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







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












$begingroup$


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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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

votes


















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











    $endgroup$














      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053033%2frecurrence-relation-with-non-constant-coefficients-a-n-1-frac12n1a-n%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      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






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053033%2frecurrence-relation-with-non-constant-coefficients-a-n-1-frac12n1a-n%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Le Mesnil-Réaume

                      Image
                      Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
                      Read more

                      Bundesstraße 106

                      Image
                      Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                      Read more

                      Ida-Boy-Ed-Garten

                      Image
                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                      Read more