Generalized Taylor series of an exponential

**bkarpuz** · March 14th 2011, 12:04 PM

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,
$\mathrm{e}^{c}=\displaystyle\sum_{\ell=0}^{\infty} \frac{c^{\ell}}{\ell!}.$

**girdav** · March 16th 2011, 12:19 PM

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 $F^{\left( n+1\right)}(x) = <br /> \dfrac{d}{dx}\left(F^{(n)}(x)\right) = <br /> \dfrac{d}{dx}\left[F(x)f_n(x)\right]$
$F^{(n+1)}(x)= F'(x)f_n(x) +F(x)f_n'(x) =F(x)f(x)f_n(x) +F(x)f_n'(x)$
hence
$F^{(n+1)}(x)<br /> =F(x)(f(x)f_n(x)+f_n'(x))=F(x)f_{n+1}(x)$ . We can now compare the two Taylor series.

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