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tentukan deret fourier dari fungsi periodik ini

We are tasked with finding the Fourier series of a given periodic function \( f(t) \). The function is defined as:

\[
f(t) = 
\begin{cases}
2(1 + t), & -1 < t < 0 \\
0, & 0 < t < 1
\end{cases}
\]
Additionally, we are given that \( f(t + 2) = f(t) \), meaning the function has a period of \( T = 2 \).

### Fourier Series General Formula

For a periodic function with period \( T \), the Fourier series is given by:

\[
f(t) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos \left( \frac{2 \pi n t}{T} \right) + b_n \sin \left( \frac{2 \pi n t}{T} \right) \right)
\]

Where the Fourier coefficients \( a_0 \), \( a_n \), and \( b_n \) are defined as follows:

\[
a_0 = \frac{2}{T} \int_0^T f(t) \, dt
\]

\[
a_n = \frac{2}{T} \int_0^T f(t) \cos \left( \frac{2 \pi n t}{T} \right) \, dt
\]

\[
b_n = \frac{2}{T} \int_0^T f(t) \sin \left( \frac{2 \pi n t}{T} \right) \, dt
\]

Given that \( T = 2 \), we will compute these coefficients over the interval \( t \in [-1, 1] \).

#### Step 1: Compute \( a_0 \)
\[
a_0 = \frac{1}{1} \left( \int_{-1}^0 2(1+t) \, dt + \int_0^1 0 \, dt \right)
\]

#### Step 2: Compute \( a_n \)
\[
a_n = \frac{1}{1} \left( \int_{-1}^0 2(1+t) \cos(n \pi t) \, dt + \int_0^1 0 \cdot \cos(n \pi t) \, dt \right)
\]

#### Step 3: Compute \( b_n \)
\[
b_n = \frac{1}{1} \left( \int_{-1}^0 2(1+t) \sin(n \pi t) \, dt + \int_0^1 0 \cdot \sin(n \pi t) \, dt \right)
\]

Let me now compute these integrals.Here are the results for the Fourier coefficients:

1. **\( a_0 \)**: 
   \[
   a_0 = 1
   \]
   This represents the average value of the function over one period.

2. **\( a_n \)** (Fourier cosine coefficients):
   \[
   a_n = \frac{2(1 - \cos(n\pi))}{\pi^2 n^2} \quad \text{for} \, n \neq 0
   \]
   \[
   a_n = 1 \quad \text{for} \, n = 0
   \]
   Since \( \cos(n\pi) = (-1)^n \), the formula simplifies based on whether \( n \) is even or odd.

3. **\( b_n \)** (Fourier sine coefficients):
   \[
   b_n = \frac{2(-\pi n + \sin(n\pi))}{\pi^2 n^2} \quad \text{for} \, n \neq 0
   \]
   Since \( \sin(n\pi) = 0 \), this simplifies to:
   \[
   b_n = \frac{-2n}{\pi n^2} = \frac{-2}{\pi n}
   \]

Thus, the Fourier series for \( f(t) \) is:

\[
f(t) = \frac{1}{2} + \sum_{n=1}^{\infty} \left( \frac{2(1 - (-1)^n)}{\pi^2 n^2} \cos(n \pi t) - \frac{2}{\pi n} \sin(n \pi t) \right)
\]

Would you like further details or explanations?

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Here are 5 related questions you might find useful:

1. How does the Fourier series converge for discontinuous functions?
2. What is the physical interpretation of Fourier coefficients in signal processing?
3. Can we find the Fourier transform for this function instead of the series?
4. How would the Fourier series change if the period \( T \) were different?
5. How do the Fourier sine and cosine terms relate to the symmetry of the function?

**Tip:** Fourier series are useful for analyzing periodic signals in fields like engineering, physics, and signal processing.

Find the Fourier series for a given periodic function f(t) defined by piecewise intervals: f(t) = 2(1 + t) for -1 < t < 0, and f(t) = 0 for 0 < t < 1. This solution explains the Fourier series formula, how to compute Fourier coefficients, and provides the final series representation. Suitable for undergraduate mathematics students.

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