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May 25, 2010 at 21:04 comment added Willie Wong Nope. The span will be the same.
May 25, 2010 at 20:48 comment added vilvarin It wouldn't even work if we extend the basis $p_j=p_(j+K)$, would it?(sorry)
May 25, 2010 at 20:19 vote accept vilvarin
May 25, 2010 at 19:50 comment added Willie Wong If you assume $f$ is in the span of $p_j$, then this is a non-issue, since the dimension is finite. If $f$ is not in the span of $p_j$, then it is also a non-issue, since trivially you cannot recover $f$ from a linear combination of $p_j$s. So I don't understand why this is a question.
May 25, 2010 at 19:14 comment added vilvarin thank you very much. The last question I have is if Fourier inversion theorem is valid for this basis. It is not true in general case, isn't it? Probably i need to check this
May 25, 2010 at 16:25 comment added Willie Wong .. I meant to write that $\langle f,p_j\rangle = \delta_{jk}a_{-1,K}$ and $a_{-1,K}p_K \neq f$ unless all other $a_{s,K} = 0$. Sorry for the typo.
May 25, 2010 at 16:24 comment added Willie Wong ... so consider $f(x) = 1$, the constant function on $[0,1]$. Compute $\langle f, p_j\rangle = \delta_{jK} a_{-1,0}$ (in your notation where numbering of $j = 1... K$. But obviously $a_{-1,0} p_0 \neq f$ unless all other $a_{s,0} = 0$. But if you have some a priori knowledge that all the functions you care about lies in the span of the $p_j$s, then you don't even need Fourier analysis: you are just working with finite dimensional linear algebra on some inner product space.
May 25, 2010 at 16:20 comment added Willie Wong It is a basis of a finite dimensional sub-space, so not all functions will be in the span of $p_j$. Recall that the Fourier coefficients can be defined as $\hat{f}(\xi) = \langle f(\cdot), \exp(2\pi i \xi \cdot)\rangle$. A general representation formula is that, for some Hilbert space $H$ with orthonormal bases $e_1,e_2, \ldots$ you can write $f = \sum \lambda_j e_j$ where $\lambda_j = \langle f,e_j\rangle$ (this essentially captures the Fourier transform and its inverse, completely analogous to the finite dimension inner product space case)...
May 25, 2010 at 14:31 comment added vilvarin I am wondering if it is possible to use basis $p_j$ instead of$\exp(2\pi ij)$ to define Fourier transform, inverse fourier transform and so on.
May 25, 2010 at 13:46 history answered Willie Wong CC BY-SA 2.5