3
$\begingroup$

This question about the intuition behind the scaling property of the Fourier transform made me wonder about the corresponding notion for a Fourier series.

The Fourier transform of $f(ax)$ is $\frac{1}{|a|} \mathcal{F(\frac{u}{a})}$. If $a>1$ then the graph of $f(ax)$ is $f$ compressed and so its Fourier transform has frequencies that are higher. However they are scaled down in magnitude.

On the other hand take $f(x)= \cos x$. Its fourier series is trivially itself, with the coefficient of $\cos x$ being $1$. Scaling it to $f(ax) = \cos ax$ where $a>1$ is an integer still has a trivial fourier series, and the coefficient of $\cos ax$ is $1$. This is unlike the Fourier transform where the "coefficients" of each frequency get scaled as the formula shows: $\frac{1}{|a|} \mathcal{F(\frac{u}{a})}$.

Is there a conceptual way to explain this discrepancy?

  • 0
    What discrepancy? If you transform $\cos ax$ you'll get the same effect. The transform *acts on* functions and the series expansion *represents* functions; these are two different concepts. The effect is seen in the former but not the latter because the former by default is a matter of change whereas the latter is by default a matter of identity.2012-02-05
  • 0
    I don't understand your comment. Can you elaborate in detail.2012-02-06

3 Answers 3