$u_{n+1} = a u_n +b u_{n-1} +c$












2












$begingroup$


Is there a closed form formula for the general term of a recurrence sequence satisfying $$u_{n+1} = a u_n +b u_{n-1} +c$$ The sequence I am interested in satisfies $u_{n+1} = 8 u_n - 3 u_{n-1} -4$.






real-analysis sequences-and-series recurrence-relations







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Is there a closed form formula for the general term of a recurrence sequence satisfying $$u_{n+1} = a u_n +b u_{n-1} +c$$ The sequence I am interested in satisfies $u_{n+1} = 8 u_n - 3 u_{n-1} -4$.






    real-analysis sequences-and-series recurrence-relations







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Is there a closed form formula for the general term of a recurrence sequence satisfying $$u_{n+1} = a u_n +b u_{n-1} +c$$ The sequence I am interested in satisfies $u_{n+1} = 8 u_n - 3 u_{n-1} -4$.






      real-analysis sequences-and-series recurrence-relations







      share|cite|improve this question











      $endgroup$




      Is there a closed form formula for the general term of a recurrence sequence satisfying $$u_{n+1} = a u_n +b u_{n-1} +c$$ The sequence I am interested in satisfies $u_{n+1} = 8 u_n - 3 u_{n-1} -4$.






      real-analysis sequences-and-series recurrence-relations




      real-analysis sequences-and-series recurrence-relations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 20 '18 at 15:04









      Ankit Kumar

      1,494221




      1,494221










      asked Dec 20 '18 at 13:59









      furyofuryo

      313




      313






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          You can solve any non-homogeneous, linear recurrence of the above form by converting it to homogeneous form.



          (I'll use $n$ instead of $n+1$, doesn't matter anyway!)



          $$u_{n}=au_{n-1}+bu_{n-2}+c$$
          $$u_{n-1}=au_{n-2}+bu_{n-3}+c$$
          Subtract them to get:-



          $$u_{n}=(a+1)u_{n-1}+(b-a)u_{n-2}-bu_{n-3}$$
          You can then use characteristic polynomial to find the answer.




          Speaking strictly for the equation you wrote, the characteristic polynomial is
          $$z^3=9z^2-11z+3$$ which has the roots
          $$z=1, 4+sqrt{13}, 4-sqrt{13}$$
          giving $$u_n=a_1+a_2{(4+sqrt{13})}^n+a_3{(4-sqrt{13})}^n$$, where $a_i$s depend on the initial condition.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So $u_n$ wouldn't converge? I find it a little bit strange.
            $endgroup$
            – furyo
            Dec 20 '18 at 16:39










          • $begingroup$
            @furyo It isn't strange at all. Stuff like this (convergence) depends on the initial conditions. Depending upon them, the series may or may not convegre!
            $endgroup$
            – Ankit Kumar
            Dec 20 '18 at 16:43










          • $begingroup$
            For instance, if you take $u_1=u_2=1$, $u_i=1$ for all $i$, which is a convergent sequence. On the other hand, if you take $u_1=1, u_2=2$, then it'll be a divergent sequence.
            $endgroup$
            – Ankit Kumar
            Dec 20 '18 at 16:44










          • $begingroup$
            I would've expected a non constant sequence increasing to $1$. Here it's either constant or divergent.
            $endgroup$
            – furyo
            Dec 20 '18 at 16:47










          • $begingroup$
            You can't get a non-constant, convergent sequence because it isn't constant $implies $ at least one of $a_2, a_3$ is non-zero $implies $ $u_n$ will be a strictly increasing/decreasing (depending upon sign of $a_i$s) function of $n$.
            $endgroup$
            – Ankit Kumar
            Dec 20 '18 at 16:54





















          0












          $begingroup$

          This is a linear difference equation so can be stated as



          $$
          u = u_h + u_p
          $$



          with



          $$
          u_h(n+1)-au_h(n)-bu_h(n-1) = 0\
          u_p(n+1)-au_p(n)-bu_p(n-1) = c\
          $$



          now making $u_h(n) = phi^n$ and substituting we have



          $$
          phi^nleft(phi-a-bphi^{-1}right) = 0
          $$



          so



          $$
          phi = frac{1}{2} left(apmsqrt{a^2+4 b}right)
          $$



          and



          $$
          u_h(n) = frac{C_1}{2^n} left(a-sqrt{a^2+4 b}right)^n+frac{C_2}{2^n} left(a+sqrt{a^2+4 b}right)^n
          $$



          and also



          $$
          u_p(n) = frac{c}{1-a-b}
          $$



          so finally



          $$
          u(n) = frac{C_1}{2^n} left(a-sqrt{a^2+4 b}right)^n+frac{C_2}{2^n} left(a+sqrt{a^2+4 b}right)^n + frac{c}{1-a-b}
          $$






          share|cite|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            You can solve any non-homogeneous, linear recurrence of the above form by converting it to homogeneous form.



            (I'll use $n$ instead of $n+1$, doesn't matter anyway!)



            $$u_{n}=au_{n-1}+bu_{n-2}+c$$
            $$u_{n-1}=au_{n-2}+bu_{n-3}+c$$
            Subtract them to get:-



            $$u_{n}=(a+1)u_{n-1}+(b-a)u_{n-2}-bu_{n-3}$$
            You can then use characteristic polynomial to find the answer.




            Speaking strictly for the equation you wrote, the characteristic polynomial is
            $$z^3=9z^2-11z+3$$ which has the roots
            $$z=1, 4+sqrt{13}, 4-sqrt{13}$$
            giving $$u_n=a_1+a_2{(4+sqrt{13})}^n+a_3{(4-sqrt{13})}^n$$, where $a_i$s depend on the initial condition.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              So $u_n$ wouldn't converge? I find it a little bit strange.
              $endgroup$
              – furyo
              Dec 20 '18 at 16:39










            • $begingroup$
              @furyo It isn't strange at all. Stuff like this (convergence) depends on the initial conditions. Depending upon them, the series may or may not convegre!
              $endgroup$
              – Ankit Kumar
              Dec 20 '18 at 16:43










            • $begingroup$
              For instance, if you take $u_1=u_2=1$, $u_i=1$ for all $i$, which is a convergent sequence. On the other hand, if you take $u_1=1, u_2=2$, then it'll be a divergent sequence.
              $endgroup$
              – Ankit Kumar
              Dec 20 '18 at 16:44










            • $begingroup$
              I would've expected a non constant sequence increasing to $1$. Here it's either constant or divergent.
              $endgroup$
              – furyo
              Dec 20 '18 at 16:47










            • $begingroup$
              You can't get a non-constant, convergent sequence because it isn't constant $implies $ at least one of $a_2, a_3$ is non-zero $implies $ $u_n$ will be a strictly increasing/decreasing (depending upon sign of $a_i$s) function of $n$.
              $endgroup$
              – Ankit Kumar
              Dec 20 '18 at 16:54


















            2












            $begingroup$

            You can solve any non-homogeneous, linear recurrence of the above form by converting it to homogeneous form.



            (I'll use $n$ instead of $n+1$, doesn't matter anyway!)



            $$u_{n}=au_{n-1}+bu_{n-2}+c$$
            $$u_{n-1}=au_{n-2}+bu_{n-3}+c$$
            Subtract them to get:-



            $$u_{n}=(a+1)u_{n-1}+(b-a)u_{n-2}-bu_{n-3}$$
            You can then use characteristic polynomial to find the answer.




            Speaking strictly for the equation you wrote, the characteristic polynomial is
            $$z^3=9z^2-11z+3$$ which has the roots
            $$z=1, 4+sqrt{13}, 4-sqrt{13}$$
            giving $$u_n=a_1+a_2{(4+sqrt{13})}^n+a_3{(4-sqrt{13})}^n$$, where $a_i$s depend on the initial condition.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              So $u_n$ wouldn't converge? I find it a little bit strange.
              $endgroup$
              – furyo
              Dec 20 '18 at 16:39










            • $begingroup$
              @furyo It isn't strange at all. Stuff like this (convergence) depends on the initial conditions. Depending upon them, the series may or may not convegre!
              $endgroup$
              – Ankit Kumar
              Dec 20 '18 at 16:43










            • $begingroup$
              For instance, if you take $u_1=u_2=1$, $u_i=1$ for all $i$, which is a convergent sequence. On the other hand, if you take $u_1=1, u_2=2$, then it'll be a divergent sequence.
              $endgroup$
              – Ankit Kumar
              Dec 20 '18 at 16:44










            • $begingroup$
              I would've expected a non constant sequence increasing to $1$. Here it's either constant or divergent.
              $endgroup$
              – furyo
              Dec 20 '18 at 16:47










            • $begingroup$
              You can't get a non-constant, convergent sequence because it isn't constant $implies $ at least one of $a_2, a_3$ is non-zero $implies $ $u_n$ will be a strictly increasing/decreasing (depending upon sign of $a_i$s) function of $n$.
              $endgroup$
              – Ankit Kumar
              Dec 20 '18 at 16:54
















            2












            2








            2





            $begingroup$

            You can solve any non-homogeneous, linear recurrence of the above form by converting it to homogeneous form.



            (I'll use $n$ instead of $n+1$, doesn't matter anyway!)



            $$u_{n}=au_{n-1}+bu_{n-2}+c$$
            $$u_{n-1}=au_{n-2}+bu_{n-3}+c$$
            Subtract them to get:-



            $$u_{n}=(a+1)u_{n-1}+(b-a)u_{n-2}-bu_{n-3}$$
            You can then use characteristic polynomial to find the answer.




            Speaking strictly for the equation you wrote, the characteristic polynomial is
            $$z^3=9z^2-11z+3$$ which has the roots
            $$z=1, 4+sqrt{13}, 4-sqrt{13}$$
            giving $$u_n=a_1+a_2{(4+sqrt{13})}^n+a_3{(4-sqrt{13})}^n$$, where $a_i$s depend on the initial condition.






            share|cite|improve this answer









            $endgroup$



            You can solve any non-homogeneous, linear recurrence of the above form by converting it to homogeneous form.



            (I'll use $n$ instead of $n+1$, doesn't matter anyway!)



            $$u_{n}=au_{n-1}+bu_{n-2}+c$$
            $$u_{n-1}=au_{n-2}+bu_{n-3}+c$$
            Subtract them to get:-



            $$u_{n}=(a+1)u_{n-1}+(b-a)u_{n-2}-bu_{n-3}$$
            You can then use characteristic polynomial to find the answer.




            Speaking strictly for the equation you wrote, the characteristic polynomial is
            $$z^3=9z^2-11z+3$$ which has the roots
            $$z=1, 4+sqrt{13}, 4-sqrt{13}$$
            giving $$u_n=a_1+a_2{(4+sqrt{13})}^n+a_3{(4-sqrt{13})}^n$$, where $a_i$s depend on the initial condition.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 20 '18 at 14:24









            Ankit KumarAnkit Kumar

            1,494221




            1,494221












            • $begingroup$
              So $u_n$ wouldn't converge? I find it a little bit strange.
              $endgroup$
              – furyo
              Dec 20 '18 at 16:39










            • $begingroup$
              @furyo It isn't strange at all. Stuff like this (convergence) depends on the initial conditions. Depending upon them, the series may or may not convegre!
              $endgroup$
              – Ankit Kumar
              Dec 20 '18 at 16:43










            • $begingroup$
              For instance, if you take $u_1=u_2=1$, $u_i=1$ for all $i$, which is a convergent sequence. On the other hand, if you take $u_1=1, u_2=2$, then it'll be a divergent sequence.
              $endgroup$
              – Ankit Kumar
              Dec 20 '18 at 16:44










            • $begingroup$
              I would've expected a non constant sequence increasing to $1$. Here it's either constant or divergent.
              $endgroup$
              – furyo
              Dec 20 '18 at 16:47










            • $begingroup$
              You can't get a non-constant, convergent sequence because it isn't constant $implies $ at least one of $a_2, a_3$ is non-zero $implies $ $u_n$ will be a strictly increasing/decreasing (depending upon sign of $a_i$s) function of $n$.
              $endgroup$
              – Ankit Kumar
              Dec 20 '18 at 16:54




















            • $begingroup$
              So $u_n$ wouldn't converge? I find it a little bit strange.
              $endgroup$
              – furyo
              Dec 20 '18 at 16:39










            • $begingroup$
              @furyo It isn't strange at all. Stuff like this (convergence) depends on the initial conditions. Depending upon them, the series may or may not convegre!
              $endgroup$
              – Ankit Kumar
              Dec 20 '18 at 16:43










            • $begingroup$
              For instance, if you take $u_1=u_2=1$, $u_i=1$ for all $i$, which is a convergent sequence. On the other hand, if you take $u_1=1, u_2=2$, then it'll be a divergent sequence.
              $endgroup$
              – Ankit Kumar
              Dec 20 '18 at 16:44










            • $begingroup$
              I would've expected a non constant sequence increasing to $1$. Here it's either constant or divergent.
              $endgroup$
              – furyo
              Dec 20 '18 at 16:47










            • $begingroup$
              You can't get a non-constant, convergent sequence because it isn't constant $implies $ at least one of $a_2, a_3$ is non-zero $implies $ $u_n$ will be a strictly increasing/decreasing (depending upon sign of $a_i$s) function of $n$.
              $endgroup$
              – Ankit Kumar
              Dec 20 '18 at 16:54


















            $begingroup$
            So $u_n$ wouldn't converge? I find it a little bit strange.
            $endgroup$
            – furyo
            Dec 20 '18 at 16:39




            $begingroup$
            So $u_n$ wouldn't converge? I find it a little bit strange.
            $endgroup$
            – furyo
            Dec 20 '18 at 16:39












            $begingroup$
            @furyo It isn't strange at all. Stuff like this (convergence) depends on the initial conditions. Depending upon them, the series may or may not convegre!
            $endgroup$
            – Ankit Kumar
            Dec 20 '18 at 16:43




            $begingroup$
            @furyo It isn't strange at all. Stuff like this (convergence) depends on the initial conditions. Depending upon them, the series may or may not convegre!
            $endgroup$
            – Ankit Kumar
            Dec 20 '18 at 16:43












            $begingroup$
            For instance, if you take $u_1=u_2=1$, $u_i=1$ for all $i$, which is a convergent sequence. On the other hand, if you take $u_1=1, u_2=2$, then it'll be a divergent sequence.
            $endgroup$
            – Ankit Kumar
            Dec 20 '18 at 16:44




            $begingroup$
            For instance, if you take $u_1=u_2=1$, $u_i=1$ for all $i$, which is a convergent sequence. On the other hand, if you take $u_1=1, u_2=2$, then it'll be a divergent sequence.
            $endgroup$
            – Ankit Kumar
            Dec 20 '18 at 16:44












            $begingroup$
            I would've expected a non constant sequence increasing to $1$. Here it's either constant or divergent.
            $endgroup$
            – furyo
            Dec 20 '18 at 16:47




            $begingroup$
            I would've expected a non constant sequence increasing to $1$. Here it's either constant or divergent.
            $endgroup$
            – furyo
            Dec 20 '18 at 16:47












            $begingroup$
            You can't get a non-constant, convergent sequence because it isn't constant $implies $ at least one of $a_2, a_3$ is non-zero $implies $ $u_n$ will be a strictly increasing/decreasing (depending upon sign of $a_i$s) function of $n$.
            $endgroup$
            – Ankit Kumar
            Dec 20 '18 at 16:54






            $begingroup$
            You can't get a non-constant, convergent sequence because it isn't constant $implies $ at least one of $a_2, a_3$ is non-zero $implies $ $u_n$ will be a strictly increasing/decreasing (depending upon sign of $a_i$s) function of $n$.
            $endgroup$
            – Ankit Kumar
            Dec 20 '18 at 16:54













            0












            $begingroup$

            This is a linear difference equation so can be stated as



            $$
            u = u_h + u_p
            $$



            with



            $$
            u_h(n+1)-au_h(n)-bu_h(n-1) = 0\
            u_p(n+1)-au_p(n)-bu_p(n-1) = c\
            $$



            now making $u_h(n) = phi^n$ and substituting we have



            $$
            phi^nleft(phi-a-bphi^{-1}right) = 0
            $$



            so



            $$
            phi = frac{1}{2} left(apmsqrt{a^2+4 b}right)
            $$



            and



            $$
            u_h(n) = frac{C_1}{2^n} left(a-sqrt{a^2+4 b}right)^n+frac{C_2}{2^n} left(a+sqrt{a^2+4 b}right)^n
            $$



            and also



            $$
            u_p(n) = frac{c}{1-a-b}
            $$



            so finally



            $$
            u(n) = frac{C_1}{2^n} left(a-sqrt{a^2+4 b}right)^n+frac{C_2}{2^n} left(a+sqrt{a^2+4 b}right)^n + frac{c}{1-a-b}
            $$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              This is a linear difference equation so can be stated as



              $$
              u = u_h + u_p
              $$



              with



              $$
              u_h(n+1)-au_h(n)-bu_h(n-1) = 0\
              u_p(n+1)-au_p(n)-bu_p(n-1) = c\
              $$



              now making $u_h(n) = phi^n$ and substituting we have



              $$
              phi^nleft(phi-a-bphi^{-1}right) = 0
              $$



              so



              $$
              phi = frac{1}{2} left(apmsqrt{a^2+4 b}right)
              $$



              and



              $$
              u_h(n) = frac{C_1}{2^n} left(a-sqrt{a^2+4 b}right)^n+frac{C_2}{2^n} left(a+sqrt{a^2+4 b}right)^n
              $$



              and also



              $$
              u_p(n) = frac{c}{1-a-b}
              $$



              so finally



              $$
              u(n) = frac{C_1}{2^n} left(a-sqrt{a^2+4 b}right)^n+frac{C_2}{2^n} left(a+sqrt{a^2+4 b}right)^n + frac{c}{1-a-b}
              $$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                This is a linear difference equation so can be stated as



                $$
                u = u_h + u_p
                $$



                with



                $$
                u_h(n+1)-au_h(n)-bu_h(n-1) = 0\
                u_p(n+1)-au_p(n)-bu_p(n-1) = c\
                $$



                now making $u_h(n) = phi^n$ and substituting we have



                $$
                phi^nleft(phi-a-bphi^{-1}right) = 0
                $$



                so



                $$
                phi = frac{1}{2} left(apmsqrt{a^2+4 b}right)
                $$



                and



                $$
                u_h(n) = frac{C_1}{2^n} left(a-sqrt{a^2+4 b}right)^n+frac{C_2}{2^n} left(a+sqrt{a^2+4 b}right)^n
                $$



                and also



                $$
                u_p(n) = frac{c}{1-a-b}
                $$



                so finally



                $$
                u(n) = frac{C_1}{2^n} left(a-sqrt{a^2+4 b}right)^n+frac{C_2}{2^n} left(a+sqrt{a^2+4 b}right)^n + frac{c}{1-a-b}
                $$






                share|cite|improve this answer









                $endgroup$



                This is a linear difference equation so can be stated as



                $$
                u = u_h + u_p
                $$



                with



                $$
                u_h(n+1)-au_h(n)-bu_h(n-1) = 0\
                u_p(n+1)-au_p(n)-bu_p(n-1) = c\
                $$



                now making $u_h(n) = phi^n$ and substituting we have



                $$
                phi^nleft(phi-a-bphi^{-1}right) = 0
                $$



                so



                $$
                phi = frac{1}{2} left(apmsqrt{a^2+4 b}right)
                $$



                and



                $$
                u_h(n) = frac{C_1}{2^n} left(a-sqrt{a^2+4 b}right)^n+frac{C_2}{2^n} left(a+sqrt{a^2+4 b}right)^n
                $$



                and also



                $$
                u_p(n) = frac{c}{1-a-b}
                $$



                so finally



                $$
                u(n) = frac{C_1}{2^n} left(a-sqrt{a^2+4 b}right)^n+frac{C_2}{2^n} left(a+sqrt{a^2+4 b}right)^n + frac{c}{1-a-b}
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 20 '18 at 14:25









                CesareoCesareo

                8,8993516




                8,8993516






























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