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Math Help - taylor approximation

  1. #1
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    Question taylor approximation

    im trying to find the interval of which 1 - (x^2)/2 is an approximation for cos(x) with an error less than 0.0001?

    I know its got something to do with the taylor polynomial...iv tried letting it equal each other and substituting 0.0001 but it all just cancels out any ideas?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by dopi View Post
    im trying to find the interval of which 1 - (x^2)/2 is an approximation for cos(x) with an error less than 0.0001?

    I know its got something to do with the taylor polynomial...iv tried letting it equal each other and substituting 0.0001 but it all just cancels out any ideas?
    Are you looking for the interval or an interval?

    The interval is the largest interval (containing x=0) on which:

    <br />
|1-x^2/2 - \cos(x)|<0.0001<br />

    We know that this interval is symmetric about x=0 so to find the upper end point of the interval you are looking for the solution of:

    <br />
1-x^2/2 - \cos(x)=-0.0001<br />

    on [0,\pi] (the -ve sign is because \cos(x)>1-x^2/2 close to 0).

    This has to be solved numerically, when we find the interval is:

    \sim [-0.221427,0.221427].

    An interval can be found by using one of the forms of the remainder of the Taylor series. Which in this case can be slightly modified to give the result that on the interval [0,\epsilon] the error is less than:

    <br />
\epsilon^4/24<br />

    so if we set \epsilon^2/24=0.0001 and solve for \epsilon we will have found such an interval. Doing this we find \epsilon = 0.221336, and so an interval solving the problem is:

    [-0.221336, 0.221336].

    RonL
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