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Thread: Taylor series approximations

  1. #1
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    Taylor series approximations

    The question:

    Find a value for n such that the absolute error in the approximation $\displaystyle f(2) \approx p_n(2)$ is less than $\displaystyle 10^{-6}$, given that $\displaystyle |R_{n+1}(x)| = |\frac{sin(c) x^{n + 1}}{(n + 1)!}|$ about 0.

    (Note: f(x) = cos(x))

    My attempt:
    I replace sin(c) with 1, since it is an upper bound, and x with 2:

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

    We want this error to be less that $\displaystyle 10^{-6}$, so:

    $\displaystyle |\frac{2^{n + 1}}{(n + 1)!}| < 10^{-6}$

    This is equivalent to finding:

    $\displaystyle 2^{n + 1} < 10^{-6}$

    $\displaystyle (n + 1)ln(2) < -6ln(10)$
    $\displaystyle (n + 1) < \frac{-6ln(10)}{ln(2)}$
    $\displaystyle n < -20.93$

    This isn't right. The answer should be n = 14. Where have I gone wrong? Thanks.
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  2. #2
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    Quote Originally Posted by Glitch View Post
    The question:

    Find a value for n such that the absolute error in the approximation $\displaystyle f(2) \approx p_n(2)$ is less than $\displaystyle 10^{-6}$, given that $\displaystyle |R_{n+1}(x)| = |\frac{sin(c) x^{n + 1}}{(n + 1)!}|$ about 0.

    (Note: f(x) = cos(x))

    My attempt:
    I replace sin(c) with 1, since it is an upper bound, and x with 2:

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

    We want this error to be less that $\displaystyle 10^{-6}$, so:

    $\displaystyle |\frac{2^{n + 1}}{(n + 1)!}| < 10^{-6}$

    This is equivalent to finding:

    $\displaystyle 2^{n + 1} < 10^{-6}$


    Why?! Of course that then you get what you get since the LHS is the power of a number greater than 2, whereas the RHS

    is a negative power of a number greater than 1...! You forgot, or for some reason you

    thought, that you can dismiss the LHS's denominator:

    $\displaystyle \displaystyle{\frac{2^{n+1}}{(n+1)!}<10^{-6}\Longrightarrow (n+1)\ln 2<-6\ln 10+\sum\limits^{n+1}_{k=1}\ln k<-6\ln 10+(n+1)\ln (n+1)\Longrightarrow}$

    $\displaystyle \displaystyle{\Longrightarrow (n+1)[\ln 2-\ln(n+1]<-6\ln 10\Longrightarrow n+1>\frac{6\ln 10}{\ln 2-\ln(n+1)}>\frac{6\ln 10}{\ln 2} \cong 19.9\Longrightarrow}$

    $\displaystyle \displaystyle{ n>18.9}$

    So I get $\displaystyle n=19$ .

    Tonio



    $\displaystyle (n + 1)ln(2) < -6ln(10)$
    $\displaystyle (n + 1) < \frac{-6ln(10)}{ln(2)}$
    $\displaystyle n < -20.93$

    This isn't right. The answer should be n = 14. Where have I gone wrong? Thanks.
    .
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