2001