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Thread: Sequence and series help

  1. #1
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    Sequence and series help

    how would i find the function? The series is 1,2,2,3,3,3,4,4,4,4 and so on with there being n number of the term n. How would i make a function to find the nth term?

    Grateful for help.
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  2. #2
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    You can try

    $\displaystyle u_n = -E(\frac{1-\sqrt{8n+1}}{2})$

    where $\displaystyle k \leq E(k) < k+1$

    If needed I can provide you the way I found this function
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  3. #3
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    How?

    Yes, I want to know how you got it.
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  4. #4
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    Quote Originally Posted by sethu1989 View Post
    Yes, I want to know how you got it.
    $\displaystyle C_n = \left\lceil {\frac{{\sqrt {8n + 1} - 1}}{2}} \right\rceil $
    That is the ceiling function.
    The 'change' come on the triangular numbers: $\displaystyle \frac {n(n+1)}{2}$.
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  5. #5
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    Quote Originally Posted by Plato View Post
    $\displaystyle C_n = \left\lceil {\frac{{\sqrt {8n + 1} - 1}}{2}} \right\rceil $
    That is the ceiling function.
    The 'change' come on the triangular numbers: $\displaystyle \frac {n(n+1)}{2}$.
    Thanks !
    I have never heard of this 'ceiling function' before
    I found it by myself
    I took $\displaystyle D_n = -\left\lceil {\frac{1 - {\sqrt {8n + 1}}}{2}} \right\rceil $

    instead of $\displaystyle C_n = \left\lceil {\frac{{\sqrt {8n + 1} - 1}}{2}} \right\rceil $
    because it is not exactly the same (if I am not mistaken)
    $\displaystyle C_n = 1, 1, 2, 2, 2, 3, 3, 3, 3 ...$
    $\displaystyle D_n = 1,2, 2, 3, 3, 3, ...$
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  6. #6
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    Quote Originally Posted by sethu1989 View Post
    Yes, I want to know how you got it.
    Not knowing the 'ceiling function' here is how I got it
    First you have to know that
    $\displaystyle \sum_{j=1}^{k}j = \frac{k(k+1)}{2}$

    $\displaystyle u_n = k$ for every n integer such as $\displaystyle \frac{(k-1)k}{2} < n \leq \frac{k(k+1)}{2}$

    n being given, to find k you must have both inequations
    $\displaystyle (k-1)k - 2n < 0$ and $\displaystyle k(k+1) - 2n \geq 0$

    $\displaystyle k^2 - k - 2n$ and $\displaystyle k^2 + k - 2n$ are parabolas with the same discriminant 8n+1

    $\displaystyle k^2 - k - 2n < 0$ for $\displaystyle k < \frac{1+\sqrt{8n + 1}}{2}$
    $\displaystyle k^2 + k - 2n \geq 0$ for $\displaystyle k \geq \frac{-1+\sqrt{8n + 1}}{2}$

    Therefore $\displaystyle \frac{-1+\sqrt{8n + 1}}{2} \leq k < \frac{1+\sqrt{8n + 1}}{2}$

    There is only one integer k satisfying these inequations because

    $\displaystyle \frac{1+\sqrt{8n + 1}}{2} - \frac{-1+\sqrt{8n + 1}}{2} = 1$
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