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Math Help - approximate with Taylor remainder

  1. #1
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    Unhappy approximate with Taylor remainder

    Hello everyone! I am trying to prove that the following inequality is right for all n equal to or larger than 9. I tried using the triangle inequality and then mathematical induction but none of it seems to work. please help!

    l e^-1 - (sigma sign from k=0 to n) (1)^-k / k!l

    thanks in advance
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  2. #2
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    Quote Originally Posted by karma View Post
    Hello everyone! I am trying to prove that the following inequality is right for all n equal to or larger than 9. I tried using the triangle inequality and then mathematical induction but none of it seems to work. please help!

    l e^-1 - (sigma sign from k=0 to n) (1)^-k / k!l

    thanks in advance
    First question what are we trying to show that it is less than(where is the inequality)?

    \bigg| e^{-1}-\sum_{k=0}^{n}\frac{(-1)^{-k}}{k!} \bigg|

    Second what class is this for do you know Taylor's Remainder theorem?.
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  3. #3
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    Sorry!! I forgot to write that. The question is that it is smaller than or equal to 10^(3-n). I actually do not know what the taylor remainder theorem is, but yes we are doing taylor series.
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    Quote Originally Posted by karma View Post
    yes we are doing taylor series.
    Does this series look familiar to you?

    f(x)=e^{x}=\sum_{n=0}^{\infty}\frac{x^n}{n!}

    consider
    f(-1)=e^{-1}=?

    Do you know the Alternating series Test error bound?

    If so you can use it to finish this problem.
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  5. #5
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    I know that lerrorl = lxl^(n+1) / (n+1)! for a Pn(x), is that what you mean by alternating series error bound? If that is so, I dont really see how you can use that.

    But if you replace f(-1) (which I hadnt seen before) than you get an infinite sum starting from k=n, right? What do you do with the absolute value then?
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  6. #6
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    Quote Originally Posted by karma View Post
    I know that lerrorl = lxl^(n+1) / (n+1)! for a Pn(x), is that what you mean by alternating series error bound? If that is so, I dont really see how you can use that.

    But if you replace f(-1) (which I hadnt seen before) than you get an infinite sum starting from k=n, right? What do you do with the absolute value then?
    The error in an alternating series is always less than the absolute value of the next term

    \bigg| e^{-1}-\sum_{k=0}^{n}\frac{(-1)^k}{k!}\bigg| < |a_{n+1}|

    So lets find out how big the next term can be using n=9

    \bigg| \frac{(-1)^{n+1}}{(n+1)!}\bigg|=\frac{1}{10!}=\frac{1}{362  8800} \approx 2.756 \times 10^{-7} < 10^{-6}=10^{3-9}=10^{3-n}

    I hope this clears it up.

    Good luck
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    Right! That really helps. Thank you very much!=)
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