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Thread: Help Generating Functions :)

  1. #1
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    Help Generating Functions :)

    Hey,
    I have a question about creating functions:
    Given - Help Generating Functions :)-1.jpg, that - Help Generating Functions :)-2.jpg.


    1. I am right that: a0=1, a1=2, a2=2?
    2. We need to find numbers - r, s, t, so that - Help Generating Functions :)-3.jpg, for all 3≤n. Think of a7 using this formula.


    To question 2 I need the start of the solution if possible please
    Thanks
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  2. #2
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    Re: Help Generating Functions :)

    $$\dfrac{d}{dx}\left( \dfrac{1}{(1-x)^2} \right) = \dfrac{2}{(1-x)^3}$$

    So, we have:

    $$\displaystyle \dfrac{1}{(1-x)^3} = \dfrac{1}{2}\sum_{n\ge 0}(n+2)(n+1)x^n$$

    This gives:

    $$\begin{matrix} & | & 1 & x & x^2 & x^3 & x^4 & \cdots & x^n \\ \hline f(x) & | & a_0 & a_1 & a_2 & a_3 & a_4 & \cdots & a_n \\ 2xf(x) & | & 0 & 2a_0 & 2a_1 & 2a_2 & 2a_3 & \cdots & 2a_{n-1} \\ 2x^2f(x) & | & 0 & 0 & 2a_0 & 2a_1 & 2a_2 & \cdots & 2a_{n-2} \\ x^3f(x) & | & 0 & 0 & 0 & a_0 & a_1 & \cdots & a_{n-3} \\ \hline \text{sum} & | & \dfrac{(2)(1)}{2} & \dfrac{(3)(2)}{2} & \dfrac{(4)(3)}{2} & \dfrac{(5)(4)}{2} & \dfrac{(6)(5)}{2} & \cdots & \dbinom{n+2}{2} \end{matrix}$$

    So, we have:

    $a_0 = 1$

    $a_1 + 2a_0 = a_1+2 = 3$ so $a_1 = 1$

    $a_2 + 2a_1 + 2a_0 = a_2 + 4 = 6$, so $a_2 = 2$

    In general:

    $$a_n + 2 a_{n-1} + 2a_{n-2} + a_{n-3} = \dbinom{n+2}{2}$$
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  3. #3
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    Re: Help Generating Functions :)

    Quote Originally Posted by SlipEternal View Post
    $$\dfrac{d}{dx}\left( \dfrac{1}{(1-x)^2} \right) = \dfrac{2}{(1-x)^3}$$

    So, we have:

    $$\displaystyle \dfrac{1}{(1-x)^3} = \dfrac{1}{2}\sum_{n\ge 0}(n+2)(n+1)x^n$$

    This gives:

    $$\begin{matrix} & | & 1 & x & x^2 & x^3 & x^4 & \cdots & x^n \\ \hline f(x) & | & a_0 & a_1 & a_2 & a_3 & a_4 & \cdots & a_n \\ 2xf(x) & | & 0 & 2a_0 & 2a_1 & 2a_2 & 2a_3 & \cdots & 2a_{n-1} \\ 2x^2f(x) & | & 0 & 0 & 2a_0 & 2a_1 & 2a_2 & \cdots & 2a_{n-2} \\ x^3f(x) & | & 0 & 0 & 0 & a_0 & a_1 & \cdots & a_{n-3} \\ \hline \text{sum} & | & \dfrac{(2)(1)}{2} & \dfrac{(3)(2)}{2} & \dfrac{(4)(3)}{2} & \dfrac{(5)(4)}{2} & \dfrac{(6)(5)}{2} & \cdots & \dbinom{n+2}{2} \end{matrix}$$

    So, we have:

    $a_0 = 1$

    $a_1 + 2a_0 = a_1+2 = 3$ so $a_1 = 1$

    $a_2 + 2a_1 + 2a_0 = a_2 + 4 = 6$, so $a_2 = 2$

    In general:

    $$a_n + 2 a_{n-1} + 2a_{n-2} + a_{n-3} = \dbinom{n+2}{2}$$

    Hi,
    First of all thank you for the detailed explanation, so I understand that I made a mistake in a1 ...
    And can you give me help with second question?
    Thanks
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  4. #4
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    Re: Help Generating Functions :)

    oops...
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  5. #5
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    Re: Help Generating Functions :)

    Quote Originally Posted by yossa View Post
    Hi,
    First of all thank you for the detailed explanation, so I understand that I made a mistake in a1 ...
    And can you give me help with second question?
    Thanks
    Solve for $a_n$ in the last equation...
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  6. #6
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    Re: Help Generating Functions :)

    Quote Originally Posted by SlipEternal View Post
    Solve for $a_n$ in the last equation...
    that the solve to equaiton 2?
    In general:

    $$a_n + 2 a_{n-1} + 2a_{n-2} + a_{n-3} = \dbinom{n+2}{2}$$[/QUOTE]
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  7. #7
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    Re: Help Generating Functions :)

    Quote Originally Posted by yossa View Post
    that the solve to equaiton 2?
    In general:

    $$a_n + 2 a_{n-1} + 2a_{n-2} + a_{n-3} = \dbinom{n+2}{2}$$
    [/QUOTE]

    Yes, that's it.
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  8. #8
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    Re: Help Generating Functions :)

    Yes, that's it.[/QUOTE]

    Ok, only if a little refinement is possible, where do I attach importance here to a7?
    and i didnt need 3 numbers?
    r,s,t?
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  9. #9
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    Re: Help Generating Functions :)

    I don't know what you mean by "where do I attach importance here to a7".

    You are looking for:
    $a_n = D(3,n) - ra_{n-1}-sa_{n-2}-ta_{n-3}$.

    You have:
    $a_n = \dbinom{n+2}{2}-2a_{n-1}-2a_{n-2}-a_{n-3}$.

    What are $r,s,t$? I am not sure I understand what $D(3,n)$ is. I assumed it was $D(3,n) = \dbinom{n+2}{2}$.
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  10. #10
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    Re: Help Generating Functions :)

    Quote Originally Posted by SlipEternal View Post
    I don't know what you mean by "where do I attach importance here to a7".

    You are looking for:
    $a_n = D(3,n) - ra_{n-1}-sa_{n-2}-ta_{n-3}$.

    You have:
    $a_n = \dbinom{n+2}{2}-2a_{n-1}-2a_{n-2}-a_{n-3}$.

    What are $r,s,t$? I am not sure I understand what $D(3,n)$ is. I assumed it was $D(3,n) = \dbinom{n+2}{2}$.
    Hey friend,
    I will try to explain myself more, I need to find three numbers (R, S, T) where the N is greater than or equal to 3, and then calculate A 7 using this formula (here I think it's only a placement and calculation) ...
    Hope more sense.
    thanks.....
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  11. #11
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    Re: Help Generating Functions :)

    Ok, then

    $a_7 = \dbinom{9}{2}-2a_6-2a_5-a_4$
    $a_6 = \dbinom{8}{2}-2a_5-2a_4-a_3$
    $a_5 = \dbinom{7}{2}-2a_4-2a_3-a_2 = \dbinom{7}{2}-2a_4-2a_3-2$
    $a_4 = \dbinom{6}{2}-2a_3-2a_2-a_1 = \dbinom{6}{2}-2a_3-5$
    $a_3 = \dbinom{5}{2}-2a_2-2a_1-a_0 = \dbinom{5}{2}-7 = 3$

    Plugging back in:
    $a_4 = \dbinom{6}{2}-2a_3-5 = 15-6-5 = 4$
    $a_5 = \dbinom{7}{2}-2a_4-2a_3-2 = 21-8-6-2 = 5$
    $a_6 = \dbinom{8}{2}-2a_5-2a_4-a_3 = 28-10-8-3 = 7$
    $a_7 = \dbinom{9}{2}-2a_6-2a_5-a_4 = 36-14-10-4 = 8$
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