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Math Help - Help me find the formula please!

  1. #1
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    Question Help me find the formula please!

    The question is...

    1st term: -1
    2nd term: 5
    3rd term: 13
    4th term: 23
    5th term: 25
    6th term: 35
    7th term: 49

    so what could nth term be?
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by clarebear14 View Post
    The question is...

    1st term: -1
    2nd term: 5
    3rd term: 13
    4th term: 23
    5th term: 25
    6th term: 35
    7th term: 49

    so what could nth term be?
    a_n=n^2+3n-5 will generate terms 1-4 and 6,7
    ie
    a_1=-1
    a_2=5
    a_3=13
    a_4=23
    a_5=35
    a_6=49

    and so forth...

    This isn't quite the same as yours, are you sure that
    a_5=25?

    I hope this helps....
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  3. #3
    Senior Member JaneBennet's Avatar
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    Clearly the nth term is

    \frac{-(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)}{720}

    +\ \frac{5(n-1)(n-3)(n-4)(n-5)(n-6)(n-7)}{-120}

    +\ \frac{13(n-1)(n-2)(n-4)(n-5)(n-6)(n-7)}{48}

    +\ \frac{23(n-1)(n-2)(n-3)(n-5)(n-6)(n-7)}{-36}

    +\ \frac{25(n-1)(n-2)(n-3)(n-4)(n-6)(n-7)}{48}

    +\ \frac{35(n-1)(n-2)(n-3)(n-4)(n-5)(n-7)}{-120}

    +\ \frac{49(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)}{720}
    Last edited by JaneBennet; February 24th 2008 at 12:13 PM. Reason: Change x to n
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  4. #4
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    lol. lagrange ftw!

    edit: btw, the fourth term's denominator should be -36 not -9.

    f(n)=\frac{{41}}<br />
{{360}}n^6  - \frac{{323}}<br />
{{120}}n^5  + \frac{{1789}}<br />
{{72}}n^4  - \frac{{907}}<br />
{8}n^3  + \frac{{47887}}<br />
{{180}}n^2  - \frac{{4604}}<br />
{{15}}n + 133<br />
    Last edited by xifentoozlerix; February 23rd 2008 at 07:00 PM.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by clarebear14 View Post
    The question is...

    1st term: -1
    2nd term: 5
    3rd term: 13
    4th term: 23
    5th term: 25
    6th term: 35
    7th term: 49

    so what could nth term be?
    Anything that you want.

    Miss Bennet has put an interpolating polynomial through your seven points,
    she could have in fact added an eighth point with value, say 7176, and put
    a polynomial throught these eight points.

    She could also have added:

    (x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)f(x)

    where f(x) is any function on \mathbb{N} to her polynomial and it would still interpolate
    your data exactly.

    Without some background information constraining solutions, such puzzles have no unique solution.

    RonL
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  6. #6
    Super Member angel.white's Avatar
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    Quote Originally Posted by JaneBennet View Post
    Clearly the nth term is

    \frac{-(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)}{720}

    +\ \frac{5(x-1)(x-3)(x-4)(x-5)(x-6)(x-7)}{-120}

    +\ \frac{13(x-1)(x-2)(x-4)(x-5)(x-6)(x-7)}{48}

    +\ \frac{23(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)}{-9}

    +\ \frac{25(x-1)(x-2)(x-3)(x-4)(x-6)(x-7)}{48}

    +\ \frac{35(x-1)(x-2)(x-3)(x-4)(x-5)(x-7)}{-120}

    +\ \frac{49(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)}{720}
    Clearly.
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  7. #7
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by clarebear14 View Post
    The question is...

    1st term: -1
    2nd term: 5
    3rd term: 13
    4th term: 23
    5th term: 25
    6th term: 35
    7th term: 49

    so what could nth term be?
    Just to clarify, the formula I provided above is on the assumption that the terms of your sequence satisfy a polynomial equation of degree at most 6.

    In general, if the first k terms of a sequence (a_n)_{n=1}^{\infty} are given, and it is assumed that the terms of the sequence satisfy a polynomial equation of degree at most k-1, then the nth term of the sequence is

    a_n\ =\ \sum_{j=1}^k{\left(a_j\prod_{1\leq i\leq k,\,i\ne j}{\frac{n-i}{j-i}}\right)}

    Alternatively, you could also let a_n=\sum_{j=1}^k{c_jn^{k-j}} and solve for the coefficients c_j by substituting the first k given values – but this is going to take a hell of a much longer time.
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by JaneBennet View Post
    Just to clarify, the formula I provided above is on the assumption that the terms of your sequence satisfy a polynomial equation of degree at most 6.

    In general, if the first k terms of a sequence (a_n)_{n=1}^{\infty} are given, and it is assumed that the terms of the sequence satisfy a polynomial equation of degree at most k-1, then the nth term of the sequence is

    a_n\ =\ \sum_{j=1}^k{\left(a_j\prod_{1\leq i\leq k,\,i\ne j}{\frac{n-i}{j-i}}\right)}

    Alternatively, you could also let a_n=\sum_{j=1}^k{c_jn^{k-j}} and solve for the coefficients c_j by substituting the first k given values – but this is going to take a hell of a much longer time.
    This is the Lagrange interpolating polynomial. There is also a method due to Newton that uses the difference table to construct the interpolating polynomial. See here

    RonL
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