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  1. #1
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    Taylor polynomial problem

    Hi -

    I'm given a problem where I must find the "minimum order of the Taylor polynomial required to approximate the quantity with an absolute error no greater than 10^-3", where the given quantity is e^-0.5

    I've calculated the remainder to be R(-0.5)=[0.5^(n+1)]/(n+1)!, yet I do not know how to solve the inequality 10^-3>[0.5^(n+1)]/(n+1)!

    I know the answer is supposed to be n=4, but I have no clue how to solve for n. Any help???

    Thanks!!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Taylor polynomial problem

    Quote Originally Posted by leonardthecow View Post
    I've calculated the remainder to be R(-0.5)=[0.5^(n+1)]/(n+1)!, yet I do not know how to solve the inequality 10^-3>[0.5^(n+1)]/(n+1)!
    We have

    \left |{R_n(x)}\right |=\left |\dfrac{e^{\xi}}{(n+1)!}x^{n+1}\right |\quad (\xi\in (-1/2,0))

    \left |{R_n(x)}\right |\leq \dfrac{1}{(n+1)!}\;\dfrac{1}{2^{n+1}}

    \dfrac{1}{(n+1)!\;2^{n+1}}\leq 10^{-3} \Leftrightarrow 1000 \leq (n+1)!\;2^{n+1}

    Now, verify that n=4 is the smallest natural number satisfying the last inequality.
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  3. #3
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    Re: Taylor polynomial problem

    Quote Originally Posted by leonardthecow View Post
    Hi -

    I'm given a problem where I must find the "minimum order of the Taylor polynomial required to approximate the quantity with an absolute error no greater than 10^-3", where the given quantity is e^-0.5

    I've calculated the remainder to be R(-0.5)=[0.5^(n+1)]/(n+1)!, yet I do not know how to solve the inequality 10^-3>[0.5^(n+1)]/(n+1)!

    I know the answer is supposed to be n=4, but I have no clue how to solve for n. Any help???

    Thanks!!
    Did you try direct computation? If n= 1, .5^{1+1}/(1+1)!= (1/4)(1/2)= 1/8= 0.125> 0.001. If n= 2, .5^{2+1}/(2+1)!= (1/8)(1/6)= 1/48= 0.0208333...> 0.001. If n= 3, .5^{3+1}/(3+1)!= (1/16)(1/24)= 1/384= 0.002604...> 0.001. If n= 4, .5^(4+1)/(4+1)!= (1/32)(1/120)= 1/3840= 0.000260...< 0.001.
    Last edited by HallsofIvy; August 4th 2012 at 01:51 PM.
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