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Thread: Sum of infinite series

  1. #1
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    Sum of infinite series

    Hello, I'm not sure what topic this comes under so apologies for putting it in the wrong place.

    I have a series of numbers, see below. I want to basically continue this number to infinite or it get's very small and then add them all up.

    My method using Excel and what I think wikipedia is telling me to do is;
    1) Calculate the log of y where x is 1 to 20.
    2) Using Excel's Linest function to calculate the slope and intercept of this curve
    3) Then I calculate Mu by Intercept + Slope * x
    3) Calculate variance by Residual Sum of Squared / Degrees if Freedom (output from Linest array formula)
    4) Calculate the expected y using exp(Mu + Variance / 2 )

    This formula like magic enables me to predict using the same curve as my actual data what it would be going forward. I can then copy this formula down and draw a new line until it gets very small. When I plot this line against my actual data it's close enough so I'm happy I've done it correctly. Again like magic if I add a trend line on my actual data it follows my predicted line perfectly the equation being y= 0.0412e-0.142x
    I now want to add up all those predicted terms, happy to not use real data at this point. If I predict this number forward by 20 x's I get a sum of 0.272, if I go forward 40 terms I get 0.274 and for my purpose either number will do.

    I'm hoping there's another magical way to calculate this is in a simple equation.

    Anyone know?


    0.045532
    0.036191
    0.030401
    0.026179
    0.020096
    0.017043
    0.014503
    0.012486
    0.009496
    0.008054
    0.006973
    0.006197
    0.005379
    0.004996
    0.004649
    0.004347
    0.00386
    0.00361
    0.003418
    0.003233
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  2. #2
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    Re: Sum of infinite series

    Alternatively:

    Can I sum y where x = 1 to n

    y= 0.0412e-0.142x
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  3. #3
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    Re: Sum of infinite series

    Quote Originally Posted by RedundantDummy View Post
    Alternatively:

    Can I sum y where x = 1 to n

    y= 0.0412e-0.142x
    $\displaystyle{\sum_{x=1}^n}~\alpha^x = \dfrac{1-\alpha^{n+1}}{1-\alpha} - 1 = \alpha \dfrac{1-\alpha^n}{1-\alpha}$

    $e^{-bx} = \left(e^{-b}\right)^x$

    $\alpha = e^{-b}$

    $\displaystyle{\sum_{x=1}^n}~c e^{-bx} = c \displaystyle{\sum_{x=1}^n}~ \left(e^{-b}\right)^x = c e^{-b} \dfrac{1-e^{-n b}}{1-e^{-b}} = c\dfrac{1-e^{-nb}}{e^b - 1}$

    here

    $c = 0.0412$

    $b = 0.142$

    I leave you to plug in the constants.
    Thanks from topsquark
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