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Math Help - I need power series help and I don't get when to change the index

  1. #1
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    I need power series help and I don't get when to change the index

    The problem:

    arctan(x/3)

    I assume you derive it: 1/(1+(x/3))

    then put it in the proper formation: 1/(1- (-x/3))

    so then I get the sigma of -1^2n -x/3^n

    and then I have to take the integral?

    I just don't get what to do, like the proper formation. Why do I have to keep the -1 and -x/3 seperate if they are both to the 2n power?

    And in an unrelated note, when do you have to change the index? I don't get that AT ALL.


    FML. thanks
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  2. #2
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    Re: I need power series help and I don't get when to change the index

    Quote Originally Posted by skinsdomination09 View Post
    The problem:

    arctan(x/3)

    I assume you derive it: 1/(1+(x/3))

    then put it in the proper formation: 1/(1- (-x/3))

    so then I get the sigma of -1^2n -x/3^n

    and then I have to take the integra
    I really have no idea what that says.

    But look at this webpage.
    Does that help?
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  3. #3
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    Re: I need power series help and I don't get when to change the index

    Quote Originally Posted by skinsdomination09 View Post
    The problem:

    arctan(x/3)

    I assume you derive it: 1/(1+(x/3))

    then put it in the proper formation: 1/(1- (-x/3))

    so then I get the sigma of -1^2n -x/3^n

    and then I have to take the integral?

    I just don't get what to do, like the proper formation. Why do I have to keep the -1 and -x/3 seperate if they are both to the 2n power?

    And in an unrelated note, when do you have to change the index? I don't get that AT ALL.


    FML. thanks
    First of all, \displaystyle \begin{align*} \frac{1}{1 - \left( -\frac{x}{3} \right) ^2} \end{align*} is NOT the same as \displaystyle \frac{1}{1 + \left( \frac{x}{3} \right) ^2}. You should be writing it as \displaystyle \begin{align*} \frac{1}{1 - \left[ - \left( \frac{x}{3} \right) ^2 \right]} \end{align*}.

    Now the easiest thing to do is to remember that an infinite geometric series \displaystyle \sum_{n = 0}^{\infty} a\,r^n = \frac{a}{1-r} , which is convergent when |r| < 1. Can you see that what you have looks like this?
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