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Next: Even and Odd Extensions Up: Fseries_1 Previous: Example 6

Even and Odd Functions

A function $ f$ defined on $ {\mathbb{R}}$ is said to be even if $ f(-x) = f(x)$ for all $ x \in {\mathbb{R}}$. For example, $ f(x) = x^2$, $ g(x) = \cos x$ are even.

A function $ f(x)$ defined on $ {\mathbb{R}}$ is said to be odd if $ f(-x) = -f(x)$ for all $ x \in {\mathbb{R}}$. For example, $ f(x) = x^3$, $ g(x) = \sin x$ are odd.

The graph of an even function is symmetric with respect to the $ y$-axis. Therefore, for any $ a$, $ \displaystyle{\int_{-a}^{a} f(x)  dx  =  2\int_0^a f(x)  dx.
}$

The graph of an odd function is symmetric with respect to the origin. Therefore, for any $ a$, $ \displaystyle{\int_{-a}^{a} f(x)  dx  = 0.
}$

Note: There are functions that are neither odd nor even. For example, $ y(x) = ln x$, or $ g(x) = (x - 1)^2$.



One can verify the following rules of multiplication for even and odd functions:

even $ \times$ even = even
odd $ \times$ odd = even
even $ \times$ odd = odd



It follows that if $ f$ is even, then the function $ f(x) \cos \displaystyle{(\frac{n \pi x}{L})}$ is even and $ f(x) \sin \displaystyle{(\frac{n \pi x}{L})}$ is odd. Therefore, the Fourier coefficients for an even function $ f(x)$ have the form

$\displaystyle a_n = \displaystyle{ \frac{1}{L} \int_{-L}^L f(x) \cos \frac{n\pi x}{L}  dx  =  \frac{2}{L} \int_{0}^L f(x) \cos \frac{n\pi x}{L}  dx ,}$ (16)



and



$\displaystyle b_n = \displaystyle{\frac{1}{L} \int_{-L}^L f(x) \sin \frac{n\pi x}{L}  dx  = 0 .}$ (17)

Thus, the Fourier series of an even function has only cosine terms.

By an analogous argument for an odd function $ f$, we arrive to the conclusion that the Fourier series of an odd function has only sine terms.


next up previous
Next: Even and Odd Extensions Up: Fseries_1 Previous: Example 6
Matthias Neufang 2002-09-18