Using a series to approximate an integral

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July 19th 2009, 11:45 PM
yellowfive

Using a series to approximate an integral

How would I use a series to approximate the following integral?
$ \int_{0.5}^{1} \frac {cos(x)} {x^2} dx $
I think that it somehow relates to
$ cos (x) = 1-\frac {x^2}{2!}+\frac {x^4}{4!}-\frac {x^6}{6!}+...+\frac{(-1)^nx^{2n}}{(2n)!}+... $
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....
July 20th 2009, 12:09 AM
DeMath

Quote:

Originally Posted by yellowfive

How would I use a series to approximate the following integral?
$ \int_{0.5}^{1} \frac {cos(x)} {x^2} dx $
I think that it somehow relates to
$ cos (x) = 1-\frac {x^2}{2!}+\frac {x^4}{4!}-\frac {x^6}{6!}+...+\frac{(-1)^nx^{2n}}{(2n)!}+... $
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.
July 20th 2009, 06:26 AM
yellowfive

Thanks! So this becomes:
$ \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 {\frac{{{{\left( { - 1} \right)}^n}{x^{2n - 1}}}}{{\left( {2n} \right)!}}} }]_{0.5}^{1} $
How do I integrate this from here though? Do I do this to find each term of the series?
$ $ $[{\frac{{{{\left( { - 1} \right)}^0}{1^{2(0) - 1}}}}{{\left( {2(0)} \right)!}}} - {\frac{{{{\left( { - 1} \right)}^0}{0.5^{2(0) - 1}}}}{{\left( {2(0)} \right)!}}} ] + [{\frac{{{{\left( { - 1} \right)}^1}{1^{2(1) - 1}}}}{{\left( {2(1)} \right)!}}} - {\frac{{{{\left( { - 1} \right)}^1}{0.5^{2(1) - 1}}}}{{\left( {2(1)} \right)!}}} ] +... $
July 20th 2009, 06:39 AM
DeMath

Quote:

Originally Posted by yellowfive

Thanks! So this becomes:
$ \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 {\frac{{{{\left( { - 1} \right)}^n}{x^{2n - 1}}}}{{\left( {2n} \right)!}}} }]_{0.5}^{1} $
How do I integrate this from here though? Do I do this to find each term of the series?
$ $ $[{\frac{{{{\left( { - 1} \right)}^0}{1^{2(0) - 1}}}}{{\left( {2(0)} \right)!}}} - {\frac{{{{\left( { - 1} \right)}^0}{0.5^{2(0) - 1}}}}{{\left( {2(0)} \right)!}}} ] + [{\frac{{{{\left( { - 1} \right)}^1}{1^{2(1) - 1}}}}{{\left( {2(1)} \right)!}}} - {\frac{{{{\left( { - 1} \right)}^1}{0.5^{2(1) - 1}}}}{{\left( {2(1)} \right)!}}} ] +... $

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

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

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