We are now ready to find formulas for the Fourier coefficients $a_m$ and $b_m$. Using the idea outlined at the start of the previous section, the coefficient of each normalized basis element is just the “dot product” of that basis element with the original “vector”. In other words, each term in the expansion takes the form “($f$ dot $u$) $u$”, where $u$ is the normalized basis element, and “dot” now refers to an integral.

Let's start with the constant term. The normalized basis element is $\frac{1}{\sqrt{L}}$, so this term in the expansion should be \begin{equation} \left( \int_0^L \frac{1}{\sqrt{L}} \,f(x) \,dx \right) \frac{1}{\sqrt{L}} = \frac{1}{L} \int_0^L f(x) \,dx . \end{equation} Comparison with the form of the Fourier series in the theorem (in the previous section) shows that \begin{equation} a_0 = \frac{2}{L} \int_0^L f(x) \,dx . \label{a0} \end{equation} Thus, the term $a_0/2$ in the series correctly represents the average value of $f$ (on the given interval).

Proceeding similarly for the other basis elements, the term involving $\cos\left(\frac{2\pi mx}{L}\right)$ is \begin{equation} \left( \int_0^L \sqrt{\frac{2}{L}}\cos\left(\frac{2\pi mx}{L}\right) \,f(x)\,dx \right) \sqrt{\frac{2}{L}}\cos\left(\frac{2\pi mx}{L}\right) \end{equation} and comparison with the Fourier series yields \begin{equation} a_m = \frac{2}{L} \int_0^L \cos\left(\frac{2\pi mx}{L}\right) \,f(x)\,dx . \label{am} \end{equation} Similar reasoning leads to \begin{equation} b_m = \frac{2}{L} \int_0^L \sin\left(\frac{2\pi mx}{L}\right) \,f(x)\,dx . \label{bm} \end{equation}

The conventional factor of $1/2$ in the definition of $a_0$ can now be explained by noting that the factors preceding the integrals in (\ref{a0}), (\ref{am}), and (\ref{bm}) are all the same.

Expressions (\ref{a0}), (\ref{am}), and (\ref{bm}) can also be obtained by inserting the assumed form of the Fourier series into the integrals, thus determining the constant factors in front of the integrals without having first normalized the basis elements.