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Thread: Generalized Taylor series of an exponential

  1. #1
    Senior Member bkarpuz's Avatar
    Sep 2008

    Generalized Taylor series of an exponential

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

    Remark. In the case of constant function, i.e., $\displaystyle f\equiv c/(b-a)$ on $\displaystyle \mathbb{R}$ (provided that $\displaystyle b>a$) for some $\displaystyle c\in\mathbb{R}$,
    we can compute that $\displaystyle f_{k}\equiv \big(c/(b-a)\big)^{k}$ on $\displaystyle \mathbb{R}$ for all $\displaystyle k\in\mathbb{N}$, which turns out to be a well known result.
    That is,
    $\displaystyle \mathrm{e}^{c}=\displaystyle\sum_{\ell=0}^{\infty} \frac{c^{\ell}}{\ell!}.$
    Last edited by bkarpuz; Mar 14th 2011 at 04:16 PM.
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  2. #2
    Super Member girdav's Avatar
    Jul 2009
    Rouen, France
    Let $\displaystyle a\in\mathbb R$ and $\displaystyle F(x) :=\exp \left\{\int_a^x f(t)dt \right\}$. We can show that $\displaystyle F$ is analytic on the whole real line (for example by Faa-di-Bruno's formula) and we have $\displaystyle F'(x) = F(x)f(x)$ according to the chain rule.
    We show by induction that $\displaystyle F^{(n)}(x) =f_n(x)f(x)$. $\displaystyle f_1=f$ so the result is true for $\displaystyle n=1$ and if it's true for n we have $\displaystyle F^{\left( n+1\right)}(x) =
    \dfrac{d}{dx}\left(F^{(n)}(x)\right) =
    \dfrac{d}{dx}\left[F(x)f_n(x)\right] $
    $\displaystyle 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) $
    $\displaystyle F^{(n+1)}(x)
    =F(x)(f(x)f_n(x)+f_n'(x))=F(x)f_{n+1}(x)$. We can now compare the two Taylor series.
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