Find the Fourier Series for $f(x)=$ { $1$ from $-\frac{\pi}{2} < x < \frac{\pi}{2}$ , $-1$ from $\frac{\pi}{2} < x < \frac{3\pi}{2}$ }

Question

I'm trying to find the Fourier Series for the following function, but I'm having trouble at some point. I'm hoping someone can give me a hand...

$ f(x) = \begin{cases} 1 \text{,} & \text{if }\quad -\frac{\pi}{2} < x < \frac{\pi}{2}\\ -1 \text{,} & \text{if }\quad \frac{\pi}{2} < x < \frac{3\pi}{2}\\ \end{cases} $

Here is what I tried so far:

$f(x) = a_0 + \sum_{n=1}^{\inf}[a_n \cdot cos(n \cdot x) + b_n \cdot sin(n \cdot x)]$

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Part (1): Finding $a_0$

$a_0 = \frac{1}{2\pi} \cdot \int_{-\pi}^{\pi}f(x) dx$

$a_0 = \frac{1}{2\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(x)dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}f(x)dx]$

$a_0 = \frac{1}{2\pi} \cdot [(1) \cdot(\frac{\pi}{2}) - (1) \cdot(-\frac{\pi}{2}) + (-1) \cdot (\frac{3\pi}{2}) - (-1) \cdot(\frac{\pi}{2})]$

$a_0 = \frac{1}{2\pi} \cdot [\frac{\pi}{2} + \frac{\pi}{2} - \frac{3\pi}{2} + \frac{\pi}{2}]$

$a_0 = 0$

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Part (2): Finding $a_n$

$a_n = \frac{1}{\pi} \cdot \int_{-\pi}^{\pi}f(x) \cdot cos(n \cdot x) dx$

$a_n = \frac{1}{\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(x) \cdot cos(n \cdot x) dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}f(x) \cdot cos(n \cdot x) dx]$

$a_n = \frac{1}{\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(1) \cdot cos(n \cdot x) dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}(-1) \cdot cos(n \cdot x) dx]$

$a_n = \frac{1}{\pi} \cdot [\frac{1}{n} \cdot sin(n \cdot \frac{\pi}{2}) - \frac{1}{n} \cdot sin(n \cdot \frac{-\pi}{2}) + (-1) \cdot \frac{1}{n} \cdot sin(n \cdot \frac{3\pi}{2}) - (-1) \cdot \frac{1}{n} \cdot sin(n \cdot \frac{\pi}{2})]$

$a_n = \frac{1}{n\pi} \cdot [sin(n \cdot \frac{\pi}{2}) - sin(n \cdot \frac{-\pi}{2}) - sin(n \cdot \frac{3\pi}{2}) + sin(n \cdot \frac{\pi}{2})]$

since

$sin(n \cdot \frac{-\pi}{2}) = sin(n \cdot \frac{\pi}{2})$

and

$sin(n \cdot \frac{3\pi}{2}) = sin(n \cdot \frac{\pi}{2})$

then

$a_n = 0$

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Part (3): Finding $b_n$

$b_n = \frac{1}{\pi} \cdot \int_{-\pi}^{\pi}f(x) \cdot sin(n \cdot x) dx$

$b_n = \frac{1}{\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(x) \cdot sin(n \cdot x) dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}f(x) \cdot sin(n \cdot x) dx]$

$b_n = \frac{1}{\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(1) \cdot sin(n \cdot x) dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}(-1) \cdot sin(n \cdot x) dx]$

$b_n = \frac{1}{\pi} \cdot [\frac{1}{n} \cdot -cos(n \cdot \frac{\pi}{2}) - \frac{1}{n} \cdot -cos(n \cdot \frac{-\pi}{2}) + (-1) \cdot \frac{1}{n} \cdot -cos(n \cdot \frac{\pi}{2}) - (-1) \cdot \frac{1}{n} \cdot -cos(n \cdot \frac{3\pi}{2})]$

$b_n = \frac{1}{n\pi} \cdot [-cos(n \cdot \frac{\pi}{2}) + cos(n \cdot \frac{-\pi}{2}) + cos(n \cdot \frac{\pi}{2}) - cos(n \cdot \frac{\pi}{2})]$

since

$cos(n \cdot \frac{-\pi}{2}) = cos(n \cdot \frac{3\pi}{2})$

then

$b_n = 0$

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I can't figure out where did I missed something. I was able to do a similar problem, but in the case where $f(x)$ is defined in two sets from $-\pi < x < 0$ and $0 < x < \pi$, but in such case, some terms got simplified because of the presence of the $0$. But here, this doesn't happen...

Answer 1

Thanks to MyGlasses I was able to finish the exercise. I'll leave the full answer here for future reference.

$ f(x) = \begin{cases} 1 \text{,} & \text{if }\quad -\frac{\pi}{2} < x < \frac{\pi}{2}\\ -1 \text{,} & \text{if }\quad \frac{\pi}{2} < x < \frac{3\pi}{2}\\ \end{cases} $

$f(x) = a_0 + \sum_{n=1}^{\inf}[a_n \cdot cos(n \cdot x) + b_n \cdot sin(n \cdot x)]$

---

Part (1): Finding $a_0$

$a_0 = \frac{1}{2\pi} \cdot \int_{-\pi}^{\pi}f(x) dx$

$a_0 = \frac{1}{2\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(x)dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}f(x)dx]$

$a_0 = \frac{1}{2\pi} \cdot [(1) \cdot(\frac{\pi}{2}) - (1) \cdot(-\frac{\pi}{2}) + (-1) \cdot (\frac{3\pi}{2}) - (-1) \cdot(\frac{\pi}{2})]$

$a_0 = \frac{1}{2\pi} \cdot [\frac{\pi}{2} + \frac{\pi}{2} - \frac{3\pi}{2} + \frac{\pi}{2}]$

$a_0 = 0$

---

Part (2): Finding $a_n$

$a_n = \frac{1}{\pi} \cdot \int_{-\pi}^{\pi}f(x) \cdot cos(n \cdot x) dx$

$a_n = \frac{1}{\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(x) \cdot cos(n \cdot x) dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}f(x) \cdot cos(n \cdot x) dx]$

$a_n = \frac{1}{\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(1) \cdot cos(n \cdot x) dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}(-1) \cdot cos(n \cdot x) dx]$

$a_n = \frac{1}{\pi} \cdot [\frac{1}{n} \cdot sin(n \cdot \frac{\pi}{2}) - \frac{1}{n} \cdot sin(n \cdot \frac{-\pi}{2}) + (-1) \cdot \frac{1}{n} \cdot sin(n \cdot \frac{3\pi}{2}) - (-1) \cdot \frac{1}{n} \cdot sin(n \cdot \frac{\pi}{2})]$

$a_n = \frac{1}{n\pi} \cdot [sin(n \cdot \frac{\pi}{2}) - sin(n \cdot \frac{-\pi}{2}) - sin(n \cdot \frac{3\pi}{2}) + sin(n \cdot \frac{\pi}{2})]$

since

$sin(-x) = -sin(x) \implies sin(n \cdot \frac{-\pi}{2}) = -sin(n \cdot \frac{\pi}{2})$

and

$sin(n \cdot \frac{3\pi}{2}) = sin(n \cdot 2 \cdot \pi - n \cdot \frac{\pi}{2}) = sin(n \cdot 2 \cdot \pi) \cdot cos(n \cdot \frac{\pi}{2}) - cos(n \cdot 2 \cdot \pi) \cdot sin(n \cdot \frac{\pi}{2}) = -sin(n \cdot \frac{\pi}{2}) $

then

$a_n = \frac{1}{n\pi} \cdot [sin(n \cdot \frac{\pi}{2}) - (-sin(n \cdot \frac{\pi}{2})) - (-sin(n \cdot \frac{\pi}{2})) + sin(n \cdot \frac{\pi}{2})]$

$a_n = \frac{4}{n \cdot \pi}*sin(n \cdot \frac{\pi}{2})$

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Part (3): Finding $b_n$

$b_n = \frac{1}{\pi} \cdot \int_{-\pi}^{\pi}f(x) \cdot sin(n \cdot x) dx$

$b_n = \frac{1}{\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(x) \cdot sin(n \cdot x) dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}f(x) \cdot sin(n \cdot x) dx]$

$b_n = \frac{1}{\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(1) \cdot sin(n \cdot x) dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}(-1) \cdot sin(n \cdot x) dx]$

$b_n = \frac{1}{\pi} \cdot [\frac{1}{n} \cdot -cos(n \cdot \frac{\pi}{2}) - \frac{1}{n} \cdot -cos(n \cdot \frac{-\pi}{2}) + (-1) \cdot \frac{1}{n} \cdot -cos(n \cdot \frac{\pi}{2}) - (-1) \cdot \frac{1}{n} \cdot -cos(n \cdot \frac{3\pi}{2})]$

$b_n = \frac{1}{n\pi} \cdot [-cos(n \cdot \frac{\pi}{2}) + cos(n \cdot \frac{-\pi}{2}) + cos(n \cdot \frac{\pi}{2}) - cos(n \cdot \frac{3\pi}{2})]$

since

$cos(-x) = cos(x) \implies cos(n \cdot \frac{-\pi}{2}) = cos(n \cdot \frac{\pi}{2})$

and

$cos(n \cdot \frac{3 \pi}{2}) = cos(n \cdot 2\pi - n \cdot \frac{\pi}{2}) = cos(n \cdot 2\pi) \cdot cos(n \cdot \frac{\pi}{2}) - sin(n \cdot 2\pi) \cdot sin(n \cdot \frac{\pi}{2}) = cos(n \cdot \frac{\pi}{2})$

then

$b_n = \frac{1}{n\pi} \cdot [-cos(n \cdot \frac{\pi}{2}) + cos(n \cdot \frac{\pi}{2}) + cos(n \cdot \frac{\pi}{2}) - cos(n \cdot \frac{\pi}{2})]$

$b_n = 0$

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Solution plot

from matplotlib import pyplot as plt
import numpy as np

def f(x):
    return np.where(
        np.logical_and(-np.pi / 2.0 < x, x < np.pi / 2.0),
        1.0,
        -1.0
    )

def f_(x, N):
    a0 = 0.0

    r = a0
    for n in range(1,N):
        an = 4. / (n * np.pi) * np.sin(n * np.pi / 2.)
        bn = 0.0
        r += an * np.cos(n * x) + bn * np.sin(n * x)

    return r

x = np.linspace(-np.pi / 2.0, 3 * np.pi / 2.0, 100)

y = f(x)
plt.plot(x, y)

y = f_(x, N=3)
plt.plot(x, y, '--')
y = f_(x, N=10)
plt.plot(x, y, '--')
y = f_(x, N=100)
plt.plot(x, y, '--')

plt.show()