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Math Help - Using a series to approximate an integral

  1. #1
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    Using a series to approximate an integral

    How would I use a series to approximate the following integral?
    <br />
\int_{0.5}^{1} \frac {cos(x)} {x^2} dx<br />
    I think that it somehow relates to
    <br />
cos (x) = 1-\frac {x^2}{2!}+\frac {x^4}{4!}-\frac {x^6}{6!}+...+\frac{(-1)^nx^{2n}}{(2n)!}+...<br />
    but I can't figure out what to do with the  x^2 part and how to solve the integral with all the different terms to the various powers....
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by yellowfive View Post
    How would I use a series to approximate the following integral?
    <br />
\int_{0.5}^{1} \frac {cos(x)} {x^2} dx<br />
    I think that it somehow relates to
    <br />
cos (x) = 1-\frac {x^2}{2!}+\frac {x^4}{4!}-\frac {x^6}{6!}+...+\frac{(-1)^nx^{2n}}{(2n)!}+...<br />
    but I can't figure out what to do with the  x^2 part and how to solve the integral with all the different terms to the various powers....
    \int\limits_{0.5}^1 {\frac{{\cos x}}{{{x^2}}}dx}  = \int\limits_{0.5}^1 {\frac{1}{{{x^2}}}\sum\limits_{n = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}}} dx}  = \int\limits_{0.5}^1 {\sum\limits_{n = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^n}{x^{2n - 2}}}}{{\left( {2n} \right)!}}} dx}  = \sum\limits_{n = 0}^\infty  {\left\{ {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n} \right)!}}\int\limits_{0.5}^1 {{x^{2n - 2}}dx} } \right\}} .

    What you need to do now, I hope you know.
    Last edited by DeMath; July 20th 2009 at 01:38 AM. Reason: typo
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  3. #3
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    Thanks! So this becomes:
    <br />
\int\limits_{0.5}^1 {\sum\limits_{n = 0}^\infty {\frac{{{{\left(<br />
{ - 1} \right)}^n}{x^{2n - 2}}}}{{\left( {2n} \right)!}}} dx} =  [{\sum\limits_{n = 0}^\infty {\frac{{{{\left(<br />
{ - 1} \right)}^n}{x^{2n - 1}}}}{{\left( {2n} \right)!}}} }]_{0.5}^{1}<br />
    How do I integrate this from here though? Do I do this to find each term of the series?
    <br />
    [{\frac{{{{\left(<br />
{ - 1} \right)}^0}{1^{2(0) - 1}}}}{{\left( {2(0)} \right)!}}} -<br />
{\frac{{{{\left(<br />
{ - 1} \right)}^0}{0.5^{2(0) - 1}}}}{{\left( {2(0)} \right)!}}} ] +<br />
[{\frac{{{{\left(<br />
{ - 1} \right)}^1}{1^{2(1) - 1}}}}{{\left( {2(1)} \right)!}}} -<br />
{\frac{{{{\left(<br />
{ - 1} \right)}^1}{0.5^{2(1) - 1}}}}{{\left( {2(1)} \right)!}}} ] +...<br />
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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by yellowfive View Post
    Thanks! So this becomes:
    <br />
\int\limits_{0.5}^1 {\sum\limits_{n = 0}^\infty {\frac{{{{\left(<br />
{ - 1} \right)}^n}{x^{2n - 2}}}}{{\left( {2n} \right)!}}} dx} =  [{\sum\limits_{n = 0}^\infty {\frac{{{{\left(<br />
{ - 1} \right)}^n}{x^{2n - 1}}}}{{\left( {2n} \right)!}}} }]_{0.5}^{1}<br />
    How do I integrate this from here though? Do I do this to find each term of the series?
    <br />
    [{\frac{{{{\left(<br />
{ - 1} \right)}^0}{1^{2(0) - 1}}}}{{\left( {2(0)} \right)!}}} -<br />
{\frac{{{{\left(<br />
{ - 1} \right)}^0}{0.5^{2(0) - 1}}}}{{\left( {2(0)} \right)!}}} ] +<br />
[{\frac{{{{\left(<br />
{ - 1} \right)}^1}{1^{2(1) - 1}}}}{{\left( {2(1)} \right)!}}} -<br />
{\frac{{{{\left(<br />
{ - 1} \right)}^1}{0.5^{2(1) - 1}}}}{{\left( {2(1)} \right)!}}} ] +...<br />
    OMG!

    \int\limits_{0.5}^1 {\frac{{\cos x}}{{{x^2}}}dx}  = \int\limits_{0.5}^1 {\frac{1}{{{x^2}}}\sum\limits_{n = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}}} dx}  =<br />
\int\limits_{0.5}^1 {\sum\limits_{n = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^n}{x^{2n - 2}}}}{{\left( {2n} \right)!}}} dx}  = \sum\limits_{n = 0}^\infty  {\left\{ {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n} \right)!}}\int\limits_{0.5}^1 {{x^{2n - 2}}dx} } \right\}} .

    Now find this integral

    \int\limits_{0.5}^1 {{x^{2n - 2}}dx}  = \left. {\frac{1}{{2n - 1}}{x^{2n - 1}}} \right|_{0.5}^1 = \frac{1}{{2n - 1}}\left( {1 - \frac{1}{{{2^{2n - 1}}}}} \right) = \frac{{{4^n} - 2}}{{\left( {2n - 1} \right){4^n}.}}

    So we have

    \int\limits_{0.5}^1 {\frac{{\cos x}}{{{x^2}}}dx}  = \sum\limits_{n = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^n}\left( {{4^n} - 2} \right)}}<br />
{{\left( {2n} \right)!\left( {2n - 1} \right){4^n}}}} .
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