Find the Fourier series representation of the function with period 2 given by $$ f(t)= \begin{cases}3 t & 0

We first need to find the Fourier coefficients a_0, a_n, and b_n. These are given by: $$ a_0 = \frac{1}{2} \int_{-1}^{1} f(t) dt $$ $$ a_n = \int_{-1}^{1} f(t) \cos(n\pi t) dt $$ $$ b_n = \int_{-1}^{1} f(t) \sin(n\pi t) dt $$ In all cases, we will need to split the integral into two pieces since the function is defined piecewise.

In order to find the coefficients, we first need to calculate the definite integrals: $$ a_0 = \frac{1}{2} \left[ \int_{0}^{1} 3t dt + \int_{1}^{2} 3 dt \right] $$ $$ a_n = \int_{0}^{1} 3t \cos(n\pi t) dt + \int_{1}^{2} 3 \cos(n\pi t) dt $$ $$ b_n = \int_{0}^{1} 3t \sin(n\pi t) dt + \int_{1}^{2} 3 \sin(n\pi t) dt $$

Now we solve the integrals: $$ a_0 = \frac{1}{2} \left[ \frac{3}{2}t^2|_0^1 + 3t|_1^2 \right] = \frac{1}{2} \left[\frac{3}{2} + 3\right] = \frac{9}{4} $$ $$ a_n = \left[ \frac{3}{n\pi} \sin(n\pi t) t |_0^1 - \frac{3}{(n\pi)^2} \cos(n\pi t)|_0^1 \right] + \left[-\frac{3}{n\pi} \sin(n\pi t)|_1^2 \right] = -\frac{3}{(n\pi)^2} \left[ 1 - \cos(n\pi) \right] $$ $$ b_n = \left[ -\frac{3}{n\pi} \cos(n\pi t) t|_0^1 +\frac{3}{(n\pi)^2} \sin(n\pi t)|_0^1 \right] + \left[ \frac{3}{n\pi}\cos(n\pi t)|_1^2 \right]= 0 $$ Since the sine function integrated over the function's interval results in the b_n coefficients equating to zero, this means that the function is even and the Fourier series will only consist of cosine terms.

Finally, we can write the Fourier series representation of the function: $$ f(t) = \frac{9}{4} - \sum_{n=1}^{\infty} \frac{3}{(n\pi)^2} \left[ 1 - \cos(n\pi) \right] \cos(n\pi t) $$ This is the Fourier series representation of the given function with period 2.