Half range sine and cosine series: eˣ

Half range sine series on (0, L) is f(x) = Σ bₙ sin(nπx/L) with bₙ = (2/L)∫₀ᴸ f(x) sin(nπx/L) dx. Here L = 2, so bₙ = ∫₀² eˣ sin(nπx/2) dx.

Use ∫ eˣ sin kx dx = eˣ(sin kx − k cos kx)/(1 + k²) with k = nπ/2 (at x = 2: sin nπ = 0, cos nπ = (−1)ⁿ):

bₙ = [ eˣ(sin kx − k cos kx)/(1 + k²) ]₀²
   = [ −k(−1)ⁿ e² + k ] / (1 + k²)
   = k[1 − (−1)ⁿ e²]/(1 + k²)
   = 2nπ[1 − (−1)ⁿ e²] / (n²π² + 4)

Sine series

eˣ = 2π Σ (n=1 to ∞) { n[1 − (−1)ⁿ e²] / (n²π² + 4) } sin(nπx/2), 0 < x < 2

Half range cosine series on (0, L) is f(x) = a₀/2 + Σ aₙ cos(nπx/L), with a₀ = (2/L)∫₀ᴸ f dx and aₙ = (2/L)∫₀ᴸ f cos(nπx/L) dx.

Use ∫ eˣ cos kx dx = eˣ(cos kx + k sin kx)/(1 + k²), k = nπ/2:

a₀ = ∫₀² eˣ dx = e² − 1

aₙ = [ eˣ(cos kx + k sin kx)/(1 + k²) ]₀²
   = [ (−1)ⁿ e² − 1 ]/(1 + k²)
   = 4[(−1)ⁿ e² − 1] / (n²π² + 4)

Cosine series

eˣ = (e² − 1)/2 + 4 Σ (n=1 to ∞) { [(−1)ⁿ e² − 1] / (n²π² + 4) } cos(nπx/2), 0 < x < 2

Both series represent eˣ on (0, 2); the sine series converges slowly because its odd extension has jumps at the end points, while the cosine series (an even extension is continuous) converges faster. A numerical check at x = 1 gives 2.718 = e from both series.