What kind of information is available in a Fourier series expansion of a real analytic function that is not (readily) available in a power series? When would one know to work with one over the other?

Elle Najt
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1 Answers1


It depends on how the Fourier series and the power series are related.

When you ave a real-analytic $2\pi$-periodic function $f:\>{\Bbb R}\to{\Bbb C}$ then it has a well convergent Fourier series $$f(t)=\sum_{k=-\infty}^\infty c_k\>e^{ikt}\quad(t\in{\Bbb R})\ ;\tag{1}$$ and on the other hand it can be developed into a Taylor series near $0$, say, of the form $$f(t)=\sum_{j=0}^\infty a_J t^j\qquad(|t|<\rho)\ .\tag{2}$$ The two series $(1)$ and $(2)$ have not much to do with each other. After all, who would guess that $$f(t):=\sum_{k=0}^\infty {(i t)^k\over k!}$$ represents a $2\pi$-periodic function whose Fourier series is simply $f(t)=e^{it}$.

If, however, you link the Fourier series $(1)$ to the Laurent series $$g(z):=\sum_{k=-\infty}^\infty c_k z^k$$ then there is certainly a connection: The function $g$ will be analytic in an annulus ${1\over\rho}<|z|<\rho$ for some $\rho>1$, and its values on $S^1\subset{\Bbb C}$ coincide with those of $f$.

See also the following answer: Connection between Fourier transform and Taylor series

Christian Blatter
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