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Math Help - power series again

  1. #1
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    power series again

    f(x) = ln[(1+x^2)/(1-x^2)]

    = ln(1+x^2) - ln(1-x^2)

    i know log(1+x) = x^n+1 / n+1 but i dont know how i am supposed to modify the equation or whatever to get what i need...
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  2. #2
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    Quote Originally Posted by jeph View Post
    f(x) = ln[(1+x^2)/(1-x^2)]

    = ln(1+x^2) - ln(1-x^2)

    i know log(1+x) = x^n+1 / n+1 but i dont know how i am supposed to modify the equation or whatever to get what i need...
    \ln (1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} + ... \mbox{ for }-1<t \leq 1

    Now to find,
    \ln (1+x^2) use t=x^2 \mbox{ for }|x|<1

    And, \ln (1-x^2) = \ln[(1-x)\cdot(1+x)] = \ln (1-x) + \ln (1+x)

    To find,
    \ln (1-x) use t=-x \mbox{ for }|x|<1

    And combine them together.
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