Fourier Series Expansion:
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Fourier Series:
A Fourier series is a way to represent a periodic function as a sum of sine and cosine functions. It is given by the equation:

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

where a0, an, and bn are the Fourier series coefficients, ω is the angular frequency (2π/T), and T is the period of the function.

Finding the Coefficients:
To find the coefficients, we can use the formulas:

an = (2/T) * ∫[f(x) * cos(nωx) dx]
bn = (2/T) * ∫[f(x) * sin(nωx) dx]

In this case, the period T is 1, so ω = 2π.

Calculating a0:
The coefficient a0 is given by:

a0 = (1/T) * ∫[f(x) dx]

Substituting the given function, we have:

a0 = (1/1) * ∫[π * sin(π * x) dx]
= π * ∫[sin(π * x) dx]
= -π * cos(π * x) / π
= -cos(π * x)

Calculating an:
The coefficient an is given by:

an = (2/T) * ∫[f(x) * cos(nωx) dx]

Substituting the given function, we have:

an = (2/1) * ∫[π * sin(π * x) * cos(n2πx) dx]
= (2π/nπ) * ∫[sin(π * x) * cos(n2πx) dx]
= (2/n) * ∫[sin(π * x) * cos(n2πx) dx]

Calculating bn:
The coefficient bn is given by:

bn = (2/T) * ∫[f(x) * sin(nωx) dx]

Substituting the given function, we have:

bn = (2/1) * ∫[π * sin(π * x) * sin(n2πx) dx]
= (2π/nπ) * ∫[sin(π * x) * sin(n2πx) dx]
= (2/n) * ∫[sin(π * x) * sin(n2πx) dx]

Calculating the Coefficients:
To calculate the coefficients, we need to evaluate the integrals. However, since the given function is odd, the integrals involving cosine will be zero. Only the integrals involving sine will contribute.

For an = (2/n) * ∫[sin(π * x) * cos(n2πx) dx], the integral will be zero.

For bn = (2/n) * ∫[sin(π * x) * sin(n2πx) dx], the integral will be non-zero only when n = 1.

Thus,