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Math Help - closed form is needed

  1. #1
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    closed form is needed

    I've been searching the Internet for a closed of a following expression :



    $\sum_{i=1}^{x} \frac{i}{2^{i}}$

    however unsuccessful.

    Can anyone shad some light ??

    b

    pS

    even an approximation would do.


    sorry for wasting time ,wikipedia has the answer i somehow overlooked it

    sorry
    Last edited by baxy77bax; August 16th 2014 at 07:42 AM.
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  2. #2
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    Re: closed form is needed

    Quote Originally Posted by baxy77bax View Post
    I've been searching the Internet for a closed of a following expression :



    $\sum_{i=1}^{x} \frac{i}{2^{i}}$

    however unsuccessful.

    Can anyone shad some light ??

    b

    pS

    even an approximation would do.
    $\displaystyle{\sum_{i=1}^x}\dfrac{i}{2^i}=2^{-x} \left(-x+2^{x+1}-2\right)$
    Thanks from Prove It and topsquark
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  3. #3
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    Re: closed form is needed

    Quote Originally Posted by romsek View Post
    $\displaystyle{\sum_{i=1}^x}\dfrac{i}{2^i}=2^{-x} \left(-x+2^{x+1}-2\right)$
    Just out of interest romsek, how did you evaluate this?
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  4. #4
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    Re: closed form is needed

    Quote Originally Posted by Prove It View Post
    Just out of interest romsek, how did you evaluate this?
    I cheated and dumped it into Mathematica
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  5. #5
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    Re: closed form is needed

    Hello, baxy77bax!

    Find a closed form for: . \sum^x_{i=1}\frac{i}{2^i}

    \begin{array}{ccccccc}\text{We have:} & S &=& \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + \frac{5}{2^5} + \cdots + \frac{x}{2^x}\qquad\quad & [1] \\ \text{Multiply by }\frac{1}{2}\!: & \frac{1}{2}S &=& \quad\quad \frac{1}{2^2} + \frac{2}{2^3} + \frac{3}{2^4} + \frac{4}{2^5} + \cdots + \frac{x-1}{2^x} + \frac{x}{2^{x+1}} & [2] \\  \text{Subtract [1]-[2]:} & \frac{1}{2}S &=& \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \frac{1}{2^5} + \cdots + \frac{1}{2^x} - \frac{x}{2^{x+1}}  \end{array}

    Hence: . \frac{1}{2}S \;=\;\left(1 - \frac{1}{2^x}\right) - \frac{x}{2^{x+1}} \quad\Rightarrow\quad S \;=\;2 - \frac{1}{2^{x-1}} - \frac{x}{2^x}

    Therefore: . S \;=\;\frac{1}{2^x}\left(2^{x+1} - 2 - x\right)
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