Recurrence relation with non-constant coefficients...












0












$begingroup$


Is there a general form to solve a recurrence relation with non-constant coefficients?



Specifically, I'm interested in solving:



$$a_n=1+frac{1}{2^{n+1}}a_{n+1}+frac{2^{n+1}-1}{2^{n+1}}a_{n-1}$$ with $a_0=0$.



Thanks!






combinatorics discrete-mathematics recurrence-relations







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  • $begingroup$
    You also need to give $a_1$ in order to solve this, since the recurrence gives $a_{n+1}$ as a function of $a_n$ and $a_{n-1}$.
    $endgroup$
    – Kolja
    Dec 26 '18 at 15:37
















0












$begingroup$


Is there a general form to solve a recurrence relation with non-constant coefficients?



Specifically, I'm interested in solving:



$$a_n=1+frac{1}{2^{n+1}}a_{n+1}+frac{2^{n+1}-1}{2^{n+1}}a_{n-1}$$ with $a_0=0$.



Thanks!






combinatorics discrete-mathematics recurrence-relations







share|cite|improve this question









$endgroup$












  • $begingroup$
    You also need to give $a_1$ in order to solve this, since the recurrence gives $a_{n+1}$ as a function of $a_n$ and $a_{n-1}$.
    $endgroup$
    – Kolja
    Dec 26 '18 at 15:37














0












0








0





$begingroup$


Is there a general form to solve a recurrence relation with non-constant coefficients?



Specifically, I'm interested in solving:



$$a_n=1+frac{1}{2^{n+1}}a_{n+1}+frac{2^{n+1}-1}{2^{n+1}}a_{n-1}$$ with $a_0=0$.



Thanks!






combinatorics discrete-mathematics recurrence-relations







share|cite|improve this question









$endgroup$




Is there a general form to solve a recurrence relation with non-constant coefficients?



Specifically, I'm interested in solving:



$$a_n=1+frac{1}{2^{n+1}}a_{n+1}+frac{2^{n+1}-1}{2^{n+1}}a_{n-1}$$ with $a_0=0$.



Thanks!






combinatorics discrete-mathematics recurrence-relations




combinatorics discrete-mathematics recurrence-relations






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asked Dec 26 '18 at 15:24









catch22catch22

1,3561125




1,3561125












  • $begingroup$
    You also need to give $a_1$ in order to solve this, since the recurrence gives $a_{n+1}$ as a function of $a_n$ and $a_{n-1}$.
    $endgroup$
    – Kolja
    Dec 26 '18 at 15:37


















  • $begingroup$
    You also need to give $a_1$ in order to solve this, since the recurrence gives $a_{n+1}$ as a function of $a_n$ and $a_{n-1}$.
    $endgroup$
    – Kolja
    Dec 26 '18 at 15:37
















$begingroup$
You also need to give $a_1$ in order to solve this, since the recurrence gives $a_{n+1}$ as a function of $a_n$ and $a_{n-1}$.
$endgroup$
– Kolja
Dec 26 '18 at 15:37




$begingroup$
You also need to give $a_1$ in order to solve this, since the recurrence gives $a_{n+1}$ as a function of $a_n$ and $a_{n-1}$.
$endgroup$
– Kolja
Dec 26 '18 at 15:37










2 Answers
2






active

oldest

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1












$begingroup$

Let $A(x)=sum_{i=0}^infty a_nx^n$. Suppose also that we have $a_1$. Keep in mind that the recurrence relation is $a_{n+1} = 2^{n+1}a_n - 2^{n+1} - (2^{n+1}-1)a_{n-1}$.



Then
$$
A(x)=sum_{i=0}^infty a_nx^n = a_0 + a_1 x + sum_{i=2}^infty a_nx^n = a_1x + sum_{i=0}^infty a_{n+2}x^{n+2}=cdots
$$
Now we use the above recurrence relation
$$
cdots = a_1 x + sum_{i=0}^infty 2^{n+2}a_{n+1}x^{n+2} - sum_{i=0}^infty 2^{n+2}x^{n+2} - sum_{i=0}^infty (2^{n+2}a_{n}x^{n+2} + sum_{i=0}^infty a_{n}x^{n+2} = cdots
$$
$$
cdots =a_1 x + 2xsum_{i=0}^infty a_{n+1}(2x)^{n+1} - 4x^2sum_{i=0}^infty(2x)^{n} - 4x^2sum_{i=0}^infty a_n(2x)^n + x^2sum_{i=0}^infty a_nx^n
$$
The first sum is equal to $sum_{i=0}^infty a_n(2x)^n$ since $a_0=0$. The second sum is $frac{1}{1-2x}$. The third one is equal to the first one, and they are both equal to $A(2x)$. The last sum is simply $A(x)$. Plugging it all in, we get:



$$
A(x) = a_1x + 2xA(2x) - frac{4x^2}{1-2x} - 4x^2 A(2x) + x^2A(x).
$$



$$
A(x) (1-x^2) - 2xcdot A(2x)(1-2x) = a_1x - frac{4x^2}{1-2x}
$$



I'm not really sure where to go from here, but finding a closed form for $A(x)$ will give a closed form for $a_n$'s. It doesn't always have to exist. Maybe one can solve the above equation, or turn it into a differential equation. Anyway I hope this helps.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    There is no "general form", and in many cases there is no closed form at all.

    In this case, you might look at generating functions.



    If $g(x) = sum_{n=0}^infty a_n x^n$ is the generating function, by summing your recursion times $x^n$ for $n=1$ to $infty$, I get



    $$ g(x) = frac{x}{1-x} + frac{1}{x} left(gleft(frac{x}{2}right) - a_1 frac{x}{2}right) + x g(x) - frac{x}{4} gleft(frac{x}{2}right)$$



    which "simplifies" to



    $$ g(x) = frac{x^2-4}{4 x (x-1)} gleft(frac{x}{2}right) + frac{(a(1)+2) x - a(1)}{2(x-1)^2} $$






    share|cite|improve this answer











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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      active

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      active

      oldest

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      1












      $begingroup$

      Let $A(x)=sum_{i=0}^infty a_nx^n$. Suppose also that we have $a_1$. Keep in mind that the recurrence relation is $a_{n+1} = 2^{n+1}a_n - 2^{n+1} - (2^{n+1}-1)a_{n-1}$.



      Then
      $$
      A(x)=sum_{i=0}^infty a_nx^n = a_0 + a_1 x + sum_{i=2}^infty a_nx^n = a_1x + sum_{i=0}^infty a_{n+2}x^{n+2}=cdots
      $$
      Now we use the above recurrence relation
      $$
      cdots = a_1 x + sum_{i=0}^infty 2^{n+2}a_{n+1}x^{n+2} - sum_{i=0}^infty 2^{n+2}x^{n+2} - sum_{i=0}^infty (2^{n+2}a_{n}x^{n+2} + sum_{i=0}^infty a_{n}x^{n+2} = cdots
      $$
      $$
      cdots =a_1 x + 2xsum_{i=0}^infty a_{n+1}(2x)^{n+1} - 4x^2sum_{i=0}^infty(2x)^{n} - 4x^2sum_{i=0}^infty a_n(2x)^n + x^2sum_{i=0}^infty a_nx^n
      $$
      The first sum is equal to $sum_{i=0}^infty a_n(2x)^n$ since $a_0=0$. The second sum is $frac{1}{1-2x}$. The third one is equal to the first one, and they are both equal to $A(2x)$. The last sum is simply $A(x)$. Plugging it all in, we get:



      $$
      A(x) = a_1x + 2xA(2x) - frac{4x^2}{1-2x} - 4x^2 A(2x) + x^2A(x).
      $$



      $$
      A(x) (1-x^2) - 2xcdot A(2x)(1-2x) = a_1x - frac{4x^2}{1-2x}
      $$



      I'm not really sure where to go from here, but finding a closed form for $A(x)$ will give a closed form for $a_n$'s. It doesn't always have to exist. Maybe one can solve the above equation, or turn it into a differential equation. Anyway I hope this helps.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Let $A(x)=sum_{i=0}^infty a_nx^n$. Suppose also that we have $a_1$. Keep in mind that the recurrence relation is $a_{n+1} = 2^{n+1}a_n - 2^{n+1} - (2^{n+1}-1)a_{n-1}$.



        Then
        $$
        A(x)=sum_{i=0}^infty a_nx^n = a_0 + a_1 x + sum_{i=2}^infty a_nx^n = a_1x + sum_{i=0}^infty a_{n+2}x^{n+2}=cdots
        $$
        Now we use the above recurrence relation
        $$
        cdots = a_1 x + sum_{i=0}^infty 2^{n+2}a_{n+1}x^{n+2} - sum_{i=0}^infty 2^{n+2}x^{n+2} - sum_{i=0}^infty (2^{n+2}a_{n}x^{n+2} + sum_{i=0}^infty a_{n}x^{n+2} = cdots
        $$
        $$
        cdots =a_1 x + 2xsum_{i=0}^infty a_{n+1}(2x)^{n+1} - 4x^2sum_{i=0}^infty(2x)^{n} - 4x^2sum_{i=0}^infty a_n(2x)^n + x^2sum_{i=0}^infty a_nx^n
        $$
        The first sum is equal to $sum_{i=0}^infty a_n(2x)^n$ since $a_0=0$. The second sum is $frac{1}{1-2x}$. The third one is equal to the first one, and they are both equal to $A(2x)$. The last sum is simply $A(x)$. Plugging it all in, we get:



        $$
        A(x) = a_1x + 2xA(2x) - frac{4x^2}{1-2x} - 4x^2 A(2x) + x^2A(x).
        $$



        $$
        A(x) (1-x^2) - 2xcdot A(2x)(1-2x) = a_1x - frac{4x^2}{1-2x}
        $$



        I'm not really sure where to go from here, but finding a closed form for $A(x)$ will give a closed form for $a_n$'s. It doesn't always have to exist. Maybe one can solve the above equation, or turn it into a differential equation. Anyway I hope this helps.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Let $A(x)=sum_{i=0}^infty a_nx^n$. Suppose also that we have $a_1$. Keep in mind that the recurrence relation is $a_{n+1} = 2^{n+1}a_n - 2^{n+1} - (2^{n+1}-1)a_{n-1}$.



          Then
          $$
          A(x)=sum_{i=0}^infty a_nx^n = a_0 + a_1 x + sum_{i=2}^infty a_nx^n = a_1x + sum_{i=0}^infty a_{n+2}x^{n+2}=cdots
          $$
          Now we use the above recurrence relation
          $$
          cdots = a_1 x + sum_{i=0}^infty 2^{n+2}a_{n+1}x^{n+2} - sum_{i=0}^infty 2^{n+2}x^{n+2} - sum_{i=0}^infty (2^{n+2}a_{n}x^{n+2} + sum_{i=0}^infty a_{n}x^{n+2} = cdots
          $$
          $$
          cdots =a_1 x + 2xsum_{i=0}^infty a_{n+1}(2x)^{n+1} - 4x^2sum_{i=0}^infty(2x)^{n} - 4x^2sum_{i=0}^infty a_n(2x)^n + x^2sum_{i=0}^infty a_nx^n
          $$
          The first sum is equal to $sum_{i=0}^infty a_n(2x)^n$ since $a_0=0$. The second sum is $frac{1}{1-2x}$. The third one is equal to the first one, and they are both equal to $A(2x)$. The last sum is simply $A(x)$. Plugging it all in, we get:



          $$
          A(x) = a_1x + 2xA(2x) - frac{4x^2}{1-2x} - 4x^2 A(2x) + x^2A(x).
          $$



          $$
          A(x) (1-x^2) - 2xcdot A(2x)(1-2x) = a_1x - frac{4x^2}{1-2x}
          $$



          I'm not really sure where to go from here, but finding a closed form for $A(x)$ will give a closed form for $a_n$'s. It doesn't always have to exist. Maybe one can solve the above equation, or turn it into a differential equation. Anyway I hope this helps.






          share|cite|improve this answer









          $endgroup$



          Let $A(x)=sum_{i=0}^infty a_nx^n$. Suppose also that we have $a_1$. Keep in mind that the recurrence relation is $a_{n+1} = 2^{n+1}a_n - 2^{n+1} - (2^{n+1}-1)a_{n-1}$.



          Then
          $$
          A(x)=sum_{i=0}^infty a_nx^n = a_0 + a_1 x + sum_{i=2}^infty a_nx^n = a_1x + sum_{i=0}^infty a_{n+2}x^{n+2}=cdots
          $$
          Now we use the above recurrence relation
          $$
          cdots = a_1 x + sum_{i=0}^infty 2^{n+2}a_{n+1}x^{n+2} - sum_{i=0}^infty 2^{n+2}x^{n+2} - sum_{i=0}^infty (2^{n+2}a_{n}x^{n+2} + sum_{i=0}^infty a_{n}x^{n+2} = cdots
          $$
          $$
          cdots =a_1 x + 2xsum_{i=0}^infty a_{n+1}(2x)^{n+1} - 4x^2sum_{i=0}^infty(2x)^{n} - 4x^2sum_{i=0}^infty a_n(2x)^n + x^2sum_{i=0}^infty a_nx^n
          $$
          The first sum is equal to $sum_{i=0}^infty a_n(2x)^n$ since $a_0=0$. The second sum is $frac{1}{1-2x}$. The third one is equal to the first one, and they are both equal to $A(2x)$. The last sum is simply $A(x)$. Plugging it all in, we get:



          $$
          A(x) = a_1x + 2xA(2x) - frac{4x^2}{1-2x} - 4x^2 A(2x) + x^2A(x).
          $$



          $$
          A(x) (1-x^2) - 2xcdot A(2x)(1-2x) = a_1x - frac{4x^2}{1-2x}
          $$



          I'm not really sure where to go from here, but finding a closed form for $A(x)$ will give a closed form for $a_n$'s. It doesn't always have to exist. Maybe one can solve the above equation, or turn it into a differential equation. Anyway I hope this helps.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 26 '18 at 15:48









          KoljaKolja

          625310




          625310























              1












              $begingroup$

              There is no "general form", and in many cases there is no closed form at all.

              In this case, you might look at generating functions.



              If $g(x) = sum_{n=0}^infty a_n x^n$ is the generating function, by summing your recursion times $x^n$ for $n=1$ to $infty$, I get



              $$ g(x) = frac{x}{1-x} + frac{1}{x} left(gleft(frac{x}{2}right) - a_1 frac{x}{2}right) + x g(x) - frac{x}{4} gleft(frac{x}{2}right)$$



              which "simplifies" to



              $$ g(x) = frac{x^2-4}{4 x (x-1)} gleft(frac{x}{2}right) + frac{(a(1)+2) x - a(1)}{2(x-1)^2} $$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                There is no "general form", and in many cases there is no closed form at all.

                In this case, you might look at generating functions.



                If $g(x) = sum_{n=0}^infty a_n x^n$ is the generating function, by summing your recursion times $x^n$ for $n=1$ to $infty$, I get



                $$ g(x) = frac{x}{1-x} + frac{1}{x} left(gleft(frac{x}{2}right) - a_1 frac{x}{2}right) + x g(x) - frac{x}{4} gleft(frac{x}{2}right)$$



                which "simplifies" to



                $$ g(x) = frac{x^2-4}{4 x (x-1)} gleft(frac{x}{2}right) + frac{(a(1)+2) x - a(1)}{2(x-1)^2} $$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  There is no "general form", and in many cases there is no closed form at all.

                  In this case, you might look at generating functions.



                  If $g(x) = sum_{n=0}^infty a_n x^n$ is the generating function, by summing your recursion times $x^n$ for $n=1$ to $infty$, I get



                  $$ g(x) = frac{x}{1-x} + frac{1}{x} left(gleft(frac{x}{2}right) - a_1 frac{x}{2}right) + x g(x) - frac{x}{4} gleft(frac{x}{2}right)$$



                  which "simplifies" to



                  $$ g(x) = frac{x^2-4}{4 x (x-1)} gleft(frac{x}{2}right) + frac{(a(1)+2) x - a(1)}{2(x-1)^2} $$






                  share|cite|improve this answer











                  $endgroup$



                  There is no "general form", and in many cases there is no closed form at all.

                  In this case, you might look at generating functions.



                  If $g(x) = sum_{n=0}^infty a_n x^n$ is the generating function, by summing your recursion times $x^n$ for $n=1$ to $infty$, I get



                  $$ g(x) = frac{x}{1-x} + frac{1}{x} left(gleft(frac{x}{2}right) - a_1 frac{x}{2}right) + x g(x) - frac{x}{4} gleft(frac{x}{2}right)$$



                  which "simplifies" to



                  $$ g(x) = frac{x^2-4}{4 x (x-1)} gleft(frac{x}{2}right) + frac{(a(1)+2) x - a(1)}{2(x-1)^2} $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 26 '18 at 15:59

























                  answered Dec 26 '18 at 15:34









                  Robert IsraelRobert Israel

                  331k23221477




                  331k23221477






























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