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Math Help - Calculus MacLaurin series

  1. #1
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    Calculus MacLaurin series

    Using the MacLaurin Series for e^x, find the Integral of ((e^-x^2)-1))/x dx from 0 to 1/5 to within 0.000001.

    Can anyone help me solve this, or a problem similiar to this one so I can attempt to solve it on my own?

    The answer to the above question is -0.019802
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  2. #2
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    Let \displaystyle f(x)=\frac{e^{-x^2}-1}{x}. We have \displaystyle e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots, so \displaystyle e^{-x^2}=1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\dots and \displaystyle f(x)=-x+\frac{x^3}{2!}-\frac{x^5}{3!}+\frac{x^7}{4!}-\dots. The terms of this series have alternating signs and decreasing absolute values (when |x|\le1); therefore, the sum from the nth term to infinity is bounded by the nth term. For example, let \displaystyle g(x)=-x+\frac{x^3}{2!}-\frac{x^5}{3!}. Then \displaystyle |f(x)-g(x)|=\left|\frac{x^7}{4!}+\dots\right|\le\frac{x^  7}{4!}.

    Now, suppose \displaystyle\varepsilon=\sup_{x\in[0,1/5]}|f(x)-g(x)|. Then |\int_0^{1/5} f(x)\,dx-\int_0^{1/5} g(x)\,dx|\le\int_0^{1/5}|f(x)-g(x)|\,dx\le\varepsilon/5. Our goal is to make \varepsilon/5<10^{-6}. One can check that \displaystyle\frac{(1/5)^7}{5\cdot 4!}<10^{-6}, but \displaystyle\frac{(1/5)^5}{5\cdot 3!}>10^{-6}, so we have to approximate f(x) with a partial sum of three terms, i.e., with g(x).

    Calculating \int_0^{1/5}g(x)\,dx gives −0.019801778, which is −0.019802 when rounded to the 6th digit.
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  3. #3
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    I understand why e^-x^2 = 1 + x^2... but why does f(x) = -x + x^3/2!-...?
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  4. #4
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    e^{-x^2} = 1 - x^2+\dots, not 1+x^2+\dots. When we subtract 1 and divide by x, we get f(x) = -x + x^3/2!-\dots. Recall that f(x)=(e^{-x^2}-1)/x.
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