# Infinite series vs compact representation

I understand the attractiveness and usefulness of infinite-series expansions such as Taylor expansions, but I wonder if they sometimes hide important aspects of the described system.

For example, take $$F(x) = e^x$$, whose Taylor expansion is $$F(x) = \sum_n \frac{x^n}{ n!}.$$ Sure, we can approximate $$F(x)$$ using the first few terms of the Taylor expansion for low values of $$x$$, and that is often useful. However, $$F(x)$$ is actually an exponential function, so has properties that are not easily evident if F(x) is approximated by truncating a Taylor expansion-- such as the beautiful fact that $$\frac{d}{dx}F(x) = F(x)$$.

Similarly, a truncated Fourier series representation of the delta function can obscure the fact that a delta function has precisely zero width.

Though chaos theory isn't in my skill set, I imagine that a truncated infinite series representation of a chaotic system might fail to reveal the fact that infinitesimally narrow domains exist in which the system is actually stationary.

It seems that quantum field theory (QFT) makes extensive use of the Euler-Heisenberg effective Lagrangian, which uses only terms up to the second order in the field invariants. One of the dreams of some field theorists has been to show that massive particles are stationary structures that "condense" out of vacuum fields due to nonlinearities; but it seems that the existence of such structures might be completely hidden if the field Lagrangian is not complete "all the way to the end" of an infinite series representation.

I would like to know if there exists an exact Lagrangian for QFT. Presumably it would not be an infinite power series expansion; instead it would be a relatively compact expression like $$L = e^{H(f(E^2 – B^2), g(E \cdot B))}$$ where $$H$$ is a simple function and $$f$$ and $$g$$ are low-order functions, or “natural” functions like $$cos$$ or $$tanh$$, and would contain only a small handful of constants whose values need to be adjusted to fit constraints set by experimental results. A Lagrangian like this could still be represented as an infinite series, but could also be represented in a very compact non-series form; and in either case it would have only a finite number of adjustable constants.

• That seems quite broad. There are many QFTs, some of which are approximations of other QFTs. Do you have anything specific in mind? – knzhou Feb 26 at 15:39
• Are you aware that the actual Euler-Heisenberg Lagrangian for QED is a complicated integral involving hyperbolic cosines of square roots of the field invariants, not just a power series? See en.wikipedia.org/wiki/Euler–Heisenberg_Lagrangian But this compact representation is still just an approximation which takes only one-loop perturbative quantum effects into account. It has no adjustable constants because it is calculated from QED. – G. Smith Feb 26 at 17:48
• Are you asking if there are any exact results in QFT? – Qmechanic Feb 26 at 18:35
• @Qmechanic, No, I'm pretty sure that there can be exact solutions to equations that, themselves, are not exact. But when a Lagrangian is first represented as an arbitrary infinite-series function and then truncated, we can be quite sure a priori that the resulting field equations are only an approximation to reality, and usually only applicable in the case of weak fields. I'd like to know if there are plausible QFT Lagrangians that span the range from extremely weak to extremely strong fields. – S. McGrew Feb 27 at 2:30
• @DanYand, please expand your "in continuous space without relying on any series expansion" comment. – S. McGrew Feb 27 at 2:34