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Math Help - Closed forms of summations and reprospective relations

  1. #1
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    Unhappy Closed forms of summations and reprospective relations

    Hello and have a nice week !
    Im asking your help with the following exercices (tried but failed).

    So the first one
    Find the closed form of the double summation
    S(from j=1 to n) S(from i=0 to infinity) j^5/3 (1-1/2j^1/3)^I


    The second one
    (a) T(n)=T(n/4)+log4_n and T(1)=0 fond the closed form .
    (b) A flower can live for 2 years and reproduct once a year.Specifically it reproducts during its first year of life . Find a reprospective relation that describes the number of flowers in time n and then find a closed form if we know that in time 0 there is only one flower.


    Thanks in advance !!!!
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  2. #2
    Grand Panjandrum
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    Re: Closed forms of summations and reprospective relations

    Quote Originally Posted by grainofsand View Post

    So the first one
    Find the closed form of the double summation
    S(from j=1 to n) S(from i=0 to infinity) j^5/3 (1-1/2j^1/3)^I
    Put in sufficient brackets to make your expressions unambiguous!, As it stands there is no closed form as the inner sum is divergent.

    So let us assume you mean:

    S(n)=\sum_{j=1}^n \sum_{i=0}^{\infty} j^{5/3}\left(1-\frac{1}{2j^{1/3}}\right)^i

    Write it as:

    S(n)=\sum_{j=1}^n j^{5/3} \sum_{i=0}^{\infty} \left(1-\frac{1}{2j^{1/3}}\right)^i

    Now the inner sum is an infinite geometric series with sum:

    S_{inner}(j)= \frac{1}{1-(1-1/(2j^{1/3}))}=2j^{1/3}

    ...

    CB
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  3. #3
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    Re: Closed forms of summations and reprospective relations

    Quote Originally Posted by CaptainBlack View Post
    Put in sufficient brackets to make your expressions unambiguous!, As it stands there is no closed form as the inner sum is divergent.

    So let us assume you mean:

    S(n)=\sum_{j=1}^n \sum_{i=0}^{\infty} j^{5/3}\left(1-\frac{1}{2j^{1/3}}\right)^i

    Write it as:

    S(n)=\sum_{j=1}^n j^{5/3} \sum_{i=0}^{\infty} \left(1-\frac{1}{2j^{1/3}}\right)^i

    Now the inner sum is an infinite geometric series with sum:

    S_{inner}(j)= \frac{1}{1-(1-1/(2j^{1/3}))}=2j^{1/3}

    ...

    CB

    really thanks for the help !
    i solved the first one !!!!!!!
    But what about the other one with the 2 parts ????
    And again thanks in advance !
    for the second one part a i found sth but i dont think it is right (cause i dont even use the t(1)=0).
    T(n)=O((log4_n)*n^(log4_2) How can i find the closed form of the reprospective relation ?
    and what about the part b ?
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