Find the Fourier series of the function𝑓𝑥=-𝑘,  -2<𝑥<0𝑘,   0<𝑥 <2 P = 2l = 4, l = 2

The Fourier series of a function can be represented as:

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

where the coefficients a0, an, and bn are given by:

a0 = (1/L) ∫ from -L to L [f(x) dx]

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

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

Given the function f(x) = -k for -2 < x < 0 and k for 0 < x < 2, with period P = 2L = 4, so L = 2.

First, calculate a0:

a0 = (1/2) ∫ from -2 to 2 [f(x) dx] = (1/2) [∫ from -2 to 0 [-k dx] + ∫ from 0 to 2 [k dx]] = (1/2) [0 - (-2k) + 2k - 0] = (1/2) * 4k = 2k

Next, calculate an:

an = (1/2) ∫ from -2 to 2 [f(x) cos(nπx/2) dx] = (1/2) [∫ from -2 to 0 [-k cos(nπx/2) dx] + ∫ from 0 to 2 [k cos(nπx/2) dx]] = (1/2) [2k/nπ [sin(nπx/2)] from -2 to 0 - 2k/nπ [sin(nπx/2)] from 0 to 2] = 0 (since sin(nπ) = 0 for all integer n)

Finally, calculate bn:

bn = (1/2) ∫ from -2 to 2 [f(x) sin(nπx/2) dx] = (1/2) [∫ from -2 to 0 [-k sin(nπx/2) dx] + ∫ from 0 to 2 [k sin(nπx/2) dx]] = (1/2) [-2k/nπ [-cos(nπx/2)] from -2 to 0 + 2k/nπ [-cos(nπx/2)] from 0 to 2] = (1/2) [4k/nπ cos(nπ)] = 2k/nπ for n odd, 0 for n even

So, the Fourier series of the function is:

f(x) = k + Σ from n=1 to ∞ [2k/nπ sin(nπx/2) for n odd]