# DFT in complex form and Trigonometric Polynomial Interpolation: why different dimensions of basis vectors set?

I'm trying to figure out, how coefficients of Discrete Fourier Transform in complex form are converted into coefficients of Trigonometric Polynomials Interpolation:

Say, I have a function vector with 7 coordinates:

I can try to decompose it into Trigonometric Polynomials like this: $$f(x) = \frac{a_0}{2}+\sum_{l=1}^3 a_l \cos(l x) + b_l \sin(lx)$$

At the same time I can decompose it into DFT like this:

What bothers me is that 6-dimensional DFT vectors correspond to 7-dimensional Trigonometric Polynomials harmonics. I don't get, how I can establish a correspondence between trigonometric coefficients $$a_l, b_l$$ and DFT coefficients as there are 7 of first kind and 6 of the second, cause I can't use full f(x) 7-vector for DFT - I'll have to drop one coordinate. Or there's no correspondence in discrete case and it works only for seria?

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Given your spacing of the sample points and assuming $f(\pi)\ne f(-\pi)$ the "base period" of the underlying periodic function would be ${7\pi\over3}$ and not $2\pi$. – Christian Blatter Apr 23 '13 at 8:29
@ChristianBlatter You're right, my mistake. I should have taken function values in points: $$X=(−\frac{6 \pi}{7},−\frac{4 \pi}{7},−\frac{2 \pi}{7},0,\frac{2 \pi}{7},\frac{4 \pi}{7},\frac{6 \pi}{7}).$$ Then it all works. Thank you. – Bob Apr 23 '13 at 21:43