Taylor expansion of functional

Question

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.

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

score 12 · Answer 1 · answered Oct 01 '13 at 12:35

12

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 $\Delta g(x)=g(x)-g(x_0)$

answered Oct 01 '13 at 12:35

max

427

3

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? – Sidharth Ghoshal Feb 02 '19 at 18:23
I don't understand why $g_0=g(x_0)$ ect is correct. Because this would suggest that $g$ is a function, and $g_0$ is a value ($g$ evaluated at a given $x_0$), which would give $f[g_0]$ no meaning. Have I missed something? As it would seem to me $g_0(x)$ is the whole function that is taken as the expansion point. – Jack Wetherell Apr 29 '20 at 23:03
1

I think @max wanted to say that $g_0 = g_0(x)$ is the function wrt which the expansion is calculated – Diego Gomez Aug 19 '20 at 14:56
arent the units wrong? For instance, from the 1st order term, (integral of delta_f dx) must have the same units as f. How does delta_f have units of 1/x? – Mondo Duke Aug 27 '22 at 06:32

score 4 · Answer 2 · edited Aug 21 '18 at 13:18

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

Its abstract reads:

A Taylor-series expansion of the density functional F[n] is constructed from the given expansion for the energy E[v]. The formalism is used to investigate the noninteracting kinetic-energy functional for one-dimensional systems. Furthermore, a formal analysis of the Taylor expansion of the density functional for an interacting electron system at finite temperature is given and the relation between density-functional theory and the field-theoretical approach to the many-particle problem is enlightened.

You can also find a chain rule to work through the high order functional derivatives that you will need.

Taylor expansion of functional

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