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Math Help - Limits and integration

  1. #1
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    Limits and integration

    If I have shown that \sum\limits_{k = 0}^\infty  {{f_k}} is uniformly convergent on [0,1], where {f_k}\left( x \right) = {\left( {1 - {x^2}} \right)^2}{x^k}, how would I then go about showing that

    \int\limits_0^1 {\frac{{{{\left( {1 - {x^2}} \right)}^2}}}{{1 - x}}} {\rm{ dx}} = \sum\limits_{k = 1}^\infty  {\frac{8}{{k\left( {k + 2} \right)\left( {k + 4} \right)}}} ?

    All help appreciated!
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by converge2 View Post
    If I have shown that \sum\limits_{k = 0}^\infty  {{f_k}} is uniformly convergent on [0,1], where {f_k}\left( x \right) = {\left( {1 - {x^2}} \right)^2}{x^k}, how would I then go about showing that

    \int\limits_0^1 {\frac{{{{\left( {1 - {x^2}} \right)}^2}}}{{1 - x}}} {\rm{ dx}} = \sum\limits_{k = 1}^\infty  {\frac{8}{{k\left( {k + 2} \right)\left( {k + 4} \right)}}} ?

    All help appreciated!
    \int_0^1 \frac{(1-x^2)^2}{1-x} ~dx=\int_0^1 (1-x^2)^2\sum_{k=0}^\infty x^k~dx=\int_0^1 \sum_{k=0}^\infty (1-x^2)^2x^k ~dx

    Since \sum_{k=0}^\infty f_k(x) converges uniformly, we can invert the sum and the integral.

    And hence \int_0^1 \frac{(1-x^2)^2}{1-x} ~dx=\sum_{k=0}^\infty \int_0^1 (1-x^2)^2 x^k ~dx

    can you finish it ? expand (1-x^2)^2 and be careful with the case k=1.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by converge2 View Post
    If I have shown that \sum\limits_{k = 0}^\infty {{f_k}} is uniformly convergent on [0,1], where {f_k}\left( x \right) = {\left( {1 - {x^2}} \right)^2}{x^k}, how would I then go about showing that

    \int\limits_0^1 {\frac{{{{\left( {1 - {x^2}} \right)}^2}}}{{1 - x}}} {\rm{ dx}} = \sum\limits_{k = 1}^\infty {\frac{8}{{k\left( {k + 2} \right)\left( {k + 4} \right)}}} ?

    All help appreciated!
    How about expanding the integrand as a power series and the integrating term by term?

    CB
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