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Taylor series expansions assume that we can expand any "good enough" function in terms of its derivatives.

My question is, could we make something similar with integrals, the inverse operator of derivation? I mean, could we expand f as follows:?

$$f(x)=\sum_{n=0}c_n(D^{-1})^nf(0)$$

$$D^{-1}f=\int dx f(x)$$

riemannium
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  • Why not try a specific example? Let $\operatorname{f}(x) = \operatorname{e}^{-x^2}$. What do you get as your integral series? – Fly by Night Sep 13 '13 at 19:14
  • Antiderivatives of smooth functions are not unique. They belong to an equivalence class however. So if you were to define a series via antiderivatives, how would you handle which antiderivative to choose? Perhaps you could let your constant of integration be 0. However I'm not sure if this would lead anywhere productive. – Cameron Williams Sep 13 '13 at 19:15

1 Answers1

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In short "yes" you can use integrals to expand out your function. Check out Fourier series or more generally integral transforms.

In another direction, you can expand a function into a series with positive and negative powers and get a Laurent series. These are used in complex analysis to help compute tricky integrals.

Bill Cook
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