# Taylor expansion of functional

The question asks for the Taylor expansion of a functional. Thus, given a real functional $f(g(x))$, what is the Taylor expansion about a function $h(x)$. What if the function is multi-variate, e.g., $f(x_1,x_2,g(x_1,x_2))$?

I've searched the web for an answer to this, and haven't come up with anything definite. If someone could help, I'd be grateful.

• Where is your functional defined? – Davide Giraudo Jan 27 '12 at 20:53
• For simplicity consider R to R – Jorge Jan 27 '12 at 23:00
• Try to search for functional derivative. In essence, you have to write $f(h(x) + \epsilon \delta g(x))$ and then perform a normal Taylor expansion in $\epsilon$ setting $\epsilon=1$ at the end (but keeping $\delta g(x)$ small. – Fabian Jan 27 '12 at 23:09
• I was hoping for a little more elaboration. – Jorge Jan 28 '12 at 2:30

This is a physicist answer and is thus not expressed in the better mathematical "vocab", but the Taylor expansion around $x_0$ of a functional $f$ of a function $g(x)$, i.e., of $f[g(x)]$ noted $f[g]$ for simplicity is:

$f[g]=f[g_0]+\int dx \frac{\delta f[g_0]}{\delta g(x)}\Delta g(x)+\frac{1}{2!}\int dx dx^\prime \frac{\delta^2f[g_0]}{\delta g(x^\prime)\delta g(x)}\Delta g(x^\prime)\Delta g(x) + ...$

with $g_0=g(x_0)$ and $\Delta g(x)=g(x)-g(x_0)$

• What is the intuition in the "quadratic" term being $\frac{\delta^2 f}{\delta g(x') \delta g(x)}$ I would've expected to see: $\frac{\delta^2 f}{\delta^2 g(x) }$. Do you have a link of some kind on this? – frogeyedpeas Feb 2 '19 at 18:23

I think this paper may be helpful to you.

"Taylor-series expansion of density functionals," by Matthias Ernzerhof. Phys. Rev. A 50, 4593 – Published 1 December 1994