# Thread: Generalized Taylor series of an exponential

1. ## Generalized Taylor series of an exponential

Let $f\in C^{\infty}(\mathbb{R},\mathbb{R})$, and define $f_{n}:=f_{n-1}^{\prime}+f_{n-1}f$ on $\mathbb{R}$ for $n\in\mathbb{N}$ with $f_{0}\equiv1$ on $\mathbb{R}$.
Show that
$\exp\bigg\{\displaystyle\int_{a}^{b}f(u)\mathrm{d} u\bigg\}=\displaystyle\sum_{\ell=0}^{\infty}f_{\el l}(a)\frac{(b-a)^{\ell}}{\ell!}\quad\text{for}\ a,b\in\mathbb{R}.$

Remark. In the case of constant function, i.e., $f\equiv c/(b-a)$ on $\mathbb{R}$ (provided that $b>a$) for some $c\in\mathbb{R}$,
we can compute that $f_{k}\equiv \big(c/(b-a)\big)^{k}$ on $\mathbb{R}$ for all $k\in\mathbb{N}$, which turns out to be a well known result.
That is,
$\mathrm{e}^{c}=\displaystyle\sum_{\ell=0}^{\infty} \frac{c^{\ell}}{\ell!}.$

2. Let $a\in\mathbb R$ and $F(x) :=\exp \left\{\int_a^x f(t)dt \right\}$. We can show that $F$ is analytic on the whole real line (for example by Faa-di-Bruno's formula) and we have $F'(x) = F(x)f(x)$ according to the chain rule.
We show by induction that $F^{(n)}(x) =f_n(x)f(x)$. $f_1=f$ so the result is true for $n=1$ and if it's true for n we have $F^{\left( n+1\right)}(x) =
\dfrac{d}{dx}\left(F^{(n)}(x)\right) =
\dfrac{d}{dx}\left[F(x)f_n(x)\right]$

$F^{(n+1)}(x)= F'(x)f_n(x) +F(x)f_n'(x) =F(x)f(x)f_n(x) +F(x)f_n'(x)$
hence
$F^{(n+1)}(x)
=F(x)(f(x)f_n(x)+f_n'(x))=F(x)f_{n+1}(x)$
. We can now compare the two Taylor series.