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\)
Section 17.4 Derivation of Fourier Coefficients
We are now ready to find formulas for the Fourier coefficients
\(a_n\) and
\(b_n\text{.}\) Using the idea outlined at the start of
Section 14.7 , the coefficient of each basis element can be obtained from the inner product, i.e. the analogue of the “dot product” of that basis element with the original “vector”.
Let’s start with the \(a_n\) coefficients for \(n\ne0\text{.}\) The corresponding basis elements are \(\cos\left(nx\right)\text{,}\) so we compute
\begin{align}
\cos\amp\left( nx\right) \cdot f(x)\notag\\
\amp= \int_0^{2\pi} \cos\left( nx\right) f(x) \,dx\notag\\
\amp= \int_0^{2\pi} \cos\left( nx\right) \left[
\frac12 a_0 + \sum_{m=1}^{\infty} a_m \cos\left( mx\right)\right.\notag\\
\amp\left.\qquad\qquad + \sum_{m=1}^{\infty}
b_m \sin\left( mx\right) \right] \,dx\notag\\
\amp= \frac12 a_0 \cancelto{0}{\int_0^{2\pi} \cos\left( nx\right)}\,dx\notag\\
\amp\qquad\qquad + \sum_{m=1}^{\infty} a_m
\cancelto{\pi\delta_{nm}}
{\int_0^{2\pi} \cos\left( nx\right)
\cos\left( mx\right) \,dx} \notag\\
\amp\qquad\qquad + \sum_{m=1}^{\infty} b_m
\cancelto{0}
{\int_0^{2\pi} \cos\left( nx\right)
\sin\left( mx\right) \,dx} \notag\\
\amp= \sum_{m=1}^{\infty} a_m \pi \delta_{mn}\notag\\
\amp= \pi a_n\tag{17.4.1}
\end{align}
where we see the squared norm of the basis function appear in the next-to-last line. In conclusion,
\begin{equation}
a_n
= \frac{1}{\pi} \int_0^{2\pi} \cos\left( nx\right) f(x) \,dx\text{.}\tag{17.4.2}
\end{equation}
Activity 17.2 . Finding Fourier Coefficients.
Use similar reasoning to obtain the formulas
\begin{equation}
a_0 = \frac{1}{\pi} \int_0^{2\pi} \cos\left( 0x\right)\, f(x) \,dx
= \frac{1}{\pi} \int_0^{2\pi} f(x) \,dx\tag{17.4.3}
\end{equation}
and
\begin{equation}
b_n
= \frac{1}{\pi} \int_0^{2\pi} \sin\left( nx\right) \,f(x)\,dx\tag{17.4.4}
\end{equation}
Solution .
For the \(a_0\) term, the corresponding basis element is just the constant function, so we compute
\begin{align}
\cos\amp\left( 0x\right) \cdot f(x)\notag\\
\amp= \int_0^{2\pi} \cancelto{1}{\cos\left( 0x\right)} f(x) \,dx\notag\\
\amp= \int_0^{2\pi} 1 \left[
\frac12 a_0 + \sum_{m=1}^{\infty} a_m \cos\left( mx\right)\right.\notag\\
\amp\left.\qquad\qquad + \sum_{m=1}^{\infty}
b_m \sin\left( mx\right) \right] \,dx\notag\\
\amp= \frac12 a_0 \int_0^{2\pi} \,dx\notag\\
\amp\qquad\qquad + \sum_{m=1}^{\infty} a_m
\cancelto{0}
{\int_0^{2\pi}
\cos\left( mx\right) \,dx} \notag\\
\amp\qquad\qquad + \sum_{m=1}^{\infty} b_m
\cancelto{0}
{\int_0^{2\pi}
\sin\left( mx\right) \,dx} \notag\\
\amp= \pi a_0\tag{17.4.5}
\end{align}
In conclusion,
\begin{equation}
\frac12 a_0
= \frac{1}{2\pi} \int_0^{2\pi} f(x) \,dx\text{.}\tag{17.4.6}
\end{equation}
Thus, the term \(a_0/2\) in the series represents the average value of \(f\) (on the given interval).
We leave the \(b_n\) calculation for the reader.
The conventional factor of
\(\frac12\) in the definition of
\(a_0\) can now be justified by noting that the factor preceding the integral in
(17.4.3) is the same as the factors preceding the integrals in
(17.4.2) and
(17.4.4) .
