For a periodic function $f(x)$ with period $p$, Fourier transform says that it can be "expressed" by a set of infinite many number.

$$ f(x)=\sum_{k=-\infty}^\infty F[k]e^{2\pi i kx/p} $$


$$ F[k]=\frac{1}{p}\int_{x=0}^pf(x)e^{-2\pi ikx/p}dx $$

Why the information of $f(x)$ in the an interval $[0,p]$ can be expressed by a set of infinity many numbers $F[k]$? It seems that there are something in common between the interval and all the integers, what is that common thing in mathematic language? Does that mean there are as many real numbers as in the interval $[0,1]$ as there are of integers?

  • 1
    $\begingroup$ cardinal(set of interesting functions in a interval) < cardinal(set of arbitrary functions in a interval). $\endgroup$ Commented Feb 7, 2014 at 19:01

1 Answer 1


Notice that it is not every periodic function. For the series to converge to the function (to give you back the function) in some sense, you must impose more conditions, say continuous for example.

Once you impose a condition like continuity then the fact that the function is determined but countable many numbers is not so surprising anymore. After all, a continuous function is determined by its values at the rational numbers, which form a countable set (a set as numerous as the integers).

  • $\begingroup$ Thanks for reminding me that. Could you put this into a common instead of an answer? $\endgroup$ Commented Feb 7, 2014 at 19:02
  • $\begingroup$ If 4 more people upvote I get to 50 and the n I can comment. $\endgroup$
    – user127038
    Commented Feb 7, 2014 at 19:03
  • $\begingroup$ Thanks for the update, that's very helpful. Could you explain why a continuous function is determined by its values at the rational numbers? $\endgroup$ Commented Feb 7, 2014 at 19:51
  • $\begingroup$ All the irrational numbers are limit points of rational points. Since it is continuous the value at the irrationals is determined by the limits at that point along the rationals. $\endgroup$
    – user127038
    Commented Feb 7, 2014 at 20:10

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