Find the Fourier series of |x| where –𝜋 < x < 𝜋.Question 5Select one:𝜋23(𝑐𝑜𝑠𝑥 + 132𝑐𝑜𝑠3𝑥- 152𝑐𝑜𝑠5𝑥 +…)𝜋23-4𝜋(𝑐𝑜𝑠𝑥 + 132𝑐𝑜𝑠3𝑥+ 152𝑐𝑜𝑠5𝑥 +…)𝜋23+4𝜋(𝑐𝑜𝑠𝑥 + 132𝑐𝑜𝑠3𝑥- 152𝑐𝑜𝑠5𝑥 +…)𝜋23+4𝜋(𝑐𝑜𝑠𝑥- 132𝑐𝑜𝑠3𝑥- 152𝑐𝑜𝑠5𝑥 +…)

The Fourier series of a function f(x) is given by:

f(x) = a0/2 + Σ [an cos(nx) + bn sin(nx)]

where the coefficients an and bn are given by:

an = (1/π) ∫ from -π to π [f(x) cos(nx) dx]

bn = (1/π) ∫ from -π to π [f(x) sin(nx) dx]

For the function f(x) = |x|, we have:

an = (1/π) ∫ from -π to π [x cos(nx) dx] for -π < x < 0

an = (1/π) ∫ from 0 to π [x cos(nx) dx] for 0 < x < π

Since |x| is an even function, all the sine terms in the Fourier series will be zero, so bn = 0 for all n.

The a0 term is given by:

a0 = (1/π) ∫ from -π to π [x dx] = 0

The an terms for n > 0 are given by:

an = (1/π) ∫ from -π to π [x cos(nx) dx] = 0 for even n

an = -4/(πn^2) for odd n

So the Fourier series of |x| is:

|x| = Σ [(-4/(πn^2)) cos(nx)]

This corresponds to the option:

𝜋/2 - 4/𝜋 (cosx - 1/3 cos3x + 1/5 cos5x - ...)