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Math Help - [SOLVED] Sum of Series...

  1. #1
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    [SOLVED] Sum of Series...

    Find, as function of x, the sum of the infinite series and state its interval of convergence.

    x - 2x^2 + 4x^3 - 8x^4 + ... + (-1)^n 2^n x^{n+1} + ...


    Hey, I hope this is the right board..

    How would I determine the answer? I think it's a combination of power series, (like e^x + another series or something like that) but I'm just guessing.

    Thanks.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Solo View Post
    Find, as function of x, the sum of the infinite series and state its interval of convergence.

    x - 2x^2 + 4x^3 - 8x^4 + ... + (-1)^n 2^n x^{n+1} + ...


    Hey, I hope this is the right board..

    How would I determine the answer? I think it's a combination of power series, (like e^x + another series or something like that) but I'm just guessing.

    Thanks.


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

    So this is a geometric series with r=(-2x) so we get

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

    Remember that geometric series converege when r< 1
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  3. #3
    Moo
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    Hello,

    You're looking for :
    \sum_{n\geq 0} (-1)^n 2^n x^{n+1}
    Factor out x :
    =x\sum_{n\geq 0} (-1)^n 2^n x^n
    And you can see that this simplifies to :
    =x\sum_{n\geq 0} (-2x)^n


    Provided that |-2x|<1 \Rightarrow |x|<\tfrac 12, this is a convergent geometric series (multiplied by x)


    Edit : muff...too late ><

    Editē :
    So this is a geometric series with r=(-2x) so we get

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

    Remember that geometric series converege when r< 1
    Well, actually you have to state the condition over x before writing what f(x) is
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  4. #4
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    Ah yes!

    You guys are legends. Thank you very much!
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by Moo View Post
    Hello,

    You're looking for :
    \sum_{n\geq 0} (-1)^n 2^n x^{n+1}
    Factor out x :
    =x\sum_{n\geq 0} (-1)^n 2^n x^n
    And you can see that this simplifies to :
    =x\sum_{n\geq 0} (-2x)^n


    Provided that |-2x|<1 \Rightarrow |x|<\tfrac 12, this is a convergent geometric series (multiplied by x)


    Edit : muff...too late ><

    Editē :

    Well, actually you have to state the condition over x before writing what f(x) is
    Thanks for keeping me honest Moo
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