a_{1}, a_{2}, ... , a_{N} are real and a_{N} is non-zero. f(x) = a_{1} sin 2πx + a_{2} sin 4πx + a_{3} sin 6πx + ... + a_{N} 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 = e^{2πix}, then the nth derivative is 1/z^{N} 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 a_{N} 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} |a_{N}| > 2^{k} |a_{1}| + 4^{k} |a_{2}| + ... + (2N-1)^{k} |a_{N-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