Compute the Fourier series of f(x)=x+|x|,-π≤x≤π

well, you have to break f(x) into two parts

for x<0, f(x) = 0
for x >= 0, f(x) = 2x

so,
a0 = 1/π ∫[0,π] 2x dx = π

and for the others just use integration by parts for integrals of x cos(nx), etc.

Vhhjjff

To compute the Fourier series of the given function f(x) = x + |x| over the interval -π ≤ x ≤ π, we need to follow these steps:

Step 1: Find the function's period (T) and make it an even function.
Since f(x) is given over the interval -π ≤ x ≤ π, the period of f(x) is 2π. To make the function an even function (symmetric about the y-axis), we can rewrite f(x) as f(x) = (x + |x|) - π, which gives us an even periodic extension over the interval -2π ≤ x ≤ 2π.

Step 2: Express f(x) as a piecewise function by determining its expression for different intervals within the period.
For -2π ≤ x ≤ -π:
f(x) = -(x + x) - π = -2x - π

For -π ≤ x ≤ 0:
f(x) = -(x - x) - π = -π

For 0 ≤ x ≤ π:
f(x) = x + x - π = 2x - π

For π ≤ x ≤ 2π:
f(x) = -(x - 2x) - π = -x - π

Step 3: Compute the Fourier coefficients using the formula:
a0 = (1/T) * ∫[T] f(x) dx
an = (2/T) * ∫[T] f(x) * cos(nωx) dx, n ≠ 0
bn = (2/T) * ∫[T] f(x) * sin(nωx) dx, n ≠ 0

where ω = 2π/T

First, calculate a0:
a0 = (1/2π) * ∫[-π]^[π] [x + |x| - π] dx
= (1/2π) * ∫[-π]^[π] [(x + x) - π] dx
= (1/2π) * ∫[-π]^[π] (2x - π) dx
= (1/2π) * [x^2 - πx]_[-π]^[π]
= (1/2π) * [(π^2 - π^2) - (-π^2 + π^2)]
= 0

Next, compute the coefficients an:
an = (1/π) * ∫[-π]^[π] [2x - π] * cos(nωx) dx
= (1/π) * ∫[-π]^[π] (2x - π) * cos(nπx) dx
= (1/π) * [(2/(-n^2π^2)) * (π^2 - 2)] * sin(nπx) + [(2/(nπ)) * (π - x)] * cos(nπx)]_[-π]^[π]
= 0

Finally, determine the coefficients bn:
bn = (1/π) * ∫[-π]^[π] [2x - π] * sin(nωx) dx
= (1/π) * [(2/(nπ)) * (π - x)] * (-cos(nπx))_[-π]^[π]
= (1/π) * [(2/(nπ)) * (π - π) - (2/(nπ)) * (π + π)]
= 0

Since a0 = 0, an = 0, and bn = 0, the Fourier series of f(x) = x + |x| over the interval -π ≤ x ≤ π is given by

f(x) = a0/2 + Σ [an*cos(nωx) + bn*sin(nωx)]
= 0/2 + Σ (0)
= 0

Therefore, the Fourier series is f(x) = 0, which means the function f(x) can be approximated by zero.