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Math Help - Taylor Inequality

  1. #1
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    Taylor Inequality

    Suppose you used the first three nonzero terms of the power series representation to approximate e^{-(x^2)} for  0\le x \le 1/2.
    Obtain an upper bound on the error.
    All the terms, are negative, I think? The parentheses confused me. So: power series representation is 1-x^2-x^4/2-x^6/6...
    The approximation is found, I think, just by plugging in x=1/2, for which I got 0.71875.
    Then use the Taylor Inequality... R_3_(x) \le \frac{M}{(n+1)!} /x-a/^{n+1}
    {f^{4}(x) \le M
    \frac {(1/2)^6}{6} \le M, so do I set M=1/384?
    I stopped here because I wasn't sure if my value for M was right...I have the feeling it's not. Any help would be great.
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  2. #2
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    Re: Taylor Inequality

    Quote Originally Posted by bcahmel View Post
    Suppose you used the first three nonzero terms of the power series representation to approximate e^{-(x^2)} for  0\le x \le 1/2.
    Obtain an upper bound on the error.
    All the terms, are negative, I think? The parentheses confused me. So: power series representation is 1-x^2-x^4/2-x^6/6...
    The approximation is found, I think, just by plugging in x=1/2, for which I got 0.71875.
    Then use the Taylor Inequality... R_3_(x) \le \frac{M}{(n+1)!} /x-a/^{n+1}
    {f^{4}(x) \le M
    \frac {(1/2)^6}{6} \le M, so do I set M=1/384?
    I stopped here because I wasn't sure if my value for M was right...I have the feeling it's not. Any help would be great.
    e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...

    e^{-x^2} = 1 + (-x^2) + \frac{(-x^2)^2}{2!} + \frac{(-x^2)^3}{3!} + ...

    so, the first three terms of the series for e^{-x^2} are ...

    e^{-x^2} \approx 1 - x^2 + \frac{x^4}{2!}

    ... and the series is alternating.
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  3. #3
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    Re: Taylor Inequality

    ok thanks, the parentheses dont make a difference then. I still don't understand M, however...
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