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

$$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a){(x-a)}^{n}}{n!}=f(a)+f'(a)(x-a)+\frac{f''(a)(x-a)^{2}}{2}+...$$

(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:

$$F[f(x)]=a_0+a_1(f(x)-\phi(x))+a_2(f(x)-\phi(x))^2+...$$

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?

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    $\begingroup$ 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
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    $\begingroup$ 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

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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

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  • $\begingroup$ 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
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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.

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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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