Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Periodic Function.
If f(x) = \(a_0+a_1cosx+a_2cos2x+....+a_ncosnx\) where \(a_0,a_1,....,a_n\) are non zero real numbers and \(a_n > |a_0|+|a_1|+....+|a_{n-1}|\), then number of roots of f(x)=0 in \( 0 \leq x \leq 2\pi\), is
Periodic
Real Numbers
Inequality
Answer:at least 2n
B.Stat Objective Problem 710
Challenges and Thrills of Pre-College Mathematics by University Press
f is periodic with period \(2\pi\)
here \(0 <|\displaystyle\sum_{k=0}^{n-1}a_kcos(kx)| \leq \displaystyle\sum_{k=0}^{n-1}|a_k|<a_n, x\in\)set of reals
for points \(x_k=\frac{k\pi}{n}\) \(1 \leq k \leq 2n\)
\(f(x_k)=a_n[cos(k\pi)+\theta], |\theta|<1 for 1 \leq k \leq 2n\)
[ since cos \(\theta\) is periodic and f(x) is expressed for every point x=\(x_k\)]
f has at least 2n points in such a period interval where f has alternating sign.
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