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Math Help - help me estimate a number with TAYLOR method

  1. #1
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    help me estimate a number with TAYLOR method

    Assume that f is a function with |f^(n)(x)|<or= 1 for all n and all real x. (The sine and cosine functions have this property.

    Find the least integer n for which you can be sure that Pn(2) approximates f(2) within 0.001.

    Thanks for the help!
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    Quote Originally Posted by seth55 View Post
    Assume that f is a function with |f^(n)(x)|<or= 1 for all n and all real x. (The sine and cosine functions have this property.

    Find the least integer n for which you can be sure that Pn(2) approximates f(2) within 0.001.

    Thanks for the help!
    |P_n(2) - f(2)| \leq \frac{2^{n+1}}{(n+1)!}
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    Thanks for the quick response, however, I don't quite understand your answer. How do I find the least possible integer?
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    Quote Originally Posted by seth55 View Post
    Thanks for the quick response, however, I don't quite understand your answer. How do I find the least possible integer?
    Because if we can make \frac{2^{n+1}}{(n+1)!} smaller than .001 then it would mean |T_n(2) - f(2)| \leq .001.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    |P_n(2) - f(2)| \leq \frac{2^n}{n!}
    Which form or the remainder are you using? Which point are you expanding about?

    Expanding about 0, and using the Lagrange form of the remainder I get:

    |P_n(2) - f(2)| \leq \frac{2^{n+1}}{(n+1)!},

    so the n required is the smallest positive integral solution of:

    \frac{2^{n+1}}{(n+1)!}\le 0.001

    Now expanding about 2 you will requre only one term, that is P_0(2) is exact when expansion is about 2.

    RonL
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    Quote Originally Posted by CaptainBlack View Post
    Which form or the remainder are you using? Which point are you expanding about?
    I edited before you got a chance to reply.
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