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Thread: 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 $\displaystyle <10^{-4}$ for the $\displaystyle n^{th}$ Maclaurin polynomial of $\displaystyle f(x)=e^x$ for $\displaystyle x=2$.

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

    $\displaystyle
    \frac{|2|^{n+1}}{(n+1)!}<10^{-4}
    $

    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 $\displaystyle e^{x}$ is...

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

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

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

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

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

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

    Kind regards

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

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

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

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

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

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

    ... and in particular...

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

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

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

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

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

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

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

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

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

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

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

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

    Kind regards

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

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

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

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

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

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

    Producing:

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

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

    So...

    $\displaystyle n=13$

    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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