# What is 'degrees of freedom' when using Fourier series to express a periodic waveform?

We can express any desired periodic waveform using Fourier series.

In the book I am studying from it's said:

'We see that with Fourier series, we can produce any desired periodic waveform and extract its wave number content (via the $$a_n$$ and $$b_n$$). But even though we have used an infinite number of plane wave components, we still evidently do not have enough “degrees of freedom” to produce a truly localized wave packet."

I was wondering what the 'degrees of freedom' in this context mean.

For my understanding a wave packet is:

The most general solution of the wave equation can be shown to be given by any (suitably differentiable) function of the form $$\phi(x,t)=f(x\pm vt)$$ since it satisfies:

$$(\pm v)^2 𝑓''(𝑥+vt)=\frac{\partial^2\phi(x,t)}{\partial t^2}=v^2 \frac{\partial^2 \phi(x,t)}{\partial x^2}=v^2 f''(x \pm vt).$$

This implies that any appropriate initial waveform, $$f(x)$$, can be turned into solution of $$\frac{\partial^2\phi(x,t)}{\partial t^2}=v^2 \frac{\partial^2 \phi(x,t)}{\partial x^2},$$ $$f''(x \pm vt),$$ which propagates to the right (-) or left (+) with no change in shape. Such solution is called wave packet.

But the usual concept of a wave packet is a single bump. This is impossible to make using a Fourier series. You can reduce the step $$\Delta k$$ between wavenumbers of the harmonics (thus increasing the amount of numbers—degrees of freedom), which will let you increase period and thus extend the "useful length" of the single wave packet. Only in the limit $$\Delta k\to0$$ will you get the desired single bump, and this limit corresponds to the continuous Fourier transform, rather than Fourier series. Here you'll have an uncountable infinity of degrees of freedom, and this amount is able to represent the single non-periodic wave packet.