Solve the recurrence relation $a_n=a_{n-1}+2n+3a_{n-3}?$












0












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I am currently solving a recurrence relation but I got stuck since I am not even able to find the basic root of the auxiliary equations.






recurrence-relations homogeneous-equation







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  • $begingroup$
    Do you mean $a_n = a_{n-1} + 2n + 3a_{n-3}$ or $a_n = a_{n-2} + 2 a_{n-2} + 3 a_{n-3}$ (or something else entirely)?
    $endgroup$
    – Connor Harris
    Jan 17 at 14:41










  • $begingroup$
    First one!!!!!!
    $endgroup$
    – Sagar Sharma
    Jan 17 at 14:43










  • $begingroup$
    Do you have initial conditions?
    $endgroup$
    – Adrian Keister
    Jan 17 at 14:49










  • $begingroup$
    Initial conditions are varying.
    $endgroup$
    – Sagar Sharma
    Jan 17 at 14:51








  • 1




    $begingroup$
    Let $a_n=b_n+p+qn$ to eliminate $n$
    $endgroup$
    – lab bhattacharjee
    Jan 17 at 15:16
















0












$begingroup$


I am currently solving a recurrence relation but I got stuck since I am not even able to find the basic root of the auxiliary equations.






recurrence-relations homogeneous-equation







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean $a_n = a_{n-1} + 2n + 3a_{n-3}$ or $a_n = a_{n-2} + 2 a_{n-2} + 3 a_{n-3}$ (or something else entirely)?
    $endgroup$
    – Connor Harris
    Jan 17 at 14:41










  • $begingroup$
    First one!!!!!!
    $endgroup$
    – Sagar Sharma
    Jan 17 at 14:43










  • $begingroup$
    Do you have initial conditions?
    $endgroup$
    – Adrian Keister
    Jan 17 at 14:49










  • $begingroup$
    Initial conditions are varying.
    $endgroup$
    – Sagar Sharma
    Jan 17 at 14:51








  • 1




    $begingroup$
    Let $a_n=b_n+p+qn$ to eliminate $n$
    $endgroup$
    – lab bhattacharjee
    Jan 17 at 15:16














0












0








0


1



$begingroup$


I am currently solving a recurrence relation but I got stuck since I am not even able to find the basic root of the auxiliary equations.






recurrence-relations homogeneous-equation







share|cite|improve this question











$endgroup$




I am currently solving a recurrence relation but I got stuck since I am not even able to find the basic root of the auxiliary equations.






recurrence-relations homogeneous-equation




recurrence-relations homogeneous-equation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 14:48









Adrian Keister

5,26571933




5,26571933










asked Jan 17 at 14:36









Sagar SharmaSagar Sharma

143




143












  • $begingroup$
    Do you mean $a_n = a_{n-1} + 2n + 3a_{n-3}$ or $a_n = a_{n-2} + 2 a_{n-2} + 3 a_{n-3}$ (or something else entirely)?
    $endgroup$
    – Connor Harris
    Jan 17 at 14:41










  • $begingroup$
    First one!!!!!!
    $endgroup$
    – Sagar Sharma
    Jan 17 at 14:43










  • $begingroup$
    Do you have initial conditions?
    $endgroup$
    – Adrian Keister
    Jan 17 at 14:49










  • $begingroup$
    Initial conditions are varying.
    $endgroup$
    – Sagar Sharma
    Jan 17 at 14:51








  • 1




    $begingroup$
    Let $a_n=b_n+p+qn$ to eliminate $n$
    $endgroup$
    – lab bhattacharjee
    Jan 17 at 15:16


















  • $begingroup$
    Do you mean $a_n = a_{n-1} + 2n + 3a_{n-3}$ or $a_n = a_{n-2} + 2 a_{n-2} + 3 a_{n-3}$ (or something else entirely)?
    $endgroup$
    – Connor Harris
    Jan 17 at 14:41










  • $begingroup$
    First one!!!!!!
    $endgroup$
    – Sagar Sharma
    Jan 17 at 14:43










  • $begingroup$
    Do you have initial conditions?
    $endgroup$
    – Adrian Keister
    Jan 17 at 14:49










  • $begingroup$
    Initial conditions are varying.
    $endgroup$
    – Sagar Sharma
    Jan 17 at 14:51








  • 1




    $begingroup$
    Let $a_n=b_n+p+qn$ to eliminate $n$
    $endgroup$
    – lab bhattacharjee
    Jan 17 at 15:16
















$begingroup$
Do you mean $a_n = a_{n-1} + 2n + 3a_{n-3}$ or $a_n = a_{n-2} + 2 a_{n-2} + 3 a_{n-3}$ (or something else entirely)?
$endgroup$
– Connor Harris
Jan 17 at 14:41




$begingroup$
Do you mean $a_n = a_{n-1} + 2n + 3a_{n-3}$ or $a_n = a_{n-2} + 2 a_{n-2} + 3 a_{n-3}$ (or something else entirely)?
$endgroup$
– Connor Harris
Jan 17 at 14:41












$begingroup$
First one!!!!!!
$endgroup$
– Sagar Sharma
Jan 17 at 14:43




$begingroup$
First one!!!!!!
$endgroup$
– Sagar Sharma
Jan 17 at 14:43












$begingroup$
Do you have initial conditions?
$endgroup$
– Adrian Keister
Jan 17 at 14:49




$begingroup$
Do you have initial conditions?
$endgroup$
– Adrian Keister
Jan 17 at 14:49












$begingroup$
Initial conditions are varying.
$endgroup$
– Sagar Sharma
Jan 17 at 14:51






$begingroup$
Initial conditions are varying.
$endgroup$
– Sagar Sharma
Jan 17 at 14:51






1




1




$begingroup$
Let $a_n=b_n+p+qn$ to eliminate $n$
$endgroup$
– lab bhattacharjee
Jan 17 at 15:16




$begingroup$
Let $a_n=b_n+p+qn$ to eliminate $n$
$endgroup$
– lab bhattacharjee
Jan 17 at 15:16










1 Answer
1






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1












$begingroup$

Let's rewrite this as
begin{align*}
a_{n+3} = a_{n+2} + 2(n+3) + 3a_n
end{align*}
The generating function for this sequence is
begin{align*}
frac{f(x) - a_0 - a_1x - a_2x^2}{x^3} = frac{f(x) - a_0 - a_1x}{x^2} + frac{2(3 - 2x)}{(1-x)^2} + 3f(x)
end{align*}
Solving for $f(x)$ gives us
begin{align*}
f(x) = color{blue}{frac{a_0(x-1)}{(3x^3+x-1)}} + color{red}{frac{a_1 x(x-1)}{(3x^3+x-1)}} - color{green}{frac{a_2x^2}{(3x^3+x-1)}} - color{orange}{frac{2(3-2x)x^3}{(x-1)^2(3x^3+x-1)}}
end{align*}
Let $r_1, r_2, r_3$ be the roots of $3x^3 + x - 1=0$. Then the series expansion of the blue term is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_0}{3}left(frac{1 - r_1}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{1 - r_2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{1 - r_3}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{p_n}x^n
end{align*}
Similarly, the red portion is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_1}{3}left(frac{r_1(1 - r_1)}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2(1 - r_2)}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3(1 - r_3)}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{q_n}x^n
end{align*}
The green portion is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_2}{3}left(frac{r_1^2}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2^2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3^2}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{r_n}x^n
end{align*}
Exercise Find the series expansion for the orange term, and call the coefficients as $s_n$.



Finally, your sequence can be expressed as
begin{align*}
a_n = p_n + q_n - r_n - s_n
end{align*}
There is some heavy simplification you can do. For example, you can express
begin{align*}
p_n + q_n - r_n=sum_{t=1}^{3}frac{1}{r_t^{n+1}}g_t(a_0, a_1, a_2, r_1, r_2, r_3)
end{align*}
For functions $g_1, g_2, g_3$, which you can pre-compute in a computer implementation.






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    1












    $begingroup$

    Let's rewrite this as
    begin{align*}
    a_{n+3} = a_{n+2} + 2(n+3) + 3a_n
    end{align*}
    The generating function for this sequence is
    begin{align*}
    frac{f(x) - a_0 - a_1x - a_2x^2}{x^3} = frac{f(x) - a_0 - a_1x}{x^2} + frac{2(3 - 2x)}{(1-x)^2} + 3f(x)
    end{align*}
    Solving for $f(x)$ gives us
    begin{align*}
    f(x) = color{blue}{frac{a_0(x-1)}{(3x^3+x-1)}} + color{red}{frac{a_1 x(x-1)}{(3x^3+x-1)}} - color{green}{frac{a_2x^2}{(3x^3+x-1)}} - color{orange}{frac{2(3-2x)x^3}{(x-1)^2(3x^3+x-1)}}
    end{align*}
    Let $r_1, r_2, r_3$ be the roots of $3x^3 + x - 1=0$. Then the series expansion of the blue term is
    begin{align*}
    sum_{n=0}^{infty}underbrace{left{frac{a_0}{3}left(frac{1 - r_1}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{1 - r_2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{1 - r_3}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{p_n}x^n
    end{align*}
    Similarly, the red portion is
    begin{align*}
    sum_{n=0}^{infty}underbrace{left{frac{a_1}{3}left(frac{r_1(1 - r_1)}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2(1 - r_2)}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3(1 - r_3)}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{q_n}x^n
    end{align*}
    The green portion is
    begin{align*}
    sum_{n=0}^{infty}underbrace{left{frac{a_2}{3}left(frac{r_1^2}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2^2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3^2}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{r_n}x^n
    end{align*}
    Exercise Find the series expansion for the orange term, and call the coefficients as $s_n$.



    Finally, your sequence can be expressed as
    begin{align*}
    a_n = p_n + q_n - r_n - s_n
    end{align*}
    There is some heavy simplification you can do. For example, you can express
    begin{align*}
    p_n + q_n - r_n=sum_{t=1}^{3}frac{1}{r_t^{n+1}}g_t(a_0, a_1, a_2, r_1, r_2, r_3)
    end{align*}
    For functions $g_1, g_2, g_3$, which you can pre-compute in a computer implementation.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Let's rewrite this as
      begin{align*}
      a_{n+3} = a_{n+2} + 2(n+3) + 3a_n
      end{align*}
      The generating function for this sequence is
      begin{align*}
      frac{f(x) - a_0 - a_1x - a_2x^2}{x^3} = frac{f(x) - a_0 - a_1x}{x^2} + frac{2(3 - 2x)}{(1-x)^2} + 3f(x)
      end{align*}
      Solving for $f(x)$ gives us
      begin{align*}
      f(x) = color{blue}{frac{a_0(x-1)}{(3x^3+x-1)}} + color{red}{frac{a_1 x(x-1)}{(3x^3+x-1)}} - color{green}{frac{a_2x^2}{(3x^3+x-1)}} - color{orange}{frac{2(3-2x)x^3}{(x-1)^2(3x^3+x-1)}}
      end{align*}
      Let $r_1, r_2, r_3$ be the roots of $3x^3 + x - 1=0$. Then the series expansion of the blue term is
      begin{align*}
      sum_{n=0}^{infty}underbrace{left{frac{a_0}{3}left(frac{1 - r_1}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{1 - r_2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{1 - r_3}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{p_n}x^n
      end{align*}
      Similarly, the red portion is
      begin{align*}
      sum_{n=0}^{infty}underbrace{left{frac{a_1}{3}left(frac{r_1(1 - r_1)}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2(1 - r_2)}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3(1 - r_3)}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{q_n}x^n
      end{align*}
      The green portion is
      begin{align*}
      sum_{n=0}^{infty}underbrace{left{frac{a_2}{3}left(frac{r_1^2}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2^2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3^2}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{r_n}x^n
      end{align*}
      Exercise Find the series expansion for the orange term, and call the coefficients as $s_n$.



      Finally, your sequence can be expressed as
      begin{align*}
      a_n = p_n + q_n - r_n - s_n
      end{align*}
      There is some heavy simplification you can do. For example, you can express
      begin{align*}
      p_n + q_n - r_n=sum_{t=1}^{3}frac{1}{r_t^{n+1}}g_t(a_0, a_1, a_2, r_1, r_2, r_3)
      end{align*}
      For functions $g_1, g_2, g_3$, which you can pre-compute in a computer implementation.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Let's rewrite this as
        begin{align*}
        a_{n+3} = a_{n+2} + 2(n+3) + 3a_n
        end{align*}
        The generating function for this sequence is
        begin{align*}
        frac{f(x) - a_0 - a_1x - a_2x^2}{x^3} = frac{f(x) - a_0 - a_1x}{x^2} + frac{2(3 - 2x)}{(1-x)^2} + 3f(x)
        end{align*}
        Solving for $f(x)$ gives us
        begin{align*}
        f(x) = color{blue}{frac{a_0(x-1)}{(3x^3+x-1)}} + color{red}{frac{a_1 x(x-1)}{(3x^3+x-1)}} - color{green}{frac{a_2x^2}{(3x^3+x-1)}} - color{orange}{frac{2(3-2x)x^3}{(x-1)^2(3x^3+x-1)}}
        end{align*}
        Let $r_1, r_2, r_3$ be the roots of $3x^3 + x - 1=0$. Then the series expansion of the blue term is
        begin{align*}
        sum_{n=0}^{infty}underbrace{left{frac{a_0}{3}left(frac{1 - r_1}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{1 - r_2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{1 - r_3}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{p_n}x^n
        end{align*}
        Similarly, the red portion is
        begin{align*}
        sum_{n=0}^{infty}underbrace{left{frac{a_1}{3}left(frac{r_1(1 - r_1)}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2(1 - r_2)}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3(1 - r_3)}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{q_n}x^n
        end{align*}
        The green portion is
        begin{align*}
        sum_{n=0}^{infty}underbrace{left{frac{a_2}{3}left(frac{r_1^2}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2^2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3^2}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{r_n}x^n
        end{align*}
        Exercise Find the series expansion for the orange term, and call the coefficients as $s_n$.



        Finally, your sequence can be expressed as
        begin{align*}
        a_n = p_n + q_n - r_n - s_n
        end{align*}
        There is some heavy simplification you can do. For example, you can express
        begin{align*}
        p_n + q_n - r_n=sum_{t=1}^{3}frac{1}{r_t^{n+1}}g_t(a_0, a_1, a_2, r_1, r_2, r_3)
        end{align*}
        For functions $g_1, g_2, g_3$, which you can pre-compute in a computer implementation.






        share|cite|improve this answer











        $endgroup$



        Let's rewrite this as
        begin{align*}
        a_{n+3} = a_{n+2} + 2(n+3) + 3a_n
        end{align*}
        The generating function for this sequence is
        begin{align*}
        frac{f(x) - a_0 - a_1x - a_2x^2}{x^3} = frac{f(x) - a_0 - a_1x}{x^2} + frac{2(3 - 2x)}{(1-x)^2} + 3f(x)
        end{align*}
        Solving for $f(x)$ gives us
        begin{align*}
        f(x) = color{blue}{frac{a_0(x-1)}{(3x^3+x-1)}} + color{red}{frac{a_1 x(x-1)}{(3x^3+x-1)}} - color{green}{frac{a_2x^2}{(3x^3+x-1)}} - color{orange}{frac{2(3-2x)x^3}{(x-1)^2(3x^3+x-1)}}
        end{align*}
        Let $r_1, r_2, r_3$ be the roots of $3x^3 + x - 1=0$. Then the series expansion of the blue term is
        begin{align*}
        sum_{n=0}^{infty}underbrace{left{frac{a_0}{3}left(frac{1 - r_1}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{1 - r_2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{1 - r_3}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{p_n}x^n
        end{align*}
        Similarly, the red portion is
        begin{align*}
        sum_{n=0}^{infty}underbrace{left{frac{a_1}{3}left(frac{r_1(1 - r_1)}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2(1 - r_2)}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3(1 - r_3)}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{q_n}x^n
        end{align*}
        The green portion is
        begin{align*}
        sum_{n=0}^{infty}underbrace{left{frac{a_2}{3}left(frac{r_1^2}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2^2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3^2}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{r_n}x^n
        end{align*}
        Exercise Find the series expansion for the orange term, and call the coefficients as $s_n$.



        Finally, your sequence can be expressed as
        begin{align*}
        a_n = p_n + q_n - r_n - s_n
        end{align*}
        There is some heavy simplification you can do. For example, you can express
        begin{align*}
        p_n + q_n - r_n=sum_{t=1}^{3}frac{1}{r_t^{n+1}}g_t(a_0, a_1, a_2, r_1, r_2, r_3)
        end{align*}
        For functions $g_1, g_2, g_3$, which you can pre-compute in a computer implementation.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 17 at 16:38

























        answered Jan 17 at 16:30









        Tom ChenTom Chen

        1,568715




        1,568715






























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