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Math Help - Manipulating power series #2

  1. #1
    Senior Member mollymcf2009's Avatar
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    Manipulating power series #2

    Find a power series representation of the function:

    f(x) = \frac{2+x}{1-x}

    Here is what I did:

    f(x) = (2+x)(\frac{1}{1-x})

    2+x \sum^{\infty}_{n=0} (x)^n

    = 2 + \sum^{\infty}_{n=0} x^{2n}

    Which is not right. The correct answer is:

    2 + 3 \sum^{\infty}_{n=1}x^n

    But I don't understand how you would get that answer.
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  2. #2
    o_O
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    You messed up your brackets:

    \left( 2+x\right) \left(\frac{1}{1-x}\right) = \left(2 + x\right) \sum_{n = 0}^{\infty} x^n = 2\sum_{n = 0}^{\infty}x^n + x\sum_{n = 0}^{\infty}x^n

    Now you have to modify your indices so you can add them together:
    = 2\sum_{n = 0}^{\infty} x^n + \sum_{n=0}^{\infty}x^{n+1} = 2\left(1 + \sum_{n = 1}^{\infty} x^n\right) + \sum_{n=1}^{\infty}x^{n} = \cdots
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  3. #3
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    Quote Originally Posted by mollymcf2009 View Post
    Find a power series representation of the function:

    f(x) = \frac{2+x}{1-x}

    Here is what I did:

    f(x) = (2+x)(\frac{1}{1-x})

    2+x \sum^{\infty}_{n=0} (x)^n

    = 2 + \sum^{\infty}_{n=0} x^{2n}

    Which is not right. The correct answer is:

    2 + 3 \sum^{\infty}_{n=1}x^n

    But I don't understand how you would get that answer.
    It's probably easier to notice that

    \frac{2+x}{1-x} = -1 + \frac{3}{1-x}

    use the usual power series for \frac{1}{1-x} multiply by 3 and substract 1.
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  4. #4
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by o_O View Post
    You messed up your brackets:

    \left( 2+x\right) \left(\frac{1}{1-x}\right) = \left(2 + x\right) \sum_{n = 0}^{\infty} x^n = 2\sum_{n = 0}^{\infty}x^n + x\sum_{n = 0}^{\infty}x^n

    Now you have to modify your indices so you can add them together:
    = 2\sum_{n = 0}^{\infty} x^n + \sum_{n=0}^{\infty}x^{n+1} = 2\left(1 + \sum_{n = 1}^{\infty} x^n\right) + \sum_{n=1}^{\infty}x^{n} = \cdots
    That definitely helps, but I still don't see how you end up with 2+3..
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  5. #5
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by danny arrigo View Post
    It's probably easier to notice that

    \frac{2+x}{1-x} = -1 + \frac{3}{1-x}

    use the usual power series for \frac{1}{1-x} multiply by 3 and substract 1.

    THANK YOU!!! That is the simple explanation that I needed!
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