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Math Help - Calculus - Maclaurin Polynomial

  1. #1
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    Calculus - Maclaurin Polynomial

    It asks me to show that the nth Maclaurin Polynomial for exp(x) is </= to exp(x) for x>0. I understand this as a theory, as you can only go so far with the Maclaurin Polynomial, whearas exp(x) goes infinitely far. I think this is what the question is getting at, and would much appreciate any help on how I would start to prove this.
    Many thanks,
    Eloise
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by EKohler View Post
    It asks me to show that the nth Maclaurin Polynomial for exp(x) is </= to exp(x) for x>0. I understand this as a theory, as you can only go so far with the Maclaurin Polynomial, whearas exp(x) goes infinitely far. I think this is what the question is getting at, and would much appreciate any help on how I would start to prove this.
    Many thanks,
    Eloise
    I believe that what it is getting at is

    \lim_{N\to\infty}\sum_{n=0}^{N}\frac{x^n}{n!}=e^x

    So \sum_{n=0}^{N_1}\frac{x^n}{n!}\approx{e^x}\quad\fo  rall{N_1}\in\mathbb{R}

    Now consider that since x>0 the truncated terms of the series (in other words the N_{1}+1 to infinitieth) terms will be greater than zero. Therefore taking them away will detract from the value of the summation

    Therefore

    \lim_{N\to\infty}\sum_{n=0}^{N}\frac{x^n}{n!}=e^x

    And

    \sum_{n=0}^{N_1\in\mathbb{R}}\frac{x^n}{n!}<e^x\qu  ad\forall{x>0}

    Now combinging these and not knowing N's value gives

    \boxed{\sum_{n=0}^{N}\frac{x^n}{n!}\leq{e^x}\quad\  forall{x>0}}
    Last edited by Mathstud28; November 3rd 2008 at 12:43 PM. Reason: Typo
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  3. #3
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    Thanks so much. that totally explained it in a manner that made sense.
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