For a function $f(x)$, it is possible to write it as a taylor series centered around a point $x=a$:


(Of course, there's a lot more mathematical nuance to Taylor Series expansion, I just want to lay it out loosely here as a basis for my intuition.)

I'm wondering if there's anyway to apply this to functionals, that is, a functional $F[f(x)]$ that maps the function $f(x)$ to an output. Is there a way to "rewrite" a functional as a series such as this:


Where $\phi(x)$ is a function that acts analogously to the point $x=a$ in a Taylor expansion.

(Again, I'm using all of my terminology and notation pretty loosely here. I'm not going for robust mathematical rigoorousness; I just want to express my intuition behind this idea.)

Is this "Functional expanded as a series" idea a thing? What is it called? Does it have any applications?

  • 1
    $\begingroup$ A functional is a linear transform. In a fair and just world, the first derivative (with respect to what, exactly?) should be constant. This seems like a relatively uninteresting Taylor series. I suppose that if your function space were separable, you could write a functional out as sum over a basis, but this expansion wouldn't be unique. This comes down to Gram-Schmidt. $\endgroup$
    – Xander Henderson
    Commented Mar 28, 2018 at 2:47
  • 3
    $\begingroup$ Functionals need not be linear. They simply map from a space to the complex numbers. $\endgroup$
    – Mark Viola
    Commented Mar 28, 2018 at 2:54
  • $\begingroup$ @MarkViola or to a vector of complex numbers, if it's a vector-valued functional, like $F[x(t)] \rightarrow \vec{v}$ or a function if it's a function-valued functional, like $F[x(t)] \rightarrow f(t)$, or a functional if it's a functional-valued functional, like $F[x(t)] \rightarrow G[y(r)]$, the possibilities are endless! $\endgroup$ Commented Jan 3, 2022 at 6:01

3 Answers 3


Indeed there is. This is used in calculus of variation. Commonly up to and including order two. See https://en.wikipedia.org/wiki/Functional_derivative

  • $\begingroup$ I've looked over this article before, and it doesn't seem to make any mention of this. Could you tell me where exactly it does? $\endgroup$
    – Sam
    Commented Mar 29, 2018 at 13:22
  • $\begingroup$ Yes, those Wikipedia article only gives a glimpse. What you want to look for are texts mentioning first and second variations (corresponding to first and second derivatives). See here for example: math.uconn.edu/~gordina/NelsonAaronHonorsThesis2012.pdf, Then you can also look for texts within: calculus of variation for finite element theory and weak form formulations of PDEs, quantum field theory and the principle of least action, as well as anything about the Euler–Lagrange equation. $\endgroup$
    – Jap88
    Commented Mar 30, 2018 at 0:41

It seems counter intuitive, but if you forget functionals (That is functions from functions to numbers) and just look at functions from functions to functions) you can build a taylor series like idea somewhat intuitively. I'm still working out the details of it, but i have been able to use it to show for example

$$ f(x+1) = f + f' + \frac{1}{2!}f''+\frac{1}{3!}f''' + ... $$

See here: How to build taylor series for infinite dimensional objects? for how to construct such series and for a list of problems that I encounter.

Somehow, most ideas in calculus generalize very cleanly in the context of "functions of functions" but fail to generalize EASILY (not saying they dont at all) in the context of functionals.


The expansion of a functional around a function is drastically more complicated than suggested in the question and in other answers here so far.

Wheras the Taylor series of $f(x)$ around point $x_0$ is written simply as:

\begin{alignat}{2} f(x) &= \sum_{n=0}^\infty c_n(x-x_0)^n &~~~,~~~c_n = \frac{f^{(n)}(x_0)}{n!}\tag{1}~, \end{alignat}

a Volterra series expansion of $f[x(r)]$ around the function $x_0(r)$ is significantly more complicated, as it requires integration over a dummy-variable version of $r$, an additional time for each successive term in the series):

\begin{align} f[x(r)] &= \sum_{n=0}^\infty \int \int \cdots \int c_n\left(\prod_{i=0}^n \left(x(r)-x_0(r^{(i)})\right) \right) \textrm{d}r^\prime \textrm{d}r^{\prime \prime} \cdots \textrm{d}r^{(n)}\tag{2}\\ &=\int \sum_{n=0}^\infty c_n \prod_{i=0}^n \left(x(r)-x_0(r^{(i)})\right) \textrm{d}r^{(i)}\tag{3}\\ c_n &=\frac{f^{(n)}[x_0(r)]}{n!}\tag{4}\\ &= \frac{1}{n!}\frac{\delta f[x_0(r)]}{\delta f[x(r^\prime)]\delta f[x(r^{\prime\prime})]\cdots \delta f[x(r^{(n)})]}. \tag{5}\\ \end{align}

Eq. 2 has been used in practical applications, for example in this 1994 paper in which $r$ represented a 3-dimensional vector so each integral in Eq. 2 was in fact a triple integral. Chapter 5 of this classical field theory book (PDF) by Nicholas Wheeler also has a lot of information and analogies about the above expansions.

I'll write the expansion out for you one more time in a way that might make it even more clear to you how this expansion compares with the Taylor expansion with which you are probably very familiar:

$$\tag{6}{\small f[x(r)] = f[x_0(r)] + \int \frac{f^{(1)}[x_0(r)]}{1!}\left(x(r)-x_0(r^\prime\right))\textrm{d}r^\prime + \int\!\!\!\int \frac{f^{(2)}[x_0(r)]}{2!}\left(x(r)-x_0(r^\prime\right))^{(2)}\textrm{d}r^\prime\textrm{d}r{^{\prime\prime}} + \cdots,} $$

in which:

$$ \left(x(r)-x_0(r^\prime\right))^{(2)}\equiv\left(x(r)-x_0(r^\prime\right))\left(x(r)-x_0(r^{\prime\prime}\right)). $$

This expansion has been the subject of my first MathOverflow question (earlier today!) which was about how to extend the Volterra series to functionals of more arbitrary complex-valued functions much like the Laurent series extends the concept of a Taylor series to more arbitrary functions of complex variables:

I also answered today my first MathOverflow question, which was a 2.5 year old unanswered question about how to extend the Pade approximant in this way:

Some other Mathematics.SE questions which may interest you are:


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