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Math Help - Taylor's Theorem Estimation

  1. #1
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    Taylor's Theorem Estimation

    Hello again,

    This question is more toward the definition of taylor polynomials, why does it guarantee that we can use a taylor polynomial to estimate "e" to any degree of accuracy? I know that the longer the polynomial, the closer the approximation and "e" is one of the easier functions for taylor polynomails. Is there a better way to explain why? other than restating the definition of a taylor polynomial?

    - oh, I forgot to say in my last post, Thanks
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  2. #2
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    Quote Originally Posted by icedragontt View Post
    Hello again,

    This question is more toward the definition of taylor polynomials, why does it guarantee that we can use a taylor polynomial to estimate "e" to any degree of accuracy? I know that the longer the polynomial, the closer the approximation and "e" is one of the easier functions for taylor polynomails. Is there a better way to explain why? other than restating the definition of a taylor polynomial?

    - oh, I forgot to say in my last post, Thanks
    You could look at the Taylor remainder. For the nth degree polynomial

    P_n(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!}

    R_n = \frac{e^c x^{n+1}}{(n+1)!} where c is between x and 0

    and
    P_n(1) = 1 + 1+ \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!}

    R_n(1) = \frac{e^c}{(n+1)!} < \frac{3}{(n+1)!}

    Clearly as we increase n, R_n decreases give a better and better answer.
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  3. #3
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    Read this section in the wiki Taylor's theorem - Wikipedia, the free encyclopedia

    Short story is

    Since the derivative of \frac{d}{dx}e^x=e^x for all x.

    for any fixed number m we can find an N large enough so that

    \frac{m^{N+1}e^m}{(N+1)!} is as small as we want.

    This tells us the error can't be any larger than the above term so we would need to take n terms in the taylor expation to get at least this accurate.

    Note this is an upper bound on the error it may be much smaller than this.

    If we take the limit at n \to \infty and the remainder goes to 0 it tells us that the taylor polynomial does indeed converge to the function.

    I hope this helps.
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