Re: Sign of the remainder of Taylor expansion

From: Peter Spellucci (spellucci_at_fb04373.mathematik.tu-darmstadt.de)
Date: 12/21/04

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    Date: Tue, 21 Dec 2004 19:00:05 +0000 (UTC)

    In article <cq6j8m$2fe$1@dizzy.math.ohio-state.edu>,
     Dan <trautkilgor@yahoo.com> writes:
    >Let f(x) be a real valued function, with Taylor expansion (around
    >zero) :
    >
    >
    >f(x) = \sum_{k=0}^{\infty} (1/k!) * (df^k(0) / dx^k) * x^k
    >
    >
    >If we take only M+1 coefficients, we get :
    >f(x) = \sum_{k=0}^{M} (1/k!) * (df^k(0) / dx^k) * x^k + R_M(x)
    >
    >Where R_M is the Lagrange remainder. So, if we're inside the radius
    >of convergence, we get that R_M(x) --> 0 when M --> \infty.
    >
    >Are there any known condition for this convergance to be always from
    >'above' ?
    >that is, that R_M(x) < 0 for all M ? (even for a specific x)
    >
    >Note that this is weaker then demanding monotone convergance.
    >
    >(My example for f is some complicated rational function, for which I'm
    >trying to prove this,
    >and I don't have a general formula for the k'th derivative)
    >
    >Thanx,
    >Dan

    this will hardly be possible:
    the remainder term is
       df^{(M+1)}(theta*x)*x^{M+1}/((M+1)!) with 0<theta<1
    hence for M even the second factor will have a sign change at zero.
    even if you insist in x>0 this would require _all_ derivatives of f
    being positive for small positive x. for some restricted class of functions this
    is the case of course. with a rational function, using the decomposition into
    partial fractions you would have a chance to say something about derivatives of
    arbitrary order provided you either can calculate the poles exactly or at
    least can estimate their distance from zero.
    hth
    peter

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