### Instruction:
Calculate the Fourier series representation of the function f(x) = x^2 on the interval [-π, π], providing step-by-step reasoning.


### Response:

To find the Fourier series representation of a function, we need to calculate its Fourier coefficients. The general formula for the Fourier series is:

f(x) = (a_0 / 2) + Σ[a_n * cos(nx) + b_n * sin(nx)]

where n goes from 1 to infinity and a_n and b_n are the Fourier coefficients.

The formulas for calculating the Fourier coefficients are as follows:

a_0 = (1 / π) * ∫[-π, π] f(x) dx

a_n = (1 / π) * ∫[-π, π] f(x) * cos(nx) dx

b_n = (1 / π) * ∫[-π, π] f(x) * sin(nx) dx

Now let's calculate these coefficients for our function f(x) = x^2.

First, we'll find a_0:

a_0 = (1 / π) * ∫[-π, π] x^2 dx
a_0 = (1 / π) * [(x^3 / 3)|(-π, π)]
a_0 = (1 / π) * [(π^3 - (-π)^3) / 3]
a_0 = (1 / π) * [2π^3 / 3]
a_0 = (2π^2 / 3)

Next, we'll find a_n:

a_n = (1 / π) * ∫[-π, π] x^2 * cos(nx) dx

To solve this integral, we can use integration by parts. Let u = x^2 and dv = cos(nx) dx. Then du = 2x dx and v = (1/n) * sin(nx).

Using integration by parts, we get:

∫[-π, π] x^2 * cos(nx) dx = (x^2 * (1/n) * sin(nx))|(-π, π) - ∫[-π, π] 2x * (1/n) * sin(nx) dx

The first term evaluates to zero since sin(nπ) = 0 for all integer values of n. Now we need to solve the second integral:

∫[-π, π] 2x * (1/n) * sin(nx) dx

Let u = x and dv = sin(nx) dx. Then du = dx and v = (-1/n) * cos(nx).

Using integration by parts again, we get:

∫[-π, π] 2x * (1/n) * sin(nx) dx = [(-2x / n^2) * cos(nx)]|(-π, π) + ∫[-π, π] (2 / n^2) * cos(nx) dx

The first term evaluates to zero since cos(nπ) = 1 or -1 for all integer values of n, and thus 2π - (-2π) = 0. The second integral also evaluates to zero because it's an integral of an odd function over a symmetric interval.

So, a_n = 0 for all n.

Finally, let's find b_n:

b_n = (1 / π) * ∫[-π, π] x^2 * sin(nx) dx

This integral is similar to the one for a_n, but with sin(nx) instead of cos(nx). We can use the same approach as before, using integration by parts twice. After doing so, we find that this integral also evaluates to zero:

b_n = 0 for all n.

Now we have all the Fourier coefficients:

a_0 = (2π^2 / 3), a_n = 0, and b_n = 0.

So the Fourier series representation of f(x) = x^2 on the interval [-π, π] is:

f(x) = (2π^2 / 3) / 2
f(x) = π^2 / 3

This result may seem surprising, but it's because the function x^2 is an even function and has only a constant term in its Fourier series representation.