Solve the recurrence relation $f(n) = 4f(n/3)+5$ where $n=3^k, k=1,2,3…$ and $f(1)=5$












1














Solve the recurrence relation:



$f(n) = 4f(n/3)+5$ where $n=3^k, k=1,2,3...$ and $f(1)=5$



I never seen a recurrence relation like this before. What would I need to use or do to solve this?






combinatorics discrete-mathematics recurrence-relations







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  • 1




    You could, as usual, try induction.
    – Lucas Henrique
    Dec 10 '18 at 5:28










  • If you are only looking for an asymptotic bound, you can use master theorem.
    – platty
    Dec 10 '18 at 5:32










  • If $n=3^k$ then $n/3=3^{k-1}$ and it becomes a usual recurrence, provided know $f(1)$ and only want it for $f(3^k),$ $k ge 0.$
    – coffeemath
    Dec 10 '18 at 5:33


















1














Solve the recurrence relation:



$f(n) = 4f(n/3)+5$ where $n=3^k, k=1,2,3...$ and $f(1)=5$



I never seen a recurrence relation like this before. What would I need to use or do to solve this?






combinatorics discrete-mathematics recurrence-relations







share|cite|improve this question


















  • 1




    You could, as usual, try induction.
    – Lucas Henrique
    Dec 10 '18 at 5:28










  • If you are only looking for an asymptotic bound, you can use master theorem.
    – platty
    Dec 10 '18 at 5:32










  • If $n=3^k$ then $n/3=3^{k-1}$ and it becomes a usual recurrence, provided know $f(1)$ and only want it for $f(3^k),$ $k ge 0.$
    – coffeemath
    Dec 10 '18 at 5:33
















1












1








1


1





Solve the recurrence relation:



$f(n) = 4f(n/3)+5$ where $n=3^k, k=1,2,3...$ and $f(1)=5$



I never seen a recurrence relation like this before. What would I need to use or do to solve this?






combinatorics discrete-mathematics recurrence-relations







share|cite|improve this question













Solve the recurrence relation:



$f(n) = 4f(n/3)+5$ where $n=3^k, k=1,2,3...$ and $f(1)=5$



I never seen a recurrence relation like this before. What would I need to use or do to solve this?






combinatorics discrete-mathematics recurrence-relations




combinatorics discrete-mathematics recurrence-relations






share|cite|improve this question













share|cite|improve this question











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asked Dec 10 '18 at 5:26









cosmicbrownie

1016




1016








  • 1




    You could, as usual, try induction.
    – Lucas Henrique
    Dec 10 '18 at 5:28










  • If you are only looking for an asymptotic bound, you can use master theorem.
    – platty
    Dec 10 '18 at 5:32










  • If $n=3^k$ then $n/3=3^{k-1}$ and it becomes a usual recurrence, provided know $f(1)$ and only want it for $f(3^k),$ $k ge 0.$
    – coffeemath
    Dec 10 '18 at 5:33
















  • 1




    You could, as usual, try induction.
    – Lucas Henrique
    Dec 10 '18 at 5:28










  • If you are only looking for an asymptotic bound, you can use master theorem.
    – platty
    Dec 10 '18 at 5:32










  • If $n=3^k$ then $n/3=3^{k-1}$ and it becomes a usual recurrence, provided know $f(1)$ and only want it for $f(3^k),$ $k ge 0.$
    – coffeemath
    Dec 10 '18 at 5:33










1




1




You could, as usual, try induction.
– Lucas Henrique
Dec 10 '18 at 5:28




You could, as usual, try induction.
– Lucas Henrique
Dec 10 '18 at 5:28












If you are only looking for an asymptotic bound, you can use master theorem.
– platty
Dec 10 '18 at 5:32




If you are only looking for an asymptotic bound, you can use master theorem.
– platty
Dec 10 '18 at 5:32












If $n=3^k$ then $n/3=3^{k-1}$ and it becomes a usual recurrence, provided know $f(1)$ and only want it for $f(3^k),$ $k ge 0.$
– coffeemath
Dec 10 '18 at 5:33






If $n=3^k$ then $n/3=3^{k-1}$ and it becomes a usual recurrence, provided know $f(1)$ and only want it for $f(3^k),$ $k ge 0.$
– coffeemath
Dec 10 '18 at 5:33












3 Answers
3






active

oldest

votes


















1














Let $f(3^k)=g(k)+c,k=0implies?$



$5=g(k)-4g(k-1)+c-4c$



Set $c-4c=5$ so that $g(k)=4g(k-1)=4^rg(k-r)=4^kg(0)$






share|cite|improve this answer





























    0














    Also, you can find a tight result from the mater theorem and we can say $f(n) = Theta(n^{log_34})$, and as $n = 3^k$, $f(n) sim (3^k)^{log_34} = 4^k$.






    share|cite|improve this answer





























      0














      One approach which I find useful is the following:



      As already mentioned by Lucas Henrique, you can first of all define $a_k := f(3^k)$. This gives you the relation
      $$
      a_k = 4a_{k-1} + 5, quad quad a_0 = 5
      $$
      From here on, you have several options. You can use generating functions, just plug in some values and try to find a pattern or assume you have a "big" $k$ and use the above formula several times:
      begin{eqnarray}
      a_k &=& 4a_{k-1} +5\
      & = & 4 (4a_{k-2}+5)+5 = 4^2 a_{k-2} + (4+1)cdot 5\
      & = & 4^2 (4a_{k-3} +5)+(4+1) cdot5 = 4^3 a_{k-3} + (4^2 + 4 + 1)cdot 5\
      & = & dots
      end{eqnarray}
      Now you may already derive a formula from that.



      Hint: for $q neq -1$, one has $sum_{k=0}^n q^k = frac{1-q^{n+1}}{1-q}$






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1














        Let $f(3^k)=g(k)+c,k=0implies?$



        $5=g(k)-4g(k-1)+c-4c$



        Set $c-4c=5$ so that $g(k)=4g(k-1)=4^rg(k-r)=4^kg(0)$






        share|cite|improve this answer


























          1














          Let $f(3^k)=g(k)+c,k=0implies?$



          $5=g(k)-4g(k-1)+c-4c$



          Set $c-4c=5$ so that $g(k)=4g(k-1)=4^rg(k-r)=4^kg(0)$






          share|cite|improve this answer
























            1












            1








            1






            Let $f(3^k)=g(k)+c,k=0implies?$



            $5=g(k)-4g(k-1)+c-4c$



            Set $c-4c=5$ so that $g(k)=4g(k-1)=4^rg(k-r)=4^kg(0)$






            share|cite|improve this answer












            Let $f(3^k)=g(k)+c,k=0implies?$



            $5=g(k)-4g(k-1)+c-4c$



            Set $c-4c=5$ so that $g(k)=4g(k-1)=4^rg(k-r)=4^kg(0)$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 10 '18 at 5:53









            lab bhattacharjee

            223k15156274




            223k15156274























                0














                Also, you can find a tight result from the mater theorem and we can say $f(n) = Theta(n^{log_34})$, and as $n = 3^k$, $f(n) sim (3^k)^{log_34} = 4^k$.






                share|cite|improve this answer


























                  0














                  Also, you can find a tight result from the mater theorem and we can say $f(n) = Theta(n^{log_34})$, and as $n = 3^k$, $f(n) sim (3^k)^{log_34} = 4^k$.






                  share|cite|improve this answer
























                    0












                    0








                    0






                    Also, you can find a tight result from the mater theorem and we can say $f(n) = Theta(n^{log_34})$, and as $n = 3^k$, $f(n) sim (3^k)^{log_34} = 4^k$.






                    share|cite|improve this answer












                    Also, you can find a tight result from the mater theorem and we can say $f(n) = Theta(n^{log_34})$, and as $n = 3^k$, $f(n) sim (3^k)^{log_34} = 4^k$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 10 '18 at 10:09









                    OmG

                    2,190620




                    2,190620























                        0














                        One approach which I find useful is the following:



                        As already mentioned by Lucas Henrique, you can first of all define $a_k := f(3^k)$. This gives you the relation
                        $$
                        a_k = 4a_{k-1} + 5, quad quad a_0 = 5
                        $$
                        From here on, you have several options. You can use generating functions, just plug in some values and try to find a pattern or assume you have a "big" $k$ and use the above formula several times:
                        begin{eqnarray}
                        a_k &=& 4a_{k-1} +5\
                        & = & 4 (4a_{k-2}+5)+5 = 4^2 a_{k-2} + (4+1)cdot 5\
                        & = & 4^2 (4a_{k-3} +5)+(4+1) cdot5 = 4^3 a_{k-3} + (4^2 + 4 + 1)cdot 5\
                        & = & dots
                        end{eqnarray}
                        Now you may already derive a formula from that.



                        Hint: for $q neq -1$, one has $sum_{k=0}^n q^k = frac{1-q^{n+1}}{1-q}$






                        share|cite|improve this answer


























                          0














                          One approach which I find useful is the following:



                          As already mentioned by Lucas Henrique, you can first of all define $a_k := f(3^k)$. This gives you the relation
                          $$
                          a_k = 4a_{k-1} + 5, quad quad a_0 = 5
                          $$
                          From here on, you have several options. You can use generating functions, just plug in some values and try to find a pattern or assume you have a "big" $k$ and use the above formula several times:
                          begin{eqnarray}
                          a_k &=& 4a_{k-1} +5\
                          & = & 4 (4a_{k-2}+5)+5 = 4^2 a_{k-2} + (4+1)cdot 5\
                          & = & 4^2 (4a_{k-3} +5)+(4+1) cdot5 = 4^3 a_{k-3} + (4^2 + 4 + 1)cdot 5\
                          & = & dots
                          end{eqnarray}
                          Now you may already derive a formula from that.



                          Hint: for $q neq -1$, one has $sum_{k=0}^n q^k = frac{1-q^{n+1}}{1-q}$






                          share|cite|improve this answer
























                            0












                            0








                            0






                            One approach which I find useful is the following:



                            As already mentioned by Lucas Henrique, you can first of all define $a_k := f(3^k)$. This gives you the relation
                            $$
                            a_k = 4a_{k-1} + 5, quad quad a_0 = 5
                            $$
                            From here on, you have several options. You can use generating functions, just plug in some values and try to find a pattern or assume you have a "big" $k$ and use the above formula several times:
                            begin{eqnarray}
                            a_k &=& 4a_{k-1} +5\
                            & = & 4 (4a_{k-2}+5)+5 = 4^2 a_{k-2} + (4+1)cdot 5\
                            & = & 4^2 (4a_{k-3} +5)+(4+1) cdot5 = 4^3 a_{k-3} + (4^2 + 4 + 1)cdot 5\
                            & = & dots
                            end{eqnarray}
                            Now you may already derive a formula from that.



                            Hint: for $q neq -1$, one has $sum_{k=0}^n q^k = frac{1-q^{n+1}}{1-q}$






                            share|cite|improve this answer












                            One approach which I find useful is the following:



                            As already mentioned by Lucas Henrique, you can first of all define $a_k := f(3^k)$. This gives you the relation
                            $$
                            a_k = 4a_{k-1} + 5, quad quad a_0 = 5
                            $$
                            From here on, you have several options. You can use generating functions, just plug in some values and try to find a pattern or assume you have a "big" $k$ and use the above formula several times:
                            begin{eqnarray}
                            a_k &=& 4a_{k-1} +5\
                            & = & 4 (4a_{k-2}+5)+5 = 4^2 a_{k-2} + (4+1)cdot 5\
                            & = & 4^2 (4a_{k-3} +5)+(4+1) cdot5 = 4^3 a_{k-3} + (4^2 + 4 + 1)cdot 5\
                            & = & dots
                            end{eqnarray}
                            Now you may already derive a formula from that.



                            Hint: for $q neq -1$, one has $sum_{k=0}^n q^k = frac{1-q^{n+1}}{1-q}$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 10 '18 at 10:34









                            bruderjakob17

                            1587




                            1587






























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