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Math Help - Taylor series remainder

  1. #1
    Newbie pc31's Avatar
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    Taylor series remainder

    could anyone help me out with this? thankx
    use taylor approximation theorem to estimate e with error less than 10^-6
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  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
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    Hi
    Quote Originally Posted by pc31 View Post
    use taylor approximation theorem to estimate e with error less than 10^-6
    Let f:x\mapsto \exp x. f\in\mathcal{C}^{\infty}(\mathbb{R}) hence for any integer n and for all a,\,b\in \mathbb{R}, Taylor's theorem states that

    f(b)=\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(b-a)^k+R_n with R_n=\frac{f^{(n+1)}(c)}{(n+1)!}(b-a)^{n+1} for some c\in[a,b].

    As we want to evaluate \exp 1 we choose b=1. a can be any real number, let's choose a=0, it'll make things easier. For any integer k and for any real number x, f^{(k)}(x)=\exp x hence the previous equality becomes


    \exp 1=\sum_{k=0}^n\frac{\exp 0}{k!}(1-0)^k+\frac{\exp c}{(n+1)!}(1-0)^{n+1}=\underbrace{\sum_{k=0}^n\frac{1}{k!}}_{\t  ext{approximation}}+\underbrace{\frac{\exp c}{(n+1)!}}_{\text{error}} for some c\in[0,1].

    As we want an error <10^{-6}, you have to find n such that  \frac{\exp c}{(n+1)!} <10^{-6}. Can you take it from here ?
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