# With which operators does a Taylor series expansion commute?

With which operators does a Taylor series expansion commute? At first I was wondering about exponentiation, but now I am also wondering more generally.

Since an exponential itself can be written as a Taylor series, is it obvious that taking the exponential of a function could also be written as an exponential of that Taylor series? i.e., that it would have the same convergence properties?

EDIT: Suppose we use $T[f(x)]$ to denote the Taylor series expansion of a function $f(x)$, and $Exp[f(x)]$ to denote $e^{f(x)}$. My question about exponentiation was this: for what class of functions do we have $Exp[T[f(x)]] = T[Exp[f(x)]]$?

Similar questions can be asked for other operators. For example, for which class of functions does Tayler-series expansion commute with the derivative operator, the Fourier transform, etc?

• The convergence properties of the Taylor series of $g(f(x))$ do not necessarily equal those of $f(x)$. Consider for example $f(x)=x$ (converges everywhere), $g(x)=\tan(x)$ (converges in $]-\pi/2,\pi/2[$) – Wouter Apr 23 '18 at 14:44
• @Wouter, that's because the tangent function has issues at $\pi/2$ and elsewhere. OP seems primarily concerned with the exponential function, which is free of such issues. – Gerry Myerson Apr 25 '18 at 1:59

## 1 Answer

$$\exp(T(f(x)))=T(\exp(f(x)))$$ Where they both converge, these expressions are obviously equal, so the question is: do they have the same radius of convergence?

$$|x|<R_1\implies T(f(x))=f(x) \implies \exp(T(f(x)))=\exp(f(x))$$ $$|x|<R_2\implies T(\exp(f(x)))=\exp(f(x))$$ $$R_1=R_2?$$

What is the radius of convergence? It is the distance to the nearest point in the complex plane where $f$ is not holomorphic. (typically the nearest point where $f$ is singular / has a pole).

The radius of convergence of a power series ƒ centered on a point a is equal to the distance from a to the nearest point where ƒ cannot be defined in a way that makes it holomorphic.

The points where $\exp(f(x))$ is not holomorphic coincide with the points where $f(x)$ is not holomorphic, so yes, $$\exp(T(f(x)))=T(\exp(f(x)))$$ are equal where they converge, and converge on the same interval.

In general, in order to determine if $g(T(f(x)))$ and $T(g(f(x)))$ have the same convergence properties, check if $g$ can introduce new poles. The derivative operator will not do this. Of the Fourier transform, I am unsure.

• ok technically $\exp(f(x))$ has an essential singularity wherever $f(x)$ has a pole... I'll make an edit. – Wouter Apr 29 '18 at 3:43