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Trigonometric Fourier Representation of Odd and Even Functions
We begin by recalling the definition of an even and odd function.\
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An even function has the property that
\[f(-x) = f(x)\]
In other words, this function is **symmetric across the y-axis**.\
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An odd function has the property that
\[f(-x) = -f(x)\]
In other words, this function is **rotationally symmetric** about the origin. That is, in \(\mathbb{R}^2\), the function can be rotated \(180^{\circ}\) and remain unchanged. \
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Now we will consider the trigonometric Fourier Series of a function on the circle \((-\pi, \pi]\).
\[f(\theta) = a_0 + \sum_{n = 1}^{\infty}\left(a_n\cos\left(n\theta\right) +b_n\sin\left(n\theta\right)\right)\]
\[a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(\theta)d\theta \ \ \ \ \ \ \ \ \ \ \]
\[a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\cos(n\theta)d\theta\]
\[b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\sin(n\theta)d\theta\]
I will show that a function that is even will be comprised of only cosine waves, as cosine is an even function. Similarly, I will show that an odd function will be comprised of only sine waves, as sine is an odd function.\
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Let \(f(\theta)\) be an even function, defined on \((-\pi, \pi]\). Then we know that \(f(-\theta) = f(\theta)\). Now let's consider its Fourier Series representation.
\[f(\theta) = a_0 + \sum_{n = 1}^{\infty}a_n\cos(n\theta) + b_n\sin(n\theta)\]
Then by property of \(f\) being even, we know that \(f(\theta) = f(-\theta)\), so let's analyze the Fourier coefficients.
\[a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\cos(n\theta)d\theta = \frac{1}{\pi}\int_{-\pi}^{\pi}f(-\theta)\cos(-n\theta)d\theta = \frac{2}{\pi}\int_{0}^{\pi}f(\theta)\cos(n\theta)d\theta\]
Above follows directly from substituting in \(f(\theta) = f(-\theta)\) and \(\cos(n\theta) = \cos(-n\theta)\), since both are even functions.\
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\[b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\sin(n\theta)d\theta = \frac{1}{\pi}\left(\int_{0}^{\pi}f(\theta)\sin(n\theta)d\theta + \int_{-\pi}^{0}f(\theta)\sin(n\theta)d\theta\right)\]
Note that
\[\int_{-\pi}^{0}f(\theta)\sin(n\theta)d\theta = \int_{0}^{\pi}f(-\theta)\sin(-n\theta)d\theta = -\int_{0}^{\pi}f(\theta)\sin(n\theta)d\theta\]
By the property that \(f\) is even, and \(\sin\) is odd. Substituting this in, we get
\[\frac{1}{\pi}\left(\int_{0}^{\pi}f(\theta)\sin(n\theta)d\theta -\int_{0}^{\pi}f(\theta)\sin(n\theta)d\theta\right) = \frac{1}{\pi}\cdot 0 = 0\]
Therefore, we see that
\[\forall n \in \mathbb{N} \text{ and } \forall \theta \in (-\pi, \pi] \ \ \ b_n = 0\]
Finally, our Fourier Series of our even function is simply
\[f(\theta) = a_0 + \sum_{n=1}^{\infty}a_n\cos(n\theta)\]
Now we will assume \(f\) is odd and show that its Fourier series is only comprised of sine waves. So, we have that
\[f(-\theta) = -f(\theta)\]
We will consider the trigonometric Fourier Series of a function on the circle \((-\pi, \pi]\).
\[f(\theta) = a_0 + \sum_{n = 1}^{\infty}\left(a_n\cos\left(n\theta\right) +b_n\sin\left(n\theta\right)\right)\]
\[a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(\theta)d\theta \ \ \ \ \ \ \ \ \ \ \]
\[a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\cos(n\theta)d\theta\]
\[b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\sin(n\theta)d\theta\]
Let's analyze \(a_n\) more closely,
\[a_n = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(\theta)\cos(n\theta)d\theta = \frac{1}{\pi}\left(\int_{0}^{\pi}f(\theta)\cos(n\theta)d\theta + \int_{-\pi}^{0}f(\theta)\cos(n\theta)d\theta\right)\]
Like before,
\[ \int_{-\pi}^{0}f(\theta)\cos(n\theta)d\theta = \int_{0}^{\pi}f(-\theta)\cos(-n\theta)d\theta\]
Plugging that in above we have,
\[a_n = \frac{1}{\pi}\left(\int_{0}^{\pi}f(\theta)\cos(n\theta)d\theta + \int_{0}^{\pi}f(-\theta)\cos(-n\theta)d\theta\right)\]
By the property that \(f(-\theta) = -f(\theta)\) and \(\cos(-n\pi) = \cos(n\pi)\), we have
\[\int_{0}^{\pi}f(-\theta)\cos(-n\theta)d\theta = -\int_{0}^{\pi}f(\theta)\cos(n\theta)d\theta\]
Making this final substitution, we have
\[a_n = \frac{1}{\pi}\left(\int_{0}^{\pi}f(\theta)\cos(n\theta)d\theta -\int_{0}^{\pi}f(\theta)\cos(n\theta)d\theta\right) = \frac{1}{\pi}\cdot 0\]
Therefore, we have shown that
\[a_n = 0\]
For any odd function on the circle \((-\pi, \pi]\). Finally,
\[f(\theta) =\sum_{n = 1}^{\infty}b_n\sin\left(n\theta\right)\]
We see that any odd function is comprised of only sine waves in its Fourier decomposition.