Consider the following function:
f(t) = -t + 2pi; -2pi < t <0
f(t) = -t - 2pi; 0 < t < 2pi
f(t) = f (t+4pi)
Find its Fourier Series.

To find the Fourier series for the given function \( f(t) \), we will need to calculate its Fourier coefficients, which include both the \( a_n \) (cosine) and \( b_n \) (sine) terms, along with the constant \( a_0 \).

### 1. **Restating the Function:**
The function is periodic with period \( 4\pi \) since it satisfies \( f(t) = f(t + 4\pi) \). The piecewise definition of the function over one period is:

\[
f(t) = \begin{cases} 
-t + 2\pi & \text{for} \quad -2\pi < t < 0 \\
-t - 2\pi & \text{for} \quad 0 < t < 2\pi 
\end{cases}
\]

The function is odd, as can be seen from the symmetry:
\[
f(-t) = -f(t).
\]
Because the function is odd, there will be no cosine terms in the Fourier series (\( a_n = 0 \) for all \( n \)).

### 2. **Fourier Series Formula:**
For an odd function with period \( 4\pi \), the Fourier series will take the form:
\[
f(t) = \sum_{n=1}^{\infty} b_n \sin\left( \frac{n\pi t}{2\pi} \right)
\]
where the sine coefficients \( b_n \) are given by:
\[
b_n = \frac{2}{4\pi} \int_{-\pi}^{\pi} f(t) \sin\left(\frac{n\pi t}{\pi} \right) dt
\]

Consider the following function: f(t) = -t + 2pi; -2pi < t <0 f(t) = -t - 2pi; 0 < t < 2pi f(t) = f(t+4pi) Find its Fourier Series.

This problem focuses on finding the Fourier Series of a periodic, odd function defined piecewise. The solution includes key concepts of Fourier analysis and the computation of sine coefficients, covering topics suitable for undergraduate students in advanced calculus or mathematical physics.

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