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Math Help - formula for a sequence

  1. #1
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    formula for a sequence

    i am looking for an explicit formula for the sequence given by the following numbers:

    0, 1/2, 1, 2/3, 1/3, 0, 1/4, 2/4, 3/4, 1, 4/5, 3/5, 2/5, 1/5, 0, 1/6, etc..
    Last edited by kkoutsothodoros; July 4th 2009 at 10:54 AM.
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  2. #2
    Super Member malaygoel's Avatar
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    I could get upto this:

    nth term, t_n=\frac{d-m}{d}

    where
    d=\left\lceil\frac{\sqrt{8n+1}-1}{2}\right\rceil

    m=(-1)^d  \left( \frac {d(d+(-1)^d)}{2}-n \right)

    someone, please silmpify
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  3. #3
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    d_{n}=\left\lfloor\sqrt{2n}+\frac{1}{2}\right\rflo  or
    Last edited by kkoutsothodoros; July 5th 2009 at 07:01 AM. Reason: wrong notation
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  4. #4
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    we get the general term of the sequence to be:


    a_{n}= \sum_{n=1}^{n}\frac{(-1)^{d_{n}}}{d_{n}}\

    i figured this out yesterday but didn't have the time to learn LaTex. And I had Mr. Fantastic sending me demerits all day long. maybe he could have spent more time helping with this as opposed to sitting in front of his computer sending out demerits!!
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  5. #5
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    i will probably get thrown off the site now!! well it was fun learning a little bit of latex!
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  6. #6
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    oh yes, one more thing,

    a_{0}=0.
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  7. #7
    Super Member malaygoel's Avatar
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    Quote Originally Posted by kkoutsothodoros View Post
    we get the general term of the sequence to be:


    a_{n}= \sum_{n=1}^{n}\frac{(-1)^{d_{n}}}{d_{n}}\
    I am not able to follow your answer...could you please give example for some values of n.
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  8. #8
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    it does seem a little unclear. sorry, i am just learning how to use latex.

    let a_{n} = \frac{(-1)^{d_{n}}}{d_{n}}\ for n >= 2, a_{1} = 0.

    let s_{n}\ = \sum_{n=1}^{n}a_{n}.

    you should get

    s_{1} = 0
    s_{2} = .5
    s_{3} = 1
    s_{4} = 2/3
    etc.
    Last edited by kkoutsothodoros; July 5th 2009 at 08:29 AM. Reason: notation error
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  9. #9
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    so s_{n} is the general term for the sequence 0, 1/2, 2/2, 2/3, 1/3, 0/3, 1/4, 2/4, 3/4, 4/4, 4/5, 3/5, etc
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  10. #10
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    the funny thing is this sequence is just used as a counterexample in a cauchy sequence problem i was looking at. it wasn't that important to get the general term formula. it just bugged me so much that i couldn't let it go. but i guess it's useful that i learned how to get a general term for 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, etc along the way. the general term for this is:

    d_{n}=\left\lfloor\sqrt{2n}+\frac{1}{2}\right\rflo  or
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  11. #11
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    Quote Originally Posted by malaygoel View Post
    I am not able to follow your answer...could you please give example for some values of n.
    is it more clear now?
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  12. #12
    Super Member malaygoel's Avatar
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    Quote Originally Posted by kkoutsothodoros View Post
    the funny thing is this sequence is just used as a counterexample in a cauchy sequence problem i was looking at. it wasn't that important to get the general term formula. it just bugged me so much that i couldn't let it go. but i guess it's useful that i learned how to get a general term for 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, etc along the way. the general term for this is:

    d_{n}=\left\lfloor\sqrt{2n}+\frac{1}{2}\right\rflo  or
    your expression and my expession for d(given in post #2) are quite different. And they both work. Can you explain how you got your expression. mine had a simple logic.
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  13. #13
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    Consider 1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6,6,6,6,6,6,etc

    where the integer m appears m times in the sequence. using the fact that

    1+2+3+...+n=\frac{n(n+1)}{2}\ we have \frac{m(m-1)}{2}\ < n <= \frac{m(m+1)}{2}\

    this inequality leads to:

    m < \sqrt{2n}+\frac{1}{2}\ <= m + 1 which is equivalent to what we are looking for.
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  14. #14
    Super Member malaygoel's Avatar
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    Quote Originally Posted by kkoutsothodoros
    Consider 1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6,6,6,6,6,6,etc

    where the integer m appears m times in the sequence. using the fact that

    1+2+3+...+n=\frac{n(n+1)}{2}\ we have \frac{m(m-1)}{2}\ < n <= \frac{m(m+1)}{2}\
    Upto here, everything is correct.

    Now, how do you get this

    m < \sqrt{2n}+\frac{1}{2}\ <= m + 1
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  15. #15
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    sorry again, this part is a little tricky and i found an error.

    multiply \frac{m(m-1)}{2}\ < n <= \frac{m(m+1)}{2}\ by 2 to get m^2 - m < 2n <=m^2 + m. now add 1/4 to both sides and factor to get m^2 - m + 1/4< 2n + 1/4 <=m^2 + m + 1/4 or (m-1/2)^2 < 2n + 1/4 <= (m + 1/2)^2. take the square root of everything and we get (m-1/2) < \sqrt{2n + 1/4}\ <= (m + 1/2).

    add 1/2 to both sides and we get m < \sqrt{2n + 1/4}\ + 1/2 <= m + 1.

    i just checked it in Excel and the 1/4 is not affecting the result, luckily. it still works if we start with n = 0.
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