I didn't understand this recurrence relation solution.












2












$begingroup$


Recently, i was trying to solve this recurrence relation
$$ a_{n+4} = frac{-alpha(x)}{(n+4)} cdot a_{n+3} +frac{-beta(x)}{(n+3)cdot (n+4)} cdot a_{n+2} $$



But i can't solve for $a_n$



I've tried to solve using Wolfram|Alpha and i got this result.

Actually i'm really good at recurrence relations, but i can't understand how do i get this result and how to solve it.

Edit: n = 0,1,2,3....






sequences-and-series combinatorics discrete-mathematics recurrence-relations







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Murat Güven is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    2












    $begingroup$


    Recently, i was trying to solve this recurrence relation
    $$ a_{n+4} = frac{-alpha(x)}{(n+4)} cdot a_{n+3} +frac{-beta(x)}{(n+3)cdot (n+4)} cdot a_{n+2} $$



    But i can't solve for $a_n$



    I've tried to solve using Wolfram|Alpha and i got this result.

    Actually i'm really good at recurrence relations, but i can't understand how do i get this result and how to solve it.

    Edit: n = 0,1,2,3....






    sequences-and-series combinatorics discrete-mathematics recurrence-relations







    share|cite|improve this question









    New contributor




    Murat Güven is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      2












      2








      2


      0



      $begingroup$


      Recently, i was trying to solve this recurrence relation
      $$ a_{n+4} = frac{-alpha(x)}{(n+4)} cdot a_{n+3} +frac{-beta(x)}{(n+3)cdot (n+4)} cdot a_{n+2} $$



      But i can't solve for $a_n$



      I've tried to solve using Wolfram|Alpha and i got this result.

      Actually i'm really good at recurrence relations, but i can't understand how do i get this result and how to solve it.

      Edit: n = 0,1,2,3....






      sequences-and-series combinatorics discrete-mathematics recurrence-relations







      share|cite|improve this question









      New contributor




      Murat Güven is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Recently, i was trying to solve this recurrence relation
      $$ a_{n+4} = frac{-alpha(x)}{(n+4)} cdot a_{n+3} +frac{-beta(x)}{(n+3)cdot (n+4)} cdot a_{n+2} $$



      But i can't solve for $a_n$



      I've tried to solve using Wolfram|Alpha and i got this result.

      Actually i'm really good at recurrence relations, but i can't understand how do i get this result and how to solve it.

      Edit: n = 0,1,2,3....






      sequences-and-series combinatorics discrete-mathematics recurrence-relations




      sequences-and-series combinatorics discrete-mathematics recurrence-relations






      share|cite|improve this question









      New contributor




      Murat Güven is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      Murat Güven is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago







      Murat Güven













      New contributor




      Murat Güven is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 5 hours ago









      Murat GüvenMurat Güven

      204




      204




      New contributor




      Murat Güven is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Murat Güven is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Murat Güven is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          1 Answer
          1






          active

          oldest

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          5












          $begingroup$

          Hint Multiplying by $(n+4)!$ you get
          $$(n+4)!a_{n+4} = -alpha(x) *(n+3)! a_{n+3} -beta(x) *(n+2)!a_{n+2}$$



          Let $b_n=n! a_n$ then, your recurrence is
          $$b_{n+4}= -alpha(x) cdot b_{n+3} -beta(x) cdot b_{n+2}$$
          which is a standard second order recurrence.



          Solve it, and then
          $$a_n=frac{b_n}{n!}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            when i solve this with wolfram|alpha for b result is this. This result is very different from the solution above is there something else i'm missing
            $endgroup$
            – Murat Güven
            3 hours ago












          • $begingroup$
            @MuratGüven WA solves the second order recurrence, the solution is given by the roots of the characteristic equation (which is what WA has in bracket, under the extra assumption that they are different).
            $endgroup$
            – N. S.
            2 hours ago










          • $begingroup$
            why beta is in the denominator here?
            $endgroup$
            – Murat Güven
            2 hours ago












          • $begingroup$
            @MuratGüven That is irrelevant... $frac{sqrt{alpha^2-beta}}{beta}c_1$ is just a contant, so writing that or $C_1$ is the same thing...That particular form probably appears because of the way in which WA solves the recurrence, but most of those constants can be put inside $C_1,C_2$.
            $endgroup$
            – N. S.
            2 hours ago













          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          Hint Multiplying by $(n+4)!$ you get
          $$(n+4)!a_{n+4} = -alpha(x) *(n+3)! a_{n+3} -beta(x) *(n+2)!a_{n+2}$$



          Let $b_n=n! a_n$ then, your recurrence is
          $$b_{n+4}= -alpha(x) cdot b_{n+3} -beta(x) cdot b_{n+2}$$
          which is a standard second order recurrence.



          Solve it, and then
          $$a_n=frac{b_n}{n!}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            when i solve this with wolfram|alpha for b result is this. This result is very different from the solution above is there something else i'm missing
            $endgroup$
            – Murat Güven
            3 hours ago












          • $begingroup$
            @MuratGüven WA solves the second order recurrence, the solution is given by the roots of the characteristic equation (which is what WA has in bracket, under the extra assumption that they are different).
            $endgroup$
            – N. S.
            2 hours ago










          • $begingroup$
            why beta is in the denominator here?
            $endgroup$
            – Murat Güven
            2 hours ago












          • $begingroup$
            @MuratGüven That is irrelevant... $frac{sqrt{alpha^2-beta}}{beta}c_1$ is just a contant, so writing that or $C_1$ is the same thing...That particular form probably appears because of the way in which WA solves the recurrence, but most of those constants can be put inside $C_1,C_2$.
            $endgroup$
            – N. S.
            2 hours ago


















          5












          $begingroup$

          Hint Multiplying by $(n+4)!$ you get
          $$(n+4)!a_{n+4} = -alpha(x) *(n+3)! a_{n+3} -beta(x) *(n+2)!a_{n+2}$$



          Let $b_n=n! a_n$ then, your recurrence is
          $$b_{n+4}= -alpha(x) cdot b_{n+3} -beta(x) cdot b_{n+2}$$
          which is a standard second order recurrence.



          Solve it, and then
          $$a_n=frac{b_n}{n!}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            when i solve this with wolfram|alpha for b result is this. This result is very different from the solution above is there something else i'm missing
            $endgroup$
            – Murat Güven
            3 hours ago












          • $begingroup$
            @MuratGüven WA solves the second order recurrence, the solution is given by the roots of the characteristic equation (which is what WA has in bracket, under the extra assumption that they are different).
            $endgroup$
            – N. S.
            2 hours ago










          • $begingroup$
            why beta is in the denominator here?
            $endgroup$
            – Murat Güven
            2 hours ago












          • $begingroup$
            @MuratGüven That is irrelevant... $frac{sqrt{alpha^2-beta}}{beta}c_1$ is just a contant, so writing that or $C_1$ is the same thing...That particular form probably appears because of the way in which WA solves the recurrence, but most of those constants can be put inside $C_1,C_2$.
            $endgroup$
            – N. S.
            2 hours ago
















          5












          5








          5





          $begingroup$

          Hint Multiplying by $(n+4)!$ you get
          $$(n+4)!a_{n+4} = -alpha(x) *(n+3)! a_{n+3} -beta(x) *(n+2)!a_{n+2}$$



          Let $b_n=n! a_n$ then, your recurrence is
          $$b_{n+4}= -alpha(x) cdot b_{n+3} -beta(x) cdot b_{n+2}$$
          which is a standard second order recurrence.



          Solve it, and then
          $$a_n=frac{b_n}{n!}$$






          share|cite|improve this answer











          $endgroup$



          Hint Multiplying by $(n+4)!$ you get
          $$(n+4)!a_{n+4} = -alpha(x) *(n+3)! a_{n+3} -beta(x) *(n+2)!a_{n+2}$$



          Let $b_n=n! a_n$ then, your recurrence is
          $$b_{n+4}= -alpha(x) cdot b_{n+3} -beta(x) cdot b_{n+2}$$
          which is a standard second order recurrence.



          Solve it, and then
          $$a_n=frac{b_n}{n!}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago









          Murat Güven

          204




          204










          answered 4 hours ago









          N. S.N. S.

          104k7112208




          104k7112208












          • $begingroup$
            when i solve this with wolfram|alpha for b result is this. This result is very different from the solution above is there something else i'm missing
            $endgroup$
            – Murat Güven
            3 hours ago












          • $begingroup$
            @MuratGüven WA solves the second order recurrence, the solution is given by the roots of the characteristic equation (which is what WA has in bracket, under the extra assumption that they are different).
            $endgroup$
            – N. S.
            2 hours ago










          • $begingroup$
            why beta is in the denominator here?
            $endgroup$
            – Murat Güven
            2 hours ago












          • $begingroup$
            @MuratGüven That is irrelevant... $frac{sqrt{alpha^2-beta}}{beta}c_1$ is just a contant, so writing that or $C_1$ is the same thing...That particular form probably appears because of the way in which WA solves the recurrence, but most of those constants can be put inside $C_1,C_2$.
            $endgroup$
            – N. S.
            2 hours ago




















          • $begingroup$
            when i solve this with wolfram|alpha for b result is this. This result is very different from the solution above is there something else i'm missing
            $endgroup$
            – Murat Güven
            3 hours ago












          • $begingroup$
            @MuratGüven WA solves the second order recurrence, the solution is given by the roots of the characteristic equation (which is what WA has in bracket, under the extra assumption that they are different).
            $endgroup$
            – N. S.
            2 hours ago










          • $begingroup$
            why beta is in the denominator here?
            $endgroup$
            – Murat Güven
            2 hours ago












          • $begingroup$
            @MuratGüven That is irrelevant... $frac{sqrt{alpha^2-beta}}{beta}c_1$ is just a contant, so writing that or $C_1$ is the same thing...That particular form probably appears because of the way in which WA solves the recurrence, but most of those constants can be put inside $C_1,C_2$.
            $endgroup$
            – N. S.
            2 hours ago


















          $begingroup$
          when i solve this with wolfram|alpha for b result is this. This result is very different from the solution above is there something else i'm missing
          $endgroup$
          – Murat Güven
          3 hours ago






          $begingroup$
          when i solve this with wolfram|alpha for b result is this. This result is very different from the solution above is there something else i'm missing
          $endgroup$
          – Murat Güven
          3 hours ago














          $begingroup$
          @MuratGüven WA solves the second order recurrence, the solution is given by the roots of the characteristic equation (which is what WA has in bracket, under the extra assumption that they are different).
          $endgroup$
          – N. S.
          2 hours ago




          $begingroup$
          @MuratGüven WA solves the second order recurrence, the solution is given by the roots of the characteristic equation (which is what WA has in bracket, under the extra assumption that they are different).
          $endgroup$
          – N. S.
          2 hours ago












          $begingroup$
          why beta is in the denominator here?
          $endgroup$
          – Murat Güven
          2 hours ago






          $begingroup$
          why beta is in the denominator here?
          $endgroup$
          – Murat Güven
          2 hours ago














          $begingroup$
          @MuratGüven That is irrelevant... $frac{sqrt{alpha^2-beta}}{beta}c_1$ is just a contant, so writing that or $C_1$ is the same thing...That particular form probably appears because of the way in which WA solves the recurrence, but most of those constants can be put inside $C_1,C_2$.
          $endgroup$
          – N. S.
          2 hours ago






          $begingroup$
          @MuratGüven That is irrelevant... $frac{sqrt{alpha^2-beta}}{beta}c_1$ is just a contant, so writing that or $C_1$ is the same thing...That particular form probably appears because of the way in which WA solves the recurrence, but most of those constants can be put inside $C_1,C_2$.
          $endgroup$
          – N. S.
          2 hours ago












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