4 ms·
> It is clear that a non-constant polynomial cannot be expressed as a finite sum of continuous periodic functions, But it can be expressed as an infinite sum o
by losvedir 13y ago
> It is clear that a non-constant polynomial cannot be expressed as a finite sum of continuous periodic functions,
But it can be expressed as an infinite sum of continuous periodic functions, right? I seem to remember that you could use all the sine functions as bases for the vector space of functions and (almost?) any function could be expressed as an infinite sum of sines.
It's been a while since I've thought about these things, is that recollection correct?
- drob 13y agoYup! This is the basis for Fourier analysis: http://en.wikipedia.org/wiki/Fourier_analysis http://en.wikipedia.org/wiki/Fourier_analysis
- kayescercle 13y agoExcept that the author is referring to the entire domain (where a polynomial cannot be represented as a countably infinite sum of periodic functions), while a Fourier transform of a function has to be taken over a finite region like [a,b].
- deleted 13y ago[deleted]
- deleted 13y ago[deleted]
- gustavn4345 13y agoI'm pretty sure a Fourier _transform_ is over the entire domain while a Fourier _series_ is over a periodic function or finite region.
- ot 13y agoYes, but only on a finite open interval, that is (a, b) with a, b real numbers. In the post the sum of periodic functions is equal to x^2 on the whole real line. This is impossible with sums of continuous periodic functions, such as Fourier series, because continuous periodic functions are bounded, while x^2 is unbounded.
- vbuterin 13y ago> This is impossible with sums of continuous periodic functions, such as Fourier series, because continuous periodic functions are bounded, while x^2 is unbounded. But infinite sums of continuous periodic functions are not necessarily bounded: sin(x) + sin(x/2) + sin(x/4) + ... is an example. So there has to be something else to the proof.
- ot 13y agoYou are right, but I was referring to either finite sums or Fourier series, your example is not one because sum of the squares of the coefficients is infinite. I don't know what can be said about arbitrary countable/uncountable sums of continuous periodic functions.