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Regarding the Fourier series expansion of a function, why are

The two representations --

  1. Knowing the function in physical space at a finite number of points
  2. Knowing a finite number of the Fourier coefficients

exactly equivalent?

I heard that such a problem referred to as "dual problem". Can someone explain what that is? Thanks.

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I never saw such a statement, but I wonder which Fourier coefficient can be found from the condition $f(\pi/4) = 1/\sqrt{2}$? – Ilya Sep 20 '11 at 12:12
up vote 2 down vote accepted

What you write is true for band-limited periodic functions. If you know that a function is periodic with period $T$ and its Fourier representation contains no frequencies higher than $\nu$, then the function is completely determined either by the Fourier coefficients for frequencies up to $\nu$ or by as many function values at points dividing any interval of length $T$ into equal intervals (where "as many" can be taken literally if you're dealing with a complex-valued function, and has to be spelled out appropriately if you're dealing with real-valued functions). This is related to the Nyquist–Shannon sampling theorem. You can transform between these two representations of the function using the discrete Fourier transform.

I've never seen this specifically referred to as a "dual problem"; with regard to viewing the relationship between a function and its Fourier transform as a duality, this section in Wikipedia and this one may be of interest to you.

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In the Fourier analysis circles formulas $\mathcal{F}f(-s) = \mathcal{F}^{-1}f(-s)$ and $\mathcal{F}^{-1}f(-t) = \mathcal{F}f(t)$ is referred to as duality of the transforms. But also in case of Fourier series the domain is the circle and the dual group is the integers so maybe that's the reason for the dual problem. – user13838 Sep 20 '11 at 12:40

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