Taylor series expansion of function, $f$, is a vector in the vector space with basis: $\{(x-a)^0, (x-a)^1, (x-a)^3, \ldots, (x-a)^n, \ldots\}$. This vector space has a countably infinite dimension. When $f$ is expressed as linear combination of the basis vector the scalar multiple for the $n$-th basis vector is $\operatorname{Diff}_n{f}(a)/n!$

Fourier series expansion of function, $f$, is a vector in the vector space with basis: $\{\sin(1x), \cos(1x), \sin(2x), \cos(2x), \ldots, \sin(nx), \cos(nx), \ldots\}$. This vector space has a countably infinite dimension. When $f$ is expressed as linear combination of the basis vector the scalar multiple for the $n$-th basis vectors are $\operatorname{Int}\{f\cdot\sin(nx)\}$ and $\operatorname{Int}\{f\cdot\cos(nx)\}$.


  1. The vector space for the Fourier series has an inner product, $\operatorname{Int}\{f\cdot g\}$, and it's this inner product that provides the above expressions like $\operatorname{Int}\{f\cdot\sin(nx)\}$ and $\operatorname{Int}\{f\cdot\cos(nx)\}$. Is there a similar inner product based derivation of the scalar multiples for the vector space of spanned by the polynomial basis in Taylor series?

  2. What is the relationship, if any, between the vector space produced by Taylor Series and that of Fourier Series? E.g. is one a subspace of the other?

  3. When Fourier series is taught, why isn't Taylor Series re-explained in the vector space framework used for Fourier series? And would this approach not lead the discussion of the implication of the choice of basis (and perhaps the choice of inner product) for function spaces?

  4. Just as Fourier series get generalized to Fourier Transform (the summation of the series becomes an integral), is there something equivalent to Taylor series?

  5. Are there any recommended resources (books, courses, etc.) available which can help clarify my thinking regarding these issues?

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  • As for your first question: Fourier series are designed to be performed on functions defined on bounded intervals, whereas Taylor series need not be. Thus we can define an inner product to extract the coefficients and odds are that we'll be get a convergent result out of it. However since Taylor series are defined on potentially unbounded intervals, it's harder to write an inner product as just $\int f(x) x^n dx$ (or something of that form) since the integrals may diverge. We can introduce weights in the integral to get convergent results, but that is a bit artificial. – Cameron Williams Jun 19 '13 at 22:50
  • As for your second question: Taylor series rely on your function being infinitely differentiable however the kinds of functions we can write down as a Fourier series is larger than that. We can even take the Fourier series of a piecewise discontinuous function however the Fourier series will not agree with the function everywhere. – Cameron Williams Jun 19 '13 at 22:53
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    Maybe [Connection between Fourier transform and Taylor series](http://math.stackexchange.com/questions/7301/connection-between-fourier-transform-and-taylor-series?rq=1) addresses some of your concerns. – Martin Jun 19 '13 at 23:25

2 Answers2


First, a general point. The big difference between Taylor and Fourier series is that Taylor series are local and Fourier series are global. That is, Taylor series are defined in terms of and capture local behavior of a function, whereas Fourier series are defined in terms of and capture global behavior. It is not quite accurate to think of a function as being just the sum of its Taylor series; even a function which has a Taylor series need not be equal to it locally (see, for example, MO), and certainly need not have any relationship to it globally, whereas every $L^2$ function on the circle is equal to the sum of its Fourier series in the $L^2$ (not pointwise) sense.

  1. Sort of. If you take the inner product to be something like $\int fg \, dx$, then the thing whose inner product with $f$ is $f(0)$ is not quite a function but a distribution, namely the Dirac distribution at $0$. The thing whose inner product with $f$ is $f^{(n)}(0)$, at least on a suitable class of functions $f$, is related to the distributional derivatives of the Dirac distribution.

  2. It depends. First, you shouldn't be treating them as bare vector spaces: you really want the language of topological vector spaces and the rest of functional analysis. Second, it depends on where you want the Taylor series to be defined. If we're talking about functions with convergent Taylor series on $\mathbb{R}$ then such functions don't have a Fourier series.

  3. The linear algebra analogy is much weaker for Taylor series than it is for Fourier series, the problem being again that Taylor series only capture behavior at a point, and generally functions are determined by their Taylor series to a much weaker extent than functions are determined by their Fourier series. (This is in general; in complex analysis there is a much tighter relationship between Taylor series and Fourier series, as described in the link Martin gives in the comments.)

  4. You shouldn't think of Fourier transforms as a generalization of Fourier series. They're both generalized by something called Pontrjagin duality, and I'm not aware of an analogous phenomenon for Taylor series.

  5. As mentioned in 2, you might want to learn some functional analysis.

Qiaochu Yuan
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I would add a comment, not an answer, but I don't have any reputation whatsoever. I also know it's an old thread, but it may help someone nowadays.

Regarding the 4th point, I would recommend looking for the Mellin tranform. It's widely used in analytic number theory. Just as the Fourier tranform is closely related to the coeficients of a fourier series, the Mellin transform is related to the coefficient of a Taylor series, through Ramanujan's interpolation theorem.