Linear recurrence solving for tighest possible big O bounds












0












$begingroup$


I am dealing with the following linear recurrence:



X0 = 1



X1 = 2



Xn = 3Xn-1 + 2Xn-2



I have proven that this has an upper bound of O(4n)



However, I have been asked to come up with tighter bounds for this linear recurrence, but I dont know how to begin this. Does this involve solving the recurrence?






recurrence-relations







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$endgroup$












  • $begingroup$
    Do you know how to solve recurrence of this type? It is fairly straightforward
    $endgroup$
    – Shubham Johri
    Dec 15 '18 at 22:13










  • $begingroup$
    I think the best constant is $frac{3+sqrt{17}}{2} approx 3.55$ instead of $4$.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 22:28










  • $begingroup$
    Do not change your question after it has been answered.
    $endgroup$
    – Joel Reyes Noche
    Dec 18 '18 at 13:13
















0












$begingroup$


I am dealing with the following linear recurrence:



X0 = 1



X1 = 2



Xn = 3Xn-1 + 2Xn-2



I have proven that this has an upper bound of O(4n)



However, I have been asked to come up with tighter bounds for this linear recurrence, but I dont know how to begin this. Does this involve solving the recurrence?






recurrence-relations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you know how to solve recurrence of this type? It is fairly straightforward
    $endgroup$
    – Shubham Johri
    Dec 15 '18 at 22:13










  • $begingroup$
    I think the best constant is $frac{3+sqrt{17}}{2} approx 3.55$ instead of $4$.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 22:28










  • $begingroup$
    Do not change your question after it has been answered.
    $endgroup$
    – Joel Reyes Noche
    Dec 18 '18 at 13:13














0












0








0





$begingroup$


I am dealing with the following linear recurrence:



X0 = 1



X1 = 2



Xn = 3Xn-1 + 2Xn-2



I have proven that this has an upper bound of O(4n)



However, I have been asked to come up with tighter bounds for this linear recurrence, but I dont know how to begin this. Does this involve solving the recurrence?






recurrence-relations







share|cite|improve this question











$endgroup$




I am dealing with the following linear recurrence:



X0 = 1



X1 = 2



Xn = 3Xn-1 + 2Xn-2



I have proven that this has an upper bound of O(4n)



However, I have been asked to come up with tighter bounds for this linear recurrence, but I dont know how to begin this. Does this involve solving the recurrence?






recurrence-relations




recurrence-relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 13:13









Joel Reyes Noche

5,65733149




5,65733149










asked Dec 15 '18 at 22:09







user626552



















  • $begingroup$
    Do you know how to solve recurrence of this type? It is fairly straightforward
    $endgroup$
    – Shubham Johri
    Dec 15 '18 at 22:13










  • $begingroup$
    I think the best constant is $frac{3+sqrt{17}}{2} approx 3.55$ instead of $4$.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 22:28










  • $begingroup$
    Do not change your question after it has been answered.
    $endgroup$
    – Joel Reyes Noche
    Dec 18 '18 at 13:13


















  • $begingroup$
    Do you know how to solve recurrence of this type? It is fairly straightforward
    $endgroup$
    – Shubham Johri
    Dec 15 '18 at 22:13










  • $begingroup$
    I think the best constant is $frac{3+sqrt{17}}{2} approx 3.55$ instead of $4$.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 22:28










  • $begingroup$
    Do not change your question after it has been answered.
    $endgroup$
    – Joel Reyes Noche
    Dec 18 '18 at 13:13
















$begingroup$
Do you know how to solve recurrence of this type? It is fairly straightforward
$endgroup$
– Shubham Johri
Dec 15 '18 at 22:13




$begingroup$
Do you know how to solve recurrence of this type? It is fairly straightforward
$endgroup$
– Shubham Johri
Dec 15 '18 at 22:13












$begingroup$
I think the best constant is $frac{3+sqrt{17}}{2} approx 3.55$ instead of $4$.
$endgroup$
– Mindlack
Dec 15 '18 at 22:28




$begingroup$
I think the best constant is $frac{3+sqrt{17}}{2} approx 3.55$ instead of $4$.
$endgroup$
– Mindlack
Dec 15 '18 at 22:28












$begingroup$
Do not change your question after it has been answered.
$endgroup$
– Joel Reyes Noche
Dec 18 '18 at 13:13




$begingroup$
Do not change your question after it has been answered.
$endgroup$
– Joel Reyes Noche
Dec 18 '18 at 13:13










1 Answer
1






active

oldest

votes


















1












$begingroup$

Write the solution as



$$
x_n = a lambda^n
$$



If you replace that in your original expression you get



$$
a lambda^n = 3 a lambda^{n - 1} + 2 alambda^{n-2}
$$



Which reduces to



$$
lambda^2 = 3 lambda + 2
$$



Solutions are



$$
lambda = frac{3}{2} pm frac{sqrt{13}}{2}
$$



So the solution is



$$
x_n = a left( frac{3}{2} + frac{sqrt{13}}{2} right)^n + bleft( frac{3}{2} - frac{sqrt{13}}{2} right)^n
$$



The constants $a$ and $b$ you can determine by setting $n=0$ and $n=1$ and using the conditions $x_0 = 1$ and $x_1 = 2$






share|cite|improve this answer









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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Write the solution as



    $$
    x_n = a lambda^n
    $$



    If you replace that in your original expression you get



    $$
    a lambda^n = 3 a lambda^{n - 1} + 2 alambda^{n-2}
    $$



    Which reduces to



    $$
    lambda^2 = 3 lambda + 2
    $$



    Solutions are



    $$
    lambda = frac{3}{2} pm frac{sqrt{13}}{2}
    $$



    So the solution is



    $$
    x_n = a left( frac{3}{2} + frac{sqrt{13}}{2} right)^n + bleft( frac{3}{2} - frac{sqrt{13}}{2} right)^n
    $$



    The constants $a$ and $b$ you can determine by setting $n=0$ and $n=1$ and using the conditions $x_0 = 1$ and $x_1 = 2$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Write the solution as



      $$
      x_n = a lambda^n
      $$



      If you replace that in your original expression you get



      $$
      a lambda^n = 3 a lambda^{n - 1} + 2 alambda^{n-2}
      $$



      Which reduces to



      $$
      lambda^2 = 3 lambda + 2
      $$



      Solutions are



      $$
      lambda = frac{3}{2} pm frac{sqrt{13}}{2}
      $$



      So the solution is



      $$
      x_n = a left( frac{3}{2} + frac{sqrt{13}}{2} right)^n + bleft( frac{3}{2} - frac{sqrt{13}}{2} right)^n
      $$



      The constants $a$ and $b$ you can determine by setting $n=0$ and $n=1$ and using the conditions $x_0 = 1$ and $x_1 = 2$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Write the solution as



        $$
        x_n = a lambda^n
        $$



        If you replace that in your original expression you get



        $$
        a lambda^n = 3 a lambda^{n - 1} + 2 alambda^{n-2}
        $$



        Which reduces to



        $$
        lambda^2 = 3 lambda + 2
        $$



        Solutions are



        $$
        lambda = frac{3}{2} pm frac{sqrt{13}}{2}
        $$



        So the solution is



        $$
        x_n = a left( frac{3}{2} + frac{sqrt{13}}{2} right)^n + bleft( frac{3}{2} - frac{sqrt{13}}{2} right)^n
        $$



        The constants $a$ and $b$ you can determine by setting $n=0$ and $n=1$ and using the conditions $x_0 = 1$ and $x_1 = 2$






        share|cite|improve this answer









        $endgroup$



        Write the solution as



        $$
        x_n = a lambda^n
        $$



        If you replace that in your original expression you get



        $$
        a lambda^n = 3 a lambda^{n - 1} + 2 alambda^{n-2}
        $$



        Which reduces to



        $$
        lambda^2 = 3 lambda + 2
        $$



        Solutions are



        $$
        lambda = frac{3}{2} pm frac{sqrt{13}}{2}
        $$



        So the solution is



        $$
        x_n = a left( frac{3}{2} + frac{sqrt{13}}{2} right)^n + bleft( frac{3}{2} - frac{sqrt{13}}{2} right)^n
        $$



        The constants $a$ and $b$ you can determine by setting $n=0$ and $n=1$ and using the conditions $x_0 = 1$ and $x_1 = 2$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 15 '18 at 22:28









        caveraccaverac

        14.8k31130




        14.8k31130






























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