0
$\begingroup$

I am going to copy/paste an old link to a question that I found, this is exactly what happened to me (Same book and everything), I think this is best, since there is little use writting it all over again when the link has everything I need to explain myslef. Sorry in advance if this is too sloppy.

Uniqueness of Fourier Coefficients

The thing is... Even though it is answered, I can´t seem to get the response: ''If the Fourier coefficients of f are all 0 then the integral of f multiplied by any trigonometric polynomial is 0. This follows by just taking linear combinations.''

Linear combinatios of what? Maybe its obvious, but I don´t understand... Sincerely, Thanks.

  • 0
    By definition we have $\hat{f}(k)=\int_0^1f(x)e^{-2\pi ikx}dx=0$. and note that any trignometric polynomial is a linear combination of finite elements in $\{e^{-2\pi ikx}\}_{k\in\mathbb{Z}}$. – 2017-01-31
  • 0
    Oh.... because cos(x)= (e**(ix)+e**(-ix))/2. Thanks... it seems so obvious now... – 2017-01-31
  • 0
    So what it means is that $\int_0^1 f(x) p(x)dx = 0$ whenever $p(x) = \sum_{n=-N}^N c_n e^{2i \pi n x}$ or equivalently $p(x)$ is a polynomial in $\sin x$ and $\cos(x)$ – 2017-02-01

0 Answers 0