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Math Help - General Forms of Recursive Series

  1. #1
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    General Forms of Recursive Series

    Hi,

    I have a set of expressions which are related. The relationship is apparent in the multipliers in each line.

    3rd:  {\it t_1}+{\it t_3}
    4th:  {\it t_1}+{\it t_2}+{\it t_3}+{\it t_4}
    5th:  {\it t_1}+2\,{\it t_2}+2\,{\it t_3}+2\,{\it t_4}+{\it t_5}
    6th:  {\it t_1}+3\,{\it t_2}+4\,{\it t_3}+4\,{\it t_4}+3\,{\it t_5}+{\it t_6}
    7th:  {\it t_1}+4\,{\it t_2}+7\,{\it t_3}+8\,{\it t_4}+7\,{\it t_5}+4\,{\it  t_6}+{\it t_7}
    8th:  {\it t_1}+5\,{\it t_2}+11\,{\it t_3}+15\,{\it t_4}+15\,{\it t_5}+11\,{\it  t_6}+5\,{\it t_7}+{\it t_8}
    9th: {\it t_1}+6\,{\it t_2}+16\,{\it t_3}+26\,{\it t_4}+30\,{\it t_5}+26\,{\it t_6}+16\,{\it t_7}+6\,{\it t_8}+{\it t_9}
    10th: {\it t_1}+7\,{\it t_2}+22\,{\it t_3}+42\,{\it t_4}+56\,{\it t_5}+56\,{\it t_6}+42\,{\it t_7}+22\,{\it t_8}+7\,{\it t_9}+{\it t_{10}}

    eg. t1 is always multiplied by 1

    the multiplication of t2 in each case increments by 1. [0 1 2 3 4 5 6 ....]

    the multiplication of t3 in each case increments using the previous increment intervals [0 1 2 3 4 5 6 ....] to give [1 1 2 4 7 11 ...].

    the multiplication of t4 in each case increments using the previous increment intervals [1 1 2 4 7 11 ...] to give [0 1 2 4 8 15 26 ...]

    My problem is this...

    I would like to give a general solution for the Nth order case using some sort of series notation but although there is a pattern clear in the multipliers I am unsure how to go about this.

    Does anyone have any experience with this type of problem or how to go about formulating a general solution.

    Thanks
    Alex
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  2. #2
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    Hello, asouthern!

    I have a set of expressions which are related.
    The relationship is apparent in the multipliers in each line.

    . . \begin{array}{ccccc}\text{3rd:}  &\qquad\qquad\qquad\qquad\qquad t_1+ t_3 \\<br />
\text{4th:}&  \qquad\qquad\qquad\qquad\qquad t_1  + t_2 + t_3 + t_4 \\  \end{array}
    . . \begin{array}{cc}<br />
\text{5th:} & t_1 + 2t_2 + 2t_3 + 2t_4 + t_5 \\ <br />
\text{6th:} & t_1 + 3t_2 + 4t_3 + 4t_4 + 3t_5 + t_6 \\<br />
\text{7th:} & t_1 + 4t_2 + 7t_3 + 8t_4 + 7t_5 + 4t_6 + t_7\\ <br />
\text{8th:} & t_1 + 5t_2 + 11t_3 + 15t_4 + 15t_5 + 11t_6 + 5t_7 + t_8 \\<br />
\text{9th:} & t_1 + 6t_2 + 16t_3 + 26t_4 + 30t_5 + 26t_6 + 16t_7 + 6t_8 + t_9 \\ <br />
\text{10th:} &t_1+7t_2+22t_3+42t_4+56t_5+56t_6+42t_7+22t_8+7t_9  +t_{10} \end{array}
    I see an even stronger pattern, but . . .


    You have a variation of Pascal's Triangle . . .

    . . \begin{array}{cccccccccccc}<br />
3 &&1\;\;0\;\;1\\<br />
4 && 1\;\;1\;\;1\;\;1\\<br />
5 && 1\;\;2\;\;2\;\;2\;\;1\\ <br />
6 && 1\;\;3\;\;4\;\;4\;\;3\;\;1\\<br />
7 && 1\;\;4\;\;7\;\;8\;\;7\;\;4\;\;1\\<br />
8 && 1\;5\;11\;15\;15\;11\;5\;1\\<br />
9 && 1\;\;6\;16\;26\;30\;26\;16\;6\;1 \\ <br />
10 && 1\;7\;22\;42\;56\;56\;42\;22\;7\;1\end{array}


    Each number is the sum of the two numbers directly above it.


    In row 3, we have: [1\:0\:1] which represents the polynomial: x^2 + 1

    Multiply by (x+1)\!:\;\;(x+1)(x^2+1) \:=\:x^3 + x^2 + x + 1
    . . and its coefficients are in row 4.

    Multiply by (x+1)\!:\;\;(x+1)(x^3+x^2+x+1) \:=\:x^4 + 2x^3 + 2x^2 + 2x + 1
    . . Its coefficients are in row 5.

    Multiply by (x+1)\!:\;\;(x+1)(x^4 + 2x^3 + 2x^2 + 2x + 1) \:=\:x^5 + 3x^4 + 4x^3 + 4x^2 + 3x + 1
    . . Ites coefficients are in row 6.


    So I see the powerful pattern embedded in the triangle,
    . . but so far, I haven't been able to express this in a closed form.

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  3. #3
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    Quote Originally Posted by asouthern View Post
    Hi,
    I have a set of expressions which are related. The relationship is apparent in the multipliers in each line.

    3rd: {\it t_1}+{\it t_3}
    4th: {\it t_1}+{\it t_2}+{\it t_3}+{\it t_4}
    5th: {\it t_1}+2\,{\it t_2}+2\,{\it t_3}+2\,{\it t_4}+{\it t_5}
    6th: {\it t_1}+3\,{\it t_2}+4\,{\it t_3}+4\,{\it t_4}+3\,{\it t_5}+{\it t_6}
    7th: {\it t_1}+4\,{\it t_2}+7\,{\it t_3}+8\,{\it t_4}+7\,{\it t_5}+4\,{\it t_6}+{\it t_7}
    8th: {\it t_1}+5\,{\it t_2}+11\,{\it t_3}+15\,{\it t_4}+15\,{\it t_5}+11\,{\it t_6}+5\,{\it t_7}+{\it t_8}
    9th: {\it t_1}+6\,{\it t_2}+16\,{\it t_3}+26\,{\it t_4}+30\,{\it t_5}+26\,{\it t_6}+16\,{\it t_7}+6\,{\it t_8}+{\it t_9}
    10th: {\it t_1}+7\,{\it t_2}+22\,{\it t_3}+42\,{\it t_4}+56\,{\it t_5}+56\,{\it t_6}+42\,{\it t_7}+22\,{\it t_8}+7\,{\it t_9}+{\it t_{10}}

    ...

    Alex
    From where are you getting the t_i values?
    Could you list ALL of the t values at the start?
    for example:

     t_0 = -1
     t_1 = 1
     t_2 = 3
     t_3 = 5
     t_4 = 7
     t_5 = 9

    which would give:
     S_0 = ?
     S_1 = ?
     S_2 = ?
     S_3 = t_1 + t_3 = 6
     S_4 = t_1 + t_2 + t_3 +t_4 = 16
     S_5 = t_1 + 2 t_2 +3 t_3 + 2 t_4 + t_5 = 45


    Soroban shows a modified Pascal Triangle for the coefficients,
    and I'm sure he can modify the binomial expansion to give coefficients for any n.

    Do you have method that generates the t values or are they already permanently defined?

    Is it possible that the  t values and the coefficients are related?
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  4. #4
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    The relation between the t terms

    Hi,

    Thanks for your replies, I agree the pascal triangles do seem the strongest relationship. As for the definition of the t values. Well,

    t_i is shorthand for r\cos{\theta_i}

    r is constant, the angle changes with every order N, so that,

    \theta_i = \frac{\pi\cdot(1+2(i-1))}{2\cdot{N}}

    i = 1,2...,N

    I didn't include the terms for when N =1 and N =2 as they are too simple.

    Thanks
    Alex
    Last edited by asouthern; July 25th 2009 at 07:00 AM. Reason: forgot to give the limits of i
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  5. #5
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    Also,

    In my mind the general representation could be,

    =\sum_{i = 1}^N(a_i\cdot{t_i})

    e.g.
    for N = 4,

    a_i =[1,1,1,1]

    e.g.
    for N = 5,

    a_i =[1,2,2,2,1]

    So some way to form the series a_i for any order N, would be sufficient.

    Any advice on this would be greatly appreciated.

    Thanks Again
    Alex
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