The coefficients

The idea behind the derivation of the Fourier coefficients is as follows: we start from the Fourier series expansion of a periodic function, $f(x)$, as in Eq. (1.3) and multiply the whole expression by a function that is orthogonal to the coefficient of $a_{n}$ (if we want to derive $a_{n}$ ) or $b_{n}$ (if we want to derive $b_{n}$ ). Then, the orthogonality property [see Eq. (1.9)] suggests that upon integration of the expression on the interval $[-\pi, \pi]$, all terms but one will vanish, allowing us to isolate the coefficients $a_{n}$ or $b_{n}$. In what follows, we determine $a_{n}$ in detail and summarise the method for $a_{0}$ and $b_{n}$.

Derivation

Determining $a_{n}$ where $n \geq 1$

Step 1: Multiply (1.3) by a set of functions that is mutually orthogonal to the coefficient of $a_{n}$ [i.e. $\cos (m x)$ ]. We therefore multiply both sides of (1.3) by $\cos (m x)$ :

Step 2: Integrate (1.17) between $-\pi$ to $\pi$ :

Since $f(x)$ is an arbitrary periodic function at this point, we leave the LHS of (1.18) as it is. The first term on the RHS (in blue) will be straightforward to deal with so what we are concerned with are the summation terms that are being integrated with respect to $x$ (shown in red). Let's deal with these next. It is possible to swap the order of the integral and the summation such that:

Note: This can always be done if the series is finite. There are certain conditions for which the swapping can be done for infinite series (this is related to Fubini's theorem, which will be discussed in Topic A2 with Prof. Matar). Here, we assume we can in fact integrate the sum term by term and hence, Eq. (1.18) becomes:

Evaluating each integral on the RHS,

Next,

and,

Again, the results in Eqs. (1.22) and (1.23) may be shown using the relations in (1.7a)-(1.7c). Substituting Eqs. (1.21)-(1.23) in Eq. (1.20), yields,

Finally, noticing that $m$ is simply an index letter, we can replace $m$ by $n$ yileding the Fourier coefficient, $a_{n}$ :

Determining $a_{0}$

We integrate (1.3) between $-\pi$ to $\pi$ and, using the orthogonality relations, we can show that the terms multiplying the $a_{n}$ and $b_{n}$ coefficients vanish leaving us with:

Solving for $a_{0}$ gives:

Determining $b_{n}$ where $n \geq 1$

Similarly to the method used to obtain $a_{n}$ (for $n \geq 1$ ), to determine $b_{n}$, we multiply (1.3) by $\sin m x$ and integrate between $-\pi$ and $\pi$. Applying orthogonality relations and upon simplifying, yields

Consider a function, $f(x)$ satisfying the periodicity condition [given by Eq. (1.4) ] then, its Fourier series is given by Eq. (1.3) and the coefficients $a_{0}, a_{n}$ and $b_{n}$ are determined using Eqs. (1.25), (1.27) and (1.28), respectively. Note that Eq. (1.27) may be recovered from (1.25) if $n=0$. These results are summarised in the box below.

Definition 1.3 The Fourier series of a periodic function $f(x)$ defined on $[-\pi, \pi]$ is given as:

where the Fourier coefficients are given by:

and

Even/odd coefficients

Given a periodic function, $f(x)$, we can represent it using the Fourier series expansion by calculating the coefficients derived in the previous section. The work involved may be reduced significantly if the function $f(x)$ is odd or even. If the function $f(x)$ is odd then the Fourier series will represent an expansion in sine modes only and therefore all even coefficients, $a_{n}$ are zero-valued:

If the function $f(x)$ is even, then the Fourier series will represent a series expansion of cosine modes only and therefore all odd coefficients, $b_{n}$ are zerovalued:

If the function is neither odd nor even then we need to compute $a_{0}$ and all coefficients $a_{n}$ and $b_{n}$. We look at an example next.

Example 1.1 Find the Fourier series of $f(x)=x$ defined in $[-\pi, \pi]$.

Solution For a function $f(x)$ to have a Fourier series representation, it must be periodic. We have the following periodicity condition: $f(x+2 \pi)=f(x)$. This implies that every $2 \pi$, the function repeats it self (see Sec. 1.4 for more details on periodic extension). We need to determine the even coefficients $a_{n}$ (for $n \geq 0$ ) and the odd coefficients $b_{n}$ (for $n \geq 1$ ). We check whether the function is odd, even or neither:

Since the function is odd, the even coefficients are zero $\rightarrow a_{n}=0, \forall n$. We only need to compute the odd coefficients, $b_{n}$. Using Eq. (1.31), we have:

but, $\cos (n \pi)=(-1)^{n}$ (this is a useful result to remember) and therefore $b_{n}$ are given by:

Substituting $a_{n}=0$ (for $n \geq 0$ ) and $b_{n}=\frac{2(-1)^{n+1}}{n}$ (for $n \geq 1$ ) in Eq. (1.29), we have:

Extension to arbitrary intervals

So far, we have looked at functions which are defined on an interval $[-\pi, \pi]$ with periodic extension $2 \pi$. We want to look at situations where the function is defined on an arbitrary interval $[-L, L]$, where $L \neq \pi, L>0$, with a periodic extension of $2 L$. We achieve this through a change of interval. If we introduce a new variable,

and define a new function, $g(t)$ such that $g(t)=f(x)$, then $g(t)$ is defined on $[-\pi, \pi]$. If $f(x)$ is periodic with period $2 L$, then $g(t)$ is periodic with period $2 \pi$. If $f(x)$ is piecewise continuous with continuous derivatives, so is $g(t)$. Hence, following Definition 1.3, the Fourier series of $g(t)$ is given by:

where the Fourier coefficients are given by:

and

Now, by undoing the change of variable, the formulae given in Eqs. (1.33)(1.35) become:

where,

and

The constants $a_{n}$ and $b_{n}$ (Eqs. (1.37) and (1.38), respectively) are the Fourier coefficients of $f(x)$ on $[-L, L]$ and Eq. (1.36) is the Fourier series of $f(x)$ on $[-L, L]$. Example 1.2 Find the Fourier series of the function $f(x)$ defined as:

Solution Observe that the interval is defined on [4,6] while Eqs. (1.36)(1.38) are defined on $[-L, L]$. We still use Eqs. (1.36)-(1.38) but we shift them to reflect the interval $[4,6]$. Let us first deal with the argument of the sine and cosine functions, i.e. $\frac{n \pi x}{L}$. The constant $L$ represents half the periodic extension so $L=\frac{1}{2}(6-4)=1$. The lower limit in Eqs. (1.37)(1.38) is $-L=-1$; since the left boundary is equal to 4 , we need to shift $x$ by 5 units to the right $\Rightarrow x-5$. With $L=1$ and the aforementioned shift, the argument becomes $n \pi(x-5)$. We have the following equations:

For $n=0, \quad a_{0}=\int_{4}^{5} 1 d x+\int_{5}^{6} 2 d x=3$,

For $n \geq 1, \quad a_{n}=\int_{4}^{5} \cos [n \pi(x-5)] d x+\int_{5}^{6} 2 \cos [n \pi(x-5)] d x$,

We distinguish between even and odd $n$ by setting $2 n$ for even and $2 n-1$ for odd. This allows us to include only the nonzero terms of the Fourier expansion as follows: