Fourier Representation of Odd and Even Functions

Symmetry-Based Simplification of Trigonometric Series

We begin by recalling the formal definitions of symmetry for functions in \(\mathbb{R}^2\).

Even Functions: A function is even if \(f(-x) = f(x)\). These functions are symmetric across the y-axis.

Odd Functions: A function is odd if \(f(-x) = -f(x)\). These functions are rotationally symmetric about the origin (unchanged by a 180° rotation).

The trigonometric Fourier Series on the interval \((-\pi, \pi]\) is defined as:

f(\theta) = a_0 + \sum_{n = 1}^{\infty}\left(a_n\cos\left(n\theta\right) + b_n\sin\left(n\theta\right)\right)

Where the coefficients are calculated as:

a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\cos(n\theta)\,d\theta, \quad b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\sin(n\theta)\,d\theta

Even Functions: The Cosine Series

If \(f(\theta)\) is even, it is comprised solely of cosine waves. Since cosine is even and sine is odd, the product properties of symmetry dictate the result.

a_n = \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

For the sine coefficients \(b_n\), we analyze the integral over the full period:

\[ b_n = \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)

By substituting \(\theta \to -\theta\) in the second integral and noting that \(f\) is even while \(\sin\) is odd, the terms cancel exactly:

\int_{-\pi}^{0}f(\theta)\sin(n\theta)\,d\theta = -\int_{0}^{\pi}f(\theta)\sin(n\theta)\,d\theta \implies b_n = 0

Result: For an even function, the series reduces to:

f(\theta) = a_0 + \sum_{n=1}^{\infty}a_n\cos(n\theta)

Odd Functions: The Sine Series

Conversely, if \(f\) is odd, its Fourier representation consists only of sine waves. Following the same logic for the cosine coefficients \(a_n\):

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)

Because \(f(-\theta) = -f(\theta)\) and \(\cos(-n\theta) = \cos(n\theta)\), the integral of the product is zero:

\int_{0}^{\pi}f(-\theta)\cos(-n\theta)\,d\theta = -\int_{0}^{\pi}f(\theta)\cos(n\theta)\,d\theta \implies a_n = 0

Result: For an odd function, the series reduces to:

f(\theta) = \sum_{n = 1}^{\infty}b_n\sin\left(n\theta\right)

This symmetry property is not just a theoretical convenience; it allows engineers to skip half the calculations when processing signals with known symmetries, such as square or triangle waves.

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