Given a vector and an orthonormal basis, it is easy to determine the components of the vector in the given basis. For example, if \begin{equation} \FF = F_x \,\xhat + F_y \,\yhat + F_z \,\zhat \end{equation} then of course \begin{equation} F_x = \FF\cdot\xhat . \end{equation} Put differently, \begin{equation} \FF = (\FF\cdot\xhat)\,\xhat + (\FF\cdot\yhat)\,\yhat + (\FF\cdot\zhat)\,\zhat . \end{equation} All we need to make this idea work is an orthonormal basis, that is, a set of mutually orthogonal elements that are unit normed.

But that's exactly what we have with the trig functions considered in the previous section! Our “dot product” is the integral from $0$ to $2\pi$! And our orthogonal elements consist of the trig functions $\{\sin(m\theta),\cos(n\theta)\}$ with $m$, $n$ positive integers, together with the constant function. Furthermore, we can normalize these functions, leading to the orthonormal basis $\{\frac{1}{\sqrt{2\pi}},\frac{1}{\sqrt\pi}\cos(m\theta), \frac{1}{\sqrt\pi}\sin(m\theta)\}$ where $m$ ranges over all positive integers. This basis is infinite! But what vector space does it span?

It turns out that any periodic function can be expanded uniquely in terms of this basis. Since the period might be $L$ rather than $2\pi$, we replace the basis above by $\{\frac{1}{\sqrt L},\sqrt{\frac{2}{L}}\cos(m\theta), {\sqrt\frac{2}{L}}\sin(m\theta)\}$, and of course integrate from $0$ to $L$. We then have:

Theorem: If $f(x+L)=f(x)$, then we have \begin{equation} f = \frac12 a_0 + \sum_{m=1}^\infty a_m \cos\left(\frac{2\pi m x}{L}\right) + \sum_{m=1}^\infty b_m \sin\left(\frac{2\pi m x}{L}\right) \end{equation} where the coefficients $a_m$, $b_m$ are uniquely determined.

This series expression for $f$ is called its Fourier series. Notice that the arguments of the trig functions have been rescaled to have period $L$, and that the normalization constants have been absorbed into the coefficients.

Finally, it is important to realize that any function on an interval such as $0\le x\le L$ can be treated as being periodic! Simply repeat the function in blocks of length $L$. This periodic function may only agree with the original function on the given interval, but that's often good enough. And there is no differentiability condition in this theorem; all that is necessary is for the integrals used (in the next section) to determine the coeffecients be well-defined.