4.4 Sine and Cosine Series

4.4.1:Even Periodic Functions

You may have noticed by now that an odd function has no cosine terms in the Fourier series and an even function has no sine terms in the Fourier series. This observation is not a coincidence. Let us look at even and odd periodic function in more detail.

Recall that a function $f (t)$ is odd if $f (- t) = - f (t)$ . A function $f (t)$ is even if $f (- t) = f (t)$ . For example, $\cos (n t)$ is even and $\sin (n t)$ is odd. Similarly the function $t^{k}$ is even if $k$ is even and odd when $k$ is odd.

If $f (t)$ and $g (t)$ are both odd, then $f (t) + g (t)$ is odd. Similarly for even functions. On the other hand, if $f (t)$ is odd and $g (t)$ even, then we cannot say anything about the sum $f (t) + g (t)$ . In fact, the Fourier series of any function is a sum of an odd (the sine terms) and an even (the cosine terms) function.

In this section we consider odd and even periodic functions. We have previously defined the $2 L$ -periodic extension of a function defined on the interval $[ minus L, L ]$ . Sometimes we are only interested in the function on the range $[ 0, L ]$ and it would be convenient to have an odd (resp. even) function. If the function is odd (resp. even), all the cosine (resp. sine) terms will disappear. What we will do is take the odd (resp. even) extension of the function to $[ minus L, L ]$ and then extend periodically to a $2 L$ -periodic function.

Take a function $f (t)$ defined on $[ 0, L ]$ . On $open parenthesis minus L, L ]$ define the functions

$\begin{matrix} F_{o d d} (t) \overset{d e f}{=} {\begin{matrix} f (t) & i f & 0 \leq t \leq L, \\ - f (- t) & i f & - L < t < 0, \end{matrix} \\ F_{e v e n} (t) \overset{d e f}{=} {\begin{matrix} f (t) & i f & 0 \leq t \leq L, \\ f (- t) & i f & - L < t < 0 . \end{matrix} \end{matrix}$

Extend $F_{o d d} (t)$ and $F_{e v e n} (t)$ to be $2 L$ -periodic. Then $F_{o d d} (t)$ is called the odd periodic extension of $f (t)$ , and $F_{e v e n} (t)$ is called the even periodic extension of $f (t)$ . For the odd extension we generally assume that $f (0) = f (L) = 0$ .

Sine and Cosine Series

Let $f (t)$ be an odd $2 L$ -periodic function. We write the Fourier series for $f (t)$ . First, we compute the coefficients $a_{n}$ (including $n equals 0$ ) and get

$a_{n} = \frac{1}{L} \int_{- L}^{L} f (t) cos (\frac{n π}{L} t) d t = 0 .$

That is, there are no cosine terms in the Fourier series of an odd function. The integral is zero because $f (t) \cos (\frac{n π}{L} t)$ is an odd function (product of an odd and an even function is odd) and the integral of an odd function over a symmetric interval is always zero. The integral of an even function over a symmetric interval $[ minus L, L ]$ is twice the integral of the function over the interval $[ 0, L ]$ . The function $f (t) \sin (\frac{n π}{L} t)$ is the product of two odd functions and hence is even.

$b_{n} = \frac{1}{L} \int_{- L}^{L} f (t) \sin (\frac{n π}{L} t) d t = \frac{2}{L} \int_{0}^{L} f (t) \sin (\frac{n π}{L} t) d t .$

We now write the Fourier series of $f (t)$ as

$\sum_{n = 1}^{\infty} b_{n} \sin (\frac{n π}{L} t) .$

Similarly, if $f (t)$ is an even $2 L$ -periodic function. For the same exact reasons as above, we find that $b sub n equals 0$ and

$a_{n} = \frac{2}{L} \int_{0}^{L} f (t) \cos (\frac{n π}{L} t) d t .$

The formula still works for $n equals 0$ , in which case it becomes

$a_{0} = \frac{2}{L} \int_{0}^{L} f (t) d t .$

The Fourier series is then

$\frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos (\frac{n π}{L} t) .$

An interesting consequence is that the coefficients of the Fourier series of an odd (or even) function can be computed by just integrating over the half interval $[ 0, L ]$ . Therefore, we can compute the Fourier series of the odd (or even) extension of a function by computing certain integrals over the interval where the original function is defined.

Note

Let $f (t)$ be a piecewise smooth function defined on $[ 0, L ]$ . Then the odd periodic extension of $f (t)$ has the Fourier series

$F_{o d d} (t) = \sum_{n = 1}^{\infty} b_{n} \sin (\frac{n π}{L} t),$

where

$b_{n} = \frac{2}{L} \int_{0}^{L} f (t) \sin (\frac{n π}{L} t) d t .$

The even periodic extension of $f (t)$ has the Fourier series

$F_{e v e n} (t) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos (\frac{n π}{L} t),$

where

$a_{n} = \frac{2}{L} \int_{0}^{L} f (t) \cos (\frac{n π}{L} t) d t .$

The series $\sum_{n = 1}^{\infty} b_{n} \sin (\frac{n π}{L} t)$ is called the sine series of $f (t)$ and the series $\frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos (\frac{n π}{L} t)$ is called the cosine series of $f (t)$ . We often do not actually care what happens outside of $[ 0, L ]$ . In this case, we pick whichever series fits our problem better.

It is not necessary to start with the full Fourier series to obtain the sine and cosine series. The sine series is really the eigenfunction expansion of $f (t)$ using eigenfunctions of the eigenvalue problem $x^{″} + λ x = 0$ $x open parenthesis 0 close parenthesis equals 0$ , $x (L) = L$ . The cosine series is the eigenfunction expansion of $f (t)$ using eigenfunctions of the eigenvalue problem $x^{″} + λ x = 0$ , $x prime open parenthesis 0 close parenthesis equals 0$ , $x^{'} (L) = L$ . We could have, therefore, gotten the same formulas by defining the inner produ

$⟨ f open parenthesis t close parenthesis, g open parenthesis y close parenthesis ⟩ equals ∫ 0 L f open parenthesis t close parenthesis g open parenthesis t close parenthesis d t,$

and following the procedure of Section 4.2. This point of view is useful, as we commonly use a specific series that arose because our underlying question led to a certain eigenvalue problem. If the eigenvalue problem is not one of the three we covered so far, you can still do an eigenfunction expansion, generalizing the results of this chapter. We will deal with such a generalization in Chapter 5.

Example 2

Find the Fourier series of the even periodic extension of the function $f (t) = t^{2}$ for $0 less than or equal to t less than or equal to π$ .

Show solution

We want to write

$f (t) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos (n t),$

where

$a_{0} = \frac{2}{π} \int_{0}^{π} t^{2} d t = \frac{2 π^{2}}{3},$

and

$\begin{matrix} a_{n} = \frac{2}{π} \int_{0}^{π} t^{2} \cos (n t) d t = \frac{2}{π} {[t^{2} \frac{1}{n} \sin (n t)]}_{0}^{π} - \frac{4}{n π} \int_{0}^{π} t \sin (n t) d t \\ = \frac{4}{n^{2} π} {[t \cos (n t)]}_{0}^{π} + \frac{4}{n^{2} π} \int_{0}^{π} \cos (n t) d t = \frac{4 (- 1)^{n}}{n^{2}} . \end{matrix}$

Note that we have “detected” the continuity of the extension since the coefficients decay as $\frac{1}{n^{2}}$ . That is, the even extension of $t^{2}$ has no jump discontinuities. It does have corners, since the derivative, which is an odd function and a sine series, has jumps; it has a Fourier series whose coefficients decay only as $\frac{1}{n}$ .

Explicitly, the first few terms of the series are

$\frac{π^{2}}{3} - 4 \cos (t) + \cos (2 t) - \frac{4}{9} \cos (3 t) + \dots$

Application

Fourier series ties in to the boundary value problems we studied earlier. Let us see this connection in more detail.

Suppose we have the boundary value problem for $0 less than t less than L$ .

$x^{″} (t) + λ x (t) = f (t),$

for the Dirichlet boundary conditions $x (0) = 0, x (L) = 0$ . By using the Fredholm alternative (Theorem 4.1.2) we note that as long as $λ$ is not an eigenvalue of the underlying homogeneous problem, there exists a unique solution. Note that the eigenfunctions of this eigenvalue problem are the functions $\sin (\frac{n π}{L} t)$ . Therefore, to find the solution, we first find the Fourier sine series for $f (t)$ . We write $x$ also as a sine series, but with unknown coefficients. We substitute the series for $x$ into the equation and solve for the unknown coefficients. If we have the Neumann boundary conditions $x prime open parenthesis 0 close parenthesis equals 0$ and $x prime open parenthesis L close parenthesis equals 0$ , we do the same procedure using the cosine series.

Let us see how this method works on examples.

Example 3

Take the boundary value problem for $0 less than t less than 1$ ,

$x^{″} (t) + 2 x (t) = f (t),$

where $f (t) = t$ on $0 less than t less than 1$ , and satisfying the Dirichlet boundary conditions $x open parenthesis 0 close parenthesis equals 0$ and $x open parenthesis 1 close parenthesis equals 0$ . We write $f (t)$ as a sine series

$f (t) = \sum_{n = 1}^{\infty} c_{n} \sin (n π t) .$

Compute

$c_{n} = 2 \int_{0}^{1} t \sin (n π t) d t = \frac{2 (- 1)^{n + 1}}{n π} .$

We write $x (t)$ as

$x (t) = \sum_{n = 1}^{\infty} b_{n} \sin (n π t) .$

We plug in to obtain

$\begin{matrix} x^{″} (t) + 2 x (t) = \underset{x^{″}}{\underset{⏟}{\sum_{n = 1}^{\infty} - b_{n} n^{2} π^{2} \sin (n π t)}} + 2 \underset{x}{\underset{⏟}{\sum_{n = 1}^{\infty} b_{n} \sin (n π t)}} \\ = \sum_{n = 1}^{\infty} b_{n} (2 - n^{2} π^{2}) \sin (n π t) \\ = f (t) = \sum_{n = 1}^{\infty} \frac{2 {(- 1)}^{n + 1}}{n π} \sin (n π t) . \end{matrix}$

Therefore,

$b_{n} (2 - n^{2} π^{2}) = \frac{2 (- 1)^{n + 1}}{n π} .$

$b_{n} = \frac{2 (- 1)^{n + 1}}{n π (2 - n^{2} π^{2})} .$

That $2 minus n squared π squared$ is not zero for any $n$ , and that we can solve for $b_{n}$ , is precisely because $2$ is not an eigenvale of the problem. We have thus obtained a Fourier series for the solution

$x (t) = \sum_{n = 1}^{\infty} \frac{2 (- 1)^{n + 1}}{n π (2 - n^{2} π^{2})} \sin (n π t) .$

See Figure $2$ for a graph of the solution. Notice that because the eigenfunctions satisfy the boundary conditions, and $x$ is written in terms of the boundary conditions, then $x$ satisfies the boundary conditions.

Figure

2

: Plot of the solution of

x^{″} + 2 x = t

x open parenthesis 0 close parenthesis equals 0

x open parenthesis 1 close parenthesis equals 0

Example 4

Similarly we handle the Neumann conditions. Take the boundary value problem for $0 less than t less than 1$ ,

$x^{″} (t) + 2 x (t) = f (t),$

where again $f (t) = t$ on $0 less than t less than 1$ , but now satisfying the Neumann boundary conditions $x prime open parenthesis 0 close parenthesis equals 0$ and $x prime open parenthesis 1 close parenthesis equals 0$ . We write $f (t)$ as a cosine series

$f (t) = \frac{c_{0}}{2} + \sum_{n = 1}^{\infty} c_{n} \cos (n π t),$

where

$c_{0} = 2 \int_{0}^{1} t d t = 1$

and

$\begin{matrix} \frac{- 4}{π^{2} n^{2}} i f n o d d, \\ 0 i f n e v e n . \end{matrix}$

We also write $x (t)$ as a cosine series

$x (t) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos (n π t) .$

We plug in to obtain

$\begin{matrix} x^{″} (t) + 2 x (t) = \sum_{n = 1}^{\infty} [- a_{n} n^{2} π^{2} \cos (n π t)] + a_{0} + 2 \sum_{n = 1}^{\infty} [a_{n} \cos (n π t)] \\ = a_{0} + \sum_{n = 1}^{\infty} a_{n} (2 - n^{2} π^{2}) \cos (n π t) \\ = f (t) = \frac{1}{2} + \sum_{\underset{n o d d}{n = 1}}^{\infty} \frac{- 4}{π^{2} n^{2}} \cos (n π t) . \end{matrix}$

Therefore, $a = \frac{1}{2}$ and $a sub n equals 0$ for $n$ even ( $n greater than or equal to 2$ ) and for $n$ odd we have

$a_{n} (2 - n^{2} π^{2}) = \frac{- 4}{π^{2} n^{2}},$

$a_{n} = \frac{- 4}{n^{2} π^{2} (2 - n^{2} π^{2})} .$

The Fourier series for the solution $x (t)$ is

$x (t) = \frac{1}{4} + \sum_{\underset{n o d d}{n = 1}}^{\infty} \frac{- 4}{n^{2} π^{2} (2 - n^{2} π^{2})} \cos (n π t) .$

Adapted from Differential Equations for Engineers by Jiří Lebl (Oklahoma State University), hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 4.0. Changes were made. License: CC-BY-SA-4.0.