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  1. #1
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    help with summation of finite series

    I want to know how to calculate this finite sum

    sum of i=1 to 52 of (i^2/3)*1/60

    what method can I use for this?
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    MHF Contributor FernandoRevilla's Avatar
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    Re: help with summation of finite series

    Quote Originally Posted by Kuma View Post
    I want to know how to calculate this finite sum sum of i=1 to 52 of (i^2/3)*1/60 what method can I use for this?
    Hint Use the well known formula \sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}
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    Re: help with summation of finite series

    Hello, Kuma!

    I want to know how to calculate this finite sum

    . . sum of i=1 to 52 of (i^2/3)*1/60

    What method can I use for this?

    Is that: . \frac{1}{60}\sum^{52}_{i=1}i^{\frac{2}{3}}

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    Re: help with summation of finite series

    Gotcha, thanks guys!
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    Re: help with summation of finite series

    hmm seems I'm getting an incorrect answer. Can I split i^2/3 as (i^2)^1/3 and then use the i^2 formula? I don't know if that is the same. I figure my wrong answer is because I used the i^2 formula, plugged in n as 52 and cubed rooted that number, then divided by 60.
    Last edited by Kuma; September 24th 2011 at 09:49 AM.
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    MHF Contributor chisigma's Avatar
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    Re: help with summation of finite series

    Quote Originally Posted by Kuma View Post
    I want to know how to calculate this finite sum

    sum of i=1 to 52 of (i^2/3)*1/60

    what method can I use for this?
    If the sum to be computed is...

    S= \sum_{i=1}^{52} i^{\frac{2}{3} (1)

    .... then, if You transform the (1) in

    S=1+\sum_{i=2}^{n} i^{\frac{2}{3}} = 1 + \sum_{k=1}^{n-1}(1+k)^{\frac{2}{3}} (2)

    ..., apart the 'direct computation' of course, You can consider to use of the 'Euler-McLaurin formula'...

    \sum_{k=1}^{n-1} f(k)= \int_{0}^{n} f(t)\ dt -\frac{1}{2}\ \{f(0)+f(n)\} + \frac{1}{2}\ \{f^{'}(n)-f^{'}(0)\} -

    -\frac{1}{720}\ \{f^{(3)}(n)- f^{(3)}(0)\} + \frac{1}{30240}\ \{f^{(5)}(n)-f^{(5)}(0)\} -

    - \frac{1}{1209600}\ \{f^{(7)}(n)-f^{(7)}(0)\} +...

    ...+ \frac{B_{2k}}{(2k)!}\ \{f^{(2k-1)}(n)-f^{(2k-1)} (0)\} +... (1)

    ... where the B_{2k} are the 'Bernoulli's numbers'. In Your case is f(t)= (1+t)^{\frac{2}{3}} and n=52. The computation of the f^{(2k-1)}(*) is left to You. It is difficult to extablish for what n the Euler-McLaurin formula [which of course is 'approximative'...] is 'competitive' with the direct computation...

    Kind regards

    \chi \sigma
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    Re: help with summation of finite series

    I still don't get it...do I have to use the euler formula for this? Wlfram alpha gave me the answer but my manual calculation is incorrect.
    help? This is what I did

    1/60 [(n)(n+1)(2n+1)6]^1/3 and plug n in as 52.
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    MHF Contributor FernandoRevilla's Avatar
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    Re: help with summation of finite series

    Quote Originally Posted by Kuma View Post
    I want to know how to calculate this finite sum
    sum of i=1 to 52 of (i^2/3)*1/60 what method can I use for this?
    If you don't use \LaTeX , be careful with brackets

    sum of i=1 to 52 of (i^2/3)*1/60 means \sum_{i=1}^{52}\frac{i^2}{3}\cdot \frac{1}{60}

    sum of i=1 to 52 of ( i^(2/3) )*1/60 means \sum_{i=1}^{52}i^{\frac{2}{3}}\cdot \frac{1}{60}
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    Re: help with summation of finite series

    Yes it is the second one. I can't use the i^2 formula on it. How do I solve this?? I've been trying to figure out if fits into some sort of series like power series but all of those are infinite that involve convergence/divergence. What should I do?!
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  10. #10
    MHF Contributor chisigma's Avatar
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    Re: help with summation of finite series

    Quote Originally Posted by Kuma View Post
    Yes it is the second one. I can't use the i^2 formula on it. How do I solve this?? I've been trying to figure out if fits into some sort of series like power series but all of those are infinite that involve convergence/divergence. What should I do?!
    Why are You so discouraged?... Your task is the computation of...

    S=\sum_{i=1}^{52} i^{\frac{2}{3}} (1)

    ... You have two possibilities...

    a) if a computer is available for You the 'brute force sum' gives as result...

    S_{52,BF}= 441.492660573...

    b) if only a [scientific...] calculator and limited time is available for You, the Euler-McLaurin formula, that I described few post ago, may be useful...

    Now we write the value of S_{n, EMc} obtained with the Euler-McLaurin formula 'breaked up' to the third term...

    S_{n,EMc}= 1 + \int_{0}^{n} f(t)\ dt -\frac{1}{2}\ \{f(0)+f(n)\} +

    + \frac{1}{2}\ \{f^{'}(n)-f^{'}(0)\} - \frac{1}{720}\ \{f^{'''}(n)-f^{'''}(0)\} (2)

    ... where is f(t)=(1+t)^{\frac{2}{3}}. If You insert in (2) n=52 You obtain...

    S_{52,EMc}= 441.333255853...

    If more precision is requested, You can use more terms of the Euler-McLaurin formula. May be interesting to compare the 'brute force sum' and the 'Euler-McLaurin third degree sum' for greater values of n...

    S_{100,BF}= 1303.28975964...\ ,\ S_{100,EMc}= 1303.10747032...

    S_{1000,BF}= 60049.8503587...\ ,\ S_{1000,EMc}= 60049.6170634...

    S_{10000,BF}= 2.78518522699... 10^{6}\ ,\ S_{10000,EMc}= 2.78518496989... 10^{6}

    Kind regards

    \chi \sigma
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