Find the Fourier series of |x| where –𝜋 < x < 𝜋.

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

Because |x| is an even function, its Fourier series contains only cosine terms, so bn = 0 for all n.

The a0 term is given by:

a0 = (1/π) ∫ from -π to π |x| dx = (1/π) [∫ from -π to 0 (-x) dx + ∫ from 0 to π x dx] = (1/π) [π^2/2 + π^2/2] = π

For n ≠ 0, we have:

an = (1/π) ∫ from -π to π |x| cos(nx) dx = (1/π) [∫ from -π to 0 (-x) cos(nx) dx + ∫ from 0 to π x cos(nx) dx] = (1/π) [2 ∫ from 0 to π x cos(nx) dx] = (1/π) [2 (sin(nx)/n^2 - x sin(nx)/n)] from 0 to π = (1/π) [2 (sin(nπ)/n^2 - π sin(nπ)/n)] = 0 (since sin(nπ) = 0 for all integer n)

So the Fourier series of |x| is:

f(x) = π/2 for -π < x < π