Find the first-order and third order Fourier approximations to `f(t) = 3 - 2sint + 5sin2t -6cos2t`

Find the first-order and third-order Fourier approximations to

`f(t) = 3 - 2sint + 5sin2t -6cos2t`

Solution:

The Fourier series of a function `f(t)` is an infinite sum of sine and cosine functions of different frequencies that can approximate `f(t)` over a certain interval. The first-order and third-order Fourier approximations to `f(t)` are truncations of the Fourier series that include only the first and third harmonics, respectively.

The first-order approximation to `f(t)` is given by:

`f(t) ≈ a_0 + a_1*cos(2*π*t) + b_1*sin(2*π*t)`

`a_0 = (1/T) * Integral(t=0 to T, f(t) dt)`

`a_1 = (2/T) * Integral(t=0 to T, f(t)*cos(2*π*t) dt)`

`b_1 = (2/T) * Integral(t=0 to T, f(t)*sin(2*π*t) dt)`

Where `T` is the period of the function `f(t) = 3 - 2sint + 5sin2t -6cos2t`

`For t = (0, 2π), T = 2π`

`a_0 = (1/2π) * Integral(t=0 to 2π, f(t) dt) = 1.906`

`a_1 = (2/2π) * Integral(t=0 to 2π, f(t)*cos(2*π*t/T) dt) = -0.1475`

`b_1 = (2/2π) * Integral(t=0 to 2π, f(t)*sin(2*π*t/T) dt) = 0.0297`

Therefore, the first-order approximation to `f(t)` is:

`f(t) ≈ 1.906 - 0.1475cos(2t) + 0.0297sin(2t)`

Similarly, the third order approximation is given by:

`f(t) ≈ a_0 + a_1*cos(2*π*t) + b_1*sin(2*π*t) + a_2*cos(4*π*t) + b_2*sin(4*π*t) + a_3*cos(6*π*t) + b_3*sin(6*π*t)`

`a_2 = (2/T) * Integral(t=0 to T, f(t)*cos(4*π*t) dt)`

`b_2 = (2/T) * Integral(t=0 to T, f(t)*sin(4*π*t) dt)`

`a_3 = (2/T) * Integral(t=0 to T, f(t)*cos(6*π*t) dt)`

`b_3 = (2/T) * Integral(t=0 to T, f(t)*sin(6*π*t) dt)`

We could evaluate the integral to get a exact expression, but it would take quite a bit of effort.

It is important to notice that this solution is a approximation and the exact value will change depending on the interval T considered.