In Problems 11 through 26, the values of a period  function  in one full period are given. find its Fourier series.

f(t) = 0 for -pi<=t<-pi/2, 1 for -pi/2<=t<=pi/2, 0 for pi/2<t<=pi

To find the Fourier series of the given periodic function \( f(t) \), defined as:

- \( f(t) = 0 \) for \( -\pi \leq t < -\frac{\pi}{2} \)
- \( f(t) = 1 \) for \( -\frac{\pi}{2} \leq t \leq \frac{\pi}{2} \)
- \( f(t) = 0 \) for \( \frac{\pi}{2} < t \leq \pi \)

we first need to identify the coefficients of the Fourier series, which are given by:

\[
a_0 = \frac{1}{T} \int_{-T/2}^{T/2} f(t) \, dt
\]

\[
a_n = \frac{1}{T} \int_{-T/2}^{T/2} f(t) \cos\left(\frac{2\pi nt}{T}\right) \, dt
\]

\[
b_n = \frac{1}{T} \int_{-T/2}^{T/2} f(t) \sin\left(\frac{2\pi nt}{T}\right) \, dt
\]

where \( T = 2\pi \) for our function.

### Step 1: Calculate \( a_0 \)

\[
a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) \, dt = \frac{1}{2\pi} \left( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dt \right)
\]
Calculating the integral:

\[
= \frac{1}{2\pi} \left( t \Big|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \right) = \frac{1}{2\pi} \left( \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \right) = \frac{1}{2\pi} \cdot \pi = \frac{1}{2}
\]

### Step 2: Calculate \( a_n \)

\[
a_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) \cos(t) \, dt
\]
Since \( f(t) = 1 \) from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \):

\[
= \frac{1}{2\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(t) \, dt
\]

Calculating this integral:

\[
= \frac{1}{2\pi} \left( \sin(t) \Big|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \right) = \frac{1}{2\pi} \left( \sin\left(\frac{\pi}{2}\right) - \sin\left(-\frac{\pi}{2}\right) \right) = \frac{1}{2\pi} \left( 1 - (-1) \right) = \frac{1}{2\pi} \cdot 2 = \frac{1}{\pi}
\]

### Step 3: Calculate \( b_n \)

\[
b_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) \sin(t) \, dt
\]

Since \( f(t) = 1 \) from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \):

\[
= \frac{1}{2\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(t) \, dt
\]

Calculating this integral:

\[
= \frac{1}{2\pi} \left( -\cos(t) \Big|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \right) = \frac{1}{2\pi} \left( -\cos\left(\frac{\pi}{2}\right) + \cos\left(-\frac{\pi}{2}\right) \right) = \frac{1}{2\pi} (0 - 0) = 0
\]

### Step 4: Assemble the Fourier Series

Thus, the Fourier coefficients are:

- \( a_0 = \frac{1}{2} \)
- \( a_n = \frac{1}{\pi} \) for \( n = 1, 2, \ldots \)
- \( b_n = 0 \)

The Fourier series is given by:

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

### Conclusion

The Fourier series for the given function \( f(t) \) is:

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

If you have any questions or need further details, let me know! Here are some related questions you might find interesting:

1. How would the Fourier series change if the function were defined differently in one of the intervals?
2. Can you explain the significance of the Fourier coefficients in signal processing?
3. What is the convergence behavior of this Fourier series?
4. How would you compute the Fourier series for a piecewise linear function?
5. What role does the period of the function play in the Fourier series representation?

**Tip:** When working with Fourier series, always check the function's continuity and periodicity in the defined intervals, as this affects convergence.

In Problems 11 through 26, the values of a periodic function in one full period are given. Find its Fourier series.

Discover how to find the Fourier series for a periodic function defined with specific values in different intervals. This detailed solution covers the calculation of Fourier coefficients a_0, a_n, and b_n for the function f(t) over the interval [-π, π]. Ideal for advanced high school and college students.

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