Computing the solution of a recurrence relation











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Let $y_k=dfrac{16}{3}y_{k-1}-dfrac{5}{3}y_{k-2}$



with $y_0=1$ and $y_1=dfrac{1}{2}$ for $k geq 2, k in mathbb{N}$.



To find the solution $lbrace y_k rbrace_{k in mathbb{N_0}}$ I used:



Firstly, I computed it for $k=2$:



$y_2=dfrac{16}{3}dfrac{1}{2}-dfrac{5}{3}1=1$



What has to be done next? Induction isn't helpful here.



I don't see how can it be found for $lbrace y_k rbrace_{k in mathbb{N_0}}$.






recurrence-relations







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    There's a whole method for solving linear recurrences. Try opening your notebook
    – Jakobian
    15 hours ago















up vote
0
down vote

favorite












Let $y_k=dfrac{16}{3}y_{k-1}-dfrac{5}{3}y_{k-2}$



with $y_0=1$ and $y_1=dfrac{1}{2}$ for $k geq 2, k in mathbb{N}$.



To find the solution $lbrace y_k rbrace_{k in mathbb{N_0}}$ I used:



Firstly, I computed it for $k=2$:



$y_2=dfrac{16}{3}dfrac{1}{2}-dfrac{5}{3}1=1$



What has to be done next? Induction isn't helpful here.



I don't see how can it be found for $lbrace y_k rbrace_{k in mathbb{N_0}}$.






recurrence-relations







share|cite|improve this question




















  • 3




    There's a whole method for solving linear recurrences. Try opening your notebook
    – Jakobian
    15 hours ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $y_k=dfrac{16}{3}y_{k-1}-dfrac{5}{3}y_{k-2}$



with $y_0=1$ and $y_1=dfrac{1}{2}$ for $k geq 2, k in mathbb{N}$.



To find the solution $lbrace y_k rbrace_{k in mathbb{N_0}}$ I used:



Firstly, I computed it for $k=2$:



$y_2=dfrac{16}{3}dfrac{1}{2}-dfrac{5}{3}1=1$



What has to be done next? Induction isn't helpful here.



I don't see how can it be found for $lbrace y_k rbrace_{k in mathbb{N_0}}$.






recurrence-relations







share|cite|improve this question















Let $y_k=dfrac{16}{3}y_{k-1}-dfrac{5}{3}y_{k-2}$



with $y_0=1$ and $y_1=dfrac{1}{2}$ for $k geq 2, k in mathbb{N}$.



To find the solution $lbrace y_k rbrace_{k in mathbb{N_0}}$ I used:



Firstly, I computed it for $k=2$:



$y_2=dfrac{16}{3}dfrac{1}{2}-dfrac{5}{3}1=1$



What has to be done next? Induction isn't helpful here.



I don't see how can it be found for $lbrace y_k rbrace_{k in mathbb{N_0}}$.






recurrence-relations




recurrence-relations






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edited 14 hours ago









Yadati Kiran

348110




348110










asked 15 hours ago









Olsgur

323




323








  • 3




    There's a whole method for solving linear recurrences. Try opening your notebook
    – Jakobian
    15 hours ago














  • 3




    There's a whole method for solving linear recurrences. Try opening your notebook
    – Jakobian
    15 hours ago








3




3




There's a whole method for solving linear recurrences. Try opening your notebook
– Jakobian
15 hours ago




There's a whole method for solving linear recurrences. Try opening your notebook
– Jakobian
15 hours ago










1 Answer
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$$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



$$[1,1/2,1,9/2,67/3,2009/18,15067/27,452009/162,3390067/243,101702009/1458,762765067/2187]$$



Solution details:



Char. equation is
$${{t}^{2}}-frac{16 t}{3}+frac{5}{3}=0$$
$$t_1=5,quad t_2=frac13$$
Then solution is
$$y_k=A,5^k+B,left(frac13right)^k$$
From initial conditions
$$A,5^0+B,left(frac13right)^0=1,\
A,5^1+B,left(frac13right)^1=frac12$$
$Rightarrow$
$$A=frac{1}{28},quad B=frac{27}{28}$$
$Rightarrow$
$$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



see for example https://www.youtube.com/watch?v=aHw7hAAjbD0






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    $$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



    $$[1,1/2,1,9/2,67/3,2009/18,15067/27,452009/162,3390067/243,101702009/1458,762765067/2187]$$



    Solution details:



    Char. equation is
    $${{t}^{2}}-frac{16 t}{3}+frac{5}{3}=0$$
    $$t_1=5,quad t_2=frac13$$
    Then solution is
    $$y_k=A,5^k+B,left(frac13right)^k$$
    From initial conditions
    $$A,5^0+B,left(frac13right)^0=1,\
    A,5^1+B,left(frac13right)^1=frac12$$
    $Rightarrow$
    $$A=frac{1}{28},quad B=frac{27}{28}$$
    $Rightarrow$
    $$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



    see for example https://www.youtube.com/watch?v=aHw7hAAjbD0






    share|cite|improve this answer



























      up vote
      -1
      down vote













      $$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



      $$[1,1/2,1,9/2,67/3,2009/18,15067/27,452009/162,3390067/243,101702009/1458,762765067/2187]$$



      Solution details:



      Char. equation is
      $${{t}^{2}}-frac{16 t}{3}+frac{5}{3}=0$$
      $$t_1=5,quad t_2=frac13$$
      Then solution is
      $$y_k=A,5^k+B,left(frac13right)^k$$
      From initial conditions
      $$A,5^0+B,left(frac13right)^0=1,\
      A,5^1+B,left(frac13right)^1=frac12$$
      $Rightarrow$
      $$A=frac{1}{28},quad B=frac{27}{28}$$
      $Rightarrow$
      $$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



      see for example https://www.youtube.com/watch?v=aHw7hAAjbD0






      share|cite|improve this answer

























        up vote
        -1
        down vote










        up vote
        -1
        down vote









        $$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



        $$[1,1/2,1,9/2,67/3,2009/18,15067/27,452009/162,3390067/243,101702009/1458,762765067/2187]$$



        Solution details:



        Char. equation is
        $${{t}^{2}}-frac{16 t}{3}+frac{5}{3}=0$$
        $$t_1=5,quad t_2=frac13$$
        Then solution is
        $$y_k=A,5^k+B,left(frac13right)^k$$
        From initial conditions
        $$A,5^0+B,left(frac13right)^0=1,\
        A,5^1+B,left(frac13right)^1=frac12$$
        $Rightarrow$
        $$A=frac{1}{28},quad B=frac{27}{28}$$
        $Rightarrow$
        $$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



        see for example https://www.youtube.com/watch?v=aHw7hAAjbD0






        share|cite|improve this answer














        $$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



        $$[1,1/2,1,9/2,67/3,2009/18,15067/27,452009/162,3390067/243,101702009/1458,762765067/2187]$$



        Solution details:



        Char. equation is
        $${{t}^{2}}-frac{16 t}{3}+frac{5}{3}=0$$
        $$t_1=5,quad t_2=frac13$$
        Then solution is
        $$y_k=A,5^k+B,left(frac13right)^k$$
        From initial conditions
        $$A,5^0+B,left(frac13right)^0=1,\
        A,5^1+B,left(frac13right)^1=frac12$$
        $Rightarrow$
        $$A=frac{1}{28},quad B=frac{27}{28}$$
        $Rightarrow$
        $$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



        see for example https://www.youtube.com/watch?v=aHw7hAAjbD0







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 4 hours ago

























        answered 15 hours ago









        Aleksas Domarkas

        7085




        7085






























             

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