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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

- 0
- at least 2n
- 53361
- 5082

Periodic

Real Numbers

Inequality

But try the problem first...

Answer:at least 2n

Source

Suggested Reading

B.Stat Objective Problem 710

Challenges and Thrills of Pre-College Mathematics by University Press

First hint

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

Second Hint

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=\(x_k\)]

Final Step

f has at least 2n points in such a period interval where f has alternating sign.

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