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Math Help - MacClaurin series discrepancy

  1. #1
    Super Member redsoxfan325's Avatar
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    MacClaurin series discrepancy

    This can happen with a bunch of different series, but let's just look at a really easy one as an example.

    f(x)=e^{x+1}

    There are two ways to calculate the Maclaurin polynomial:

    1.) e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} so e^{x+1} = \sum_{n=0}^{\infty}\frac{(x+1)^n}{n!}

    2.) f^{(n)}(x) = e for all n so the polynomial is \sum_{n=0}^{\infty}\frac{ex^n}{n!}=e\sum_{n=0}^{\i  nfty}\frac{x^n}{n!}

    Clearly, these are the same if you're summing from 0 to \infty, but what if you are just trying to approximate the function? Which is a better estimate? (I would assume it's the second one.)

    Or, the true reason behind my asking this: My friend's professor asked on an exam what the coefficient on the x^2 term was, yet depending on how you calculate this, you could get different answers. In the first method you get \frac{1}{2}; in the second, you get \frac{e}{2}.

    I'm assuming that if a_n^2 is the coefficient on x^2 (in the first method) when taking the sum from 0 to n, these coefficients will form a sequence such that \lim_{n\to\infty}a_n^2 = \frac{e}{2}

    Does anyone have anything to add to this or is that reasoning all pretty much fine?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Good ol' Mathcad e/2 = 1.359

    just first seven terms of the coefficients of x^2 for

    form the sequence

    .5,1,1.25,1.333,1.354,1.358,1.359

    for what its worth
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  3. #3
    MHF Contributor Calculus26's Avatar
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    If you consider the binomial theorem

    For(x+1)^n then the coefficient is nCr = n!/[(n-2)!*2]

    In the series the coefficient is then 1/[(n-2)!*2] for each n

    If you add these coeeficients you get

    1/2 sum{n=2toinfinity}[1/(n-2)!] = e/2
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