Question

in a fourier series for f(x) = sinx in (-π π) the value of bn is? ​

Answer 1

Given f(x) = |sinx|

f(x) = |sinx| = sinx

f(-x) = |sin(-x)| = sinx

Hence f(x) = f(-x)

∴ |sinx| is even function

The Fourier series of an even function contains only cosine terms and is known as Fourier Series and is given by

f(x)=a02+∑n=1∞ancosnx

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Answer 2

Answer:

In a Fourier series for f(x) = sinx in (-π π) the value of bₙ is Zero.

Step-by-step explanation:

Given:

$f(x) = sin x$

Limits  = (-π π)

For Sinx it has a period 2π Since sin(x+2π) =sin x

It is a odd function. Therefore sin(-x) = -sin x. It vanishes at x=0 and x=π

The three properties of sinx in Fourier series is:

• Periodic :  S(x+2π) = S(x)
• Odd : S(-x) = - S(x)
• S(0) = S(π) = 0

Formula:

f(x) = $\frac{1}{2}a_{0}$ + Σ$a_{n} Cosnx$ +  Σ $b_{n} Sin nx$

Where,

• a₀ = $\frac{1}{\pi }$∫f(x) dx
• aₙ = $\frac{1}{\pi }$ ∫ f(x) cos nx dx
• bₙ = $\frac{1}{\pi }$ ∫ f(x) sin nx dx
• n= 1,2,3,......

To find the value of bₙ in the function find whether the given function is odd or even.

Given f(x) = sinx

f(-x) = sin(-x)

f(-x) = -sinx

Therefore sinx is even function. The Fourier series contains only the cosine terms because the value of bₙ is zero.

Answer :The value of bₙ = 0

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