### Instruction: Let f(x) be a non-periodic function defined on the interval [-1,1] by
f(x) = x^3 - 2x,  -1 ≤ x ≤ 1

Find the Fourier coefficients for the above function, and use these coefficients to write the Fourier series expansion for f(x).### Response: To find the Fourier series expansion for f(x), we first need to make it periodic. We can do this by extending f(x) to be an odd function on the interval [-1, 1]. Since f(x) is already odd, we can simply extend it periodically to the entire real line with a period of 2.

Now, we can find the Fourier sine series for f(x) since it is an odd function. The Fourier sine series is given by:

f(x) = ∑[(b_n)sin(nπx/L)]

where L is half the period (in this case, L = 1), and b_n is the Fourier sine coefficient, given by:

b_n = (2/L)∫[f(x)sin(nπx/L)dx] from -L to L

For our function f(x) = x^3 - 2x, we have:

b_n = 2∫[(x^3 - 2x)sin(nπx)dx] from -1 to 1

To find b_n, we need to evaluate this integral. We can do this using integration by parts:

Let u = x^3 - 2x, dv = sin(nπx)dx
du = (3x^2 - 2)dx, v = -cos(nπx)/(nπ)

Now, we can evaluate the integral:

b_n = 2[-(x^3 - 2x)cos(nπx)/(nπ) - ∫(-cos(nπx)(3x^2 - 2)dx)] from -1 to 1

Let's evaluate the second integral using integration by parts again:

Let u = 3x^2 - 2, dv = -cos(nπx)dx
du = 6xdx, v = -sin(nπx)/(nπ)

Now, we have:

b_n = 2[-(x^3 - 2x)cos(nπx)/(nπ) + (3x^2 - 2)sin(nπx)/(nπ) - ∫(6xsin(nπx)/(nπ)dx)] from -1 to 1

The last integral can be evaluated using integration by parts one more time:

Let u = 6x, dv = sin(nπx)dx
du = 6dx, v = -cos(nπx)/(nπ)

Now, we get:

b_n = 2[-(x^3 - 2x)cos(nπx)/(nπ) + (3x^2 - 2)sin(nπx)/(nπ) + 6xcos(nπx)/(n^2π^2) + 6cos(nπx)/(n^3π^3)] from -1 to 1

Evaluating the expression at the limits, we get:

b_n = 2[-2(-1)^n/(nπ) + 2(-1)^n/(nπ) + 6(-1)^n/(n^3π^3) - 6(-1)^n/(n^3π^3)]

Simplifying, we have:

b_n = 4(-1)^n/(nπ) - 12(-1)^n/(n^3π^3)

Now, we can write the Fourier sine series for f(x):

f(x) = ∑[4(-1)^n/(nπ) - 12(-1)^n/(n^3π^3)]sin(nπx)

This is the Fourier series expansion for the given function f(x) = x^3 - 2x on the interval [-1, 1].