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Math Help - Recurrence relation

  1. #1
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    Recurrence relation

    Solve the recurrence relation
    h_n = h_{n-1} + n^3
    with the initial condition  h_0 = 0, and verify your solution by induction on n.
    Last edited by Plato; October 16th 2010 at 06:15 AM. Reason: LaTeX fix
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  2. #2
    Newbie Hugal's Avatar
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    Didn't you forget some h_{n+1} ? :/

    'Cause your relation is true just for n=1 no matter the value of h_1

    Hugal
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  3. #3
    Newbie Hugal's Avatar
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    Ok ! I got what you meant. It is :

    \displaystyle h_n = h_{n-1}+n^3

    right ?

    So, now just try to figure out what the first terms of this sequence are.
    Try to find h_1, h_2, h_3, h_4 and you'll have quite a good idea of what actually the sequence is.
    After that, you'll have to prove this. Induction is an easy way to make it.
    Good luck,

    Hugal
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  4. #4
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    Quote Originally Posted by nurdynik View Post
    Solve the recurrence relation
    h_n = h_{n-1} + n^3
    with the initial condition  h_0 = 0, and verify your solution by induction on n.
    See this webpage.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by nurdynik View Post
    Solve the recurrence relation
    h_n = h_{n-1} + n^3
    with the initial condition  h_0 = 0, and verify your solution by induction on n.
    The solution is well known...

    \displaystyle h_{n} = \sum_{k=0}^{n} k^{3} = \frac{n^{2}\ (n+1)^{2}}{4}

    Kind regards

    \chi \sigma
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  6. #6
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    Hello, nurdynik!

    Solve the recurrence relation: . h_n \:=\: h_{n-1} + n^3,\;\;h_0 = 0
    and verify your solution by induction on n.

    Crank out the first few terms and you may see a pattern.

    . . \begin{array}{cc}<br />
n & h_n \\ \hline<br />
0 & 0 \\ 1 & 1 \\ 2 & 9 \\ 3 & 36 & 4 & 100 \\ 5 & 225 \\ \vdots & \vdots \end{array}


    We see that the terms of the sequence are squares,
    . . not every consecutive square,
    . . but squares of certain numbers.

    . . \begin{array}{ccccc}<br />
0 &=& 0^2 &=& (0)^2 \\<br />
1 &=& 1^2 &=& (0\!+\!1)^2 \\<br />
9 &=& 3^2 &=& (0\!+\!1\!+\!2)^2 \\<br />
36 &=& 6^2 &=& (0\!+\!1\!+\!2\!+\!3)^2\\<br />
100 &=& 10^2 &=& (0\!+\!1\!+\!2\!+\!3\!+\14)^2 \\<br />
\vdots && \vdots && \vdots \end{array}

    These are the squares of Triangular Numbers, starting with T_0 = 0.


    \text{Therefore: }\;h_n \;=\;\left[\dfrac{n(n+1)}{2}\right]^2
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