Categories

Periodic Function | TOMATO B.Stat Objective 710

Try this TOMATO problem from I.S.I. B.Stat Entrance Objective Problem based on Periodic Function. You may use sequential hints.

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Periodic Function.

Periodic Function (B.Stat Objective Question )

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

Key Concepts

Periodic

Real Numbers

Inequality

But try the problem first…

Source

B.Stat Objective Problem 710

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints

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.