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Math Help - Tn(x) Error Bound

  1. #1
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    Tn(x) Error Bound

    Good afternoon,

    I think I can finally wrap my head around Taylor (& Maclaurin) polynomials. However, I seem to really be struggling with the error bounds. I am looking for an error bound of <10^{-4} for the n^{th} Maclaurin polynomial of f(x)=e^x for x=2.

    I'm not even sure I know how to set this up. Should it be simply:

     <br />
\frac{|2|^{n+1}}{(n+1)!}<10^{-4}<br />

    If this is what it should be, can someone explain why (or point me to some "easy to digest" reading)?
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  2. #2
    MHF Contributor chisigma's Avatar
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    The McLaurin expasion for e^{x} is...

    \displaystyle e^{x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \dots + \frac{x^{n}}{n!} + R_{n} (x) (1)

    ... where R_{n} (x) is an 'error term' that has different interpretations. One of the most simple expression of R_{n} (x) has been found by the french mathematician Augustin Cauchy and is in that case...

    \displaystyle R_{n}(x) = \frac{x^{n+1}\ (1-\theta)^{n}}{n!}\ e^{\theta x} , 0 < \theta < 1 (2)

    At this point You have to do the following steps...

    a) setting in (2) x=2 find the value of \theta_{0} that maximise R_{n} ...

    b) stting in (2) \theta=\theta_{0} find the value of n for which is R_{n} < 10^{-4} ...

    Kind regards

    \chi \sigma
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  3. #3
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    I am provided with the formula:

    |T_n(x)-f(x)|\leq K\frac{|x-a|^{n+1}}{(n+1)!}

    let K\leq|f^{(n+1)}(u)}| for all u between x and a

    I am not really sure how set that up to find the correct value of n. I was able to construct a matrix that I believe provides me with the correct answer, but I would like to be able to utilize the formula directly.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Recalling that is...

    \displaystyle R_{n} (x) =  \frac{x^{n+1}\ (1-\theta)^{n}}{n!}\ e^{\theta x} , 0 < \theta < 1 (1)

    ... setting x=2 we find that R_{n} (2) has the maximum possible value for \theta=0, so that is...

    \displaystyle R_{n} (2) \le \frac{2^{n+1}}{n!}} (2)

    ... and in particular...

    \displaystyle n=2 \rightarrow R_{n} (2) \le 4

    \displaystyle n=3 \rightarrow R_{n} (2) \le 2.666666...

    \displaystyle n=4 \rightarrow R_{n} (2) \le 1,333333...

    \displaystyle n=5 \rightarrow R_{n} (2) \le .533333...

    \displaystyle n=6 \rightarrow R_{n} (2) \le .177777...

    \displaystyle n=7 \rightarrow R_{n} (2) \le 5.079365... \ 10^{-2}

    \displaystyle n=8 \rightarrow R_{n} (2) \le 1.269841... \ 10^{-2}

    \displaystyle n=9 \rightarrow R_{n} (2) \le 2.821869... \ 10^{-3}

    \displaystyle n=10 \rightarrow R_{n} (2) \le 5.643738... \ 10^{-4}

    \displaystyle n=11 \rightarrow R_{n} (2) \le 1.026134... \ 10^{-4}

    \displaystyle n=12 \rightarrow R_{n} (2) \le 1.710223... \ 10^{-5} (3)

    ... so that the required value for n seems to be 12...

    Kind regards

    \chi \sigma
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  5. #5
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    Quote Originally Posted by chisigma View Post
    Recalling that is...

    \displaystyle R_{n} (x) =  \frac{x^{n+1}\ (1-\theta)^{n}}{n!}\ e^{\theta x} , 0 < \theta < 1 (1)

    ... setting x=2 we find that R_{n} (2) has the maximum possible value for \theta=0, so that is...

    \displaystyle R_{n} (2) \le \frac{2^{n+1}}{n!}} (2)
    I find this to be equal to:

    \displaystyle R_{n} (2) \le \frac{2^{n+1}}{n!}}

    As e^{\theta 2}   =   1 for \theta   =   0

    Producing:

    \displaystyle n=12 \rightarrow R_{n} (2) \le .000103

    \displaystyle n=13 \rightarrow R_{n} (2) \le .000017

    So...

    n=13

    Am I confused by something?
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  6. #6
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    Is my assumption on the last post incorrect? I definitely don't mean to be insulting, just confused.
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