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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 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!}.
    Last edited by bkarpuz; Mar 14th 2011 at 05:16 PM.
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  2. #2
    Super Member girdav's Avatar
    Jul 2009
    Rouen, France
    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 />
    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)
    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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