Solving recurrence $X_n = 4X_{n-1}+5$












1












$begingroup$


$X_n=4X_{n-1}+5$



How come the solution of this recurrence is this?



$X_n=frac834^n+frac53$



I also have that $X_0=1$.



I am using telescoping method and I am trying to solve it like this:



$X_n= 5 + 4X_{n-1}$



$X_n= 5 + 4(5+4X_{n-2})$



$X_n= 5 + 4times5 + 4times4times X_{n-2}$



But this leads to me getting $5times4^{n-1}times4^n$.



Can some please explain this to me?






discrete-mathematics recurrence-relations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean $$x_n=4x_{n-1}+5,x_0=1$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 15 '18 at 15:48










  • $begingroup$
    Arithmetico–geometric sequence
    $endgroup$
    – Jean-Claude Arbaut
    Dec 15 '18 at 15:54


















1












$begingroup$


$X_n=4X_{n-1}+5$



How come the solution of this recurrence is this?



$X_n=frac834^n+frac53$



I also have that $X_0=1$.



I am using telescoping method and I am trying to solve it like this:



$X_n= 5 + 4X_{n-1}$



$X_n= 5 + 4(5+4X_{n-2})$



$X_n= 5 + 4times5 + 4times4times X_{n-2}$



But this leads to me getting $5times4^{n-1}times4^n$.



Can some please explain this to me?






discrete-mathematics recurrence-relations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean $$x_n=4x_{n-1}+5,x_0=1$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 15 '18 at 15:48










  • $begingroup$
    Arithmetico–geometric sequence
    $endgroup$
    – Jean-Claude Arbaut
    Dec 15 '18 at 15:54
















1












1








1





$begingroup$


$X_n=4X_{n-1}+5$



How come the solution of this recurrence is this?



$X_n=frac834^n+frac53$



I also have that $X_0=1$.



I am using telescoping method and I am trying to solve it like this:



$X_n= 5 + 4X_{n-1}$



$X_n= 5 + 4(5+4X_{n-2})$



$X_n= 5 + 4times5 + 4times4times X_{n-2}$



But this leads to me getting $5times4^{n-1}times4^n$.



Can some please explain this to me?






discrete-mathematics recurrence-relations







share|cite|improve this question











$endgroup$




$X_n=4X_{n-1}+5$



How come the solution of this recurrence is this?



$X_n=frac834^n+frac53$



I also have that $X_0=1$.



I am using telescoping method and I am trying to solve it like this:



$X_n= 5 + 4X_{n-1}$



$X_n= 5 + 4(5+4X_{n-2})$



$X_n= 5 + 4times5 + 4times4times X_{n-2}$



But this leads to me getting $5times4^{n-1}times4^n$.



Can some please explain this to me?






discrete-mathematics recurrence-relations




discrete-mathematics recurrence-relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 15:53









Jean-Claude Arbaut

14.8k63464




14.8k63464










asked Dec 15 '18 at 15:46









ponikoliponikoli

416




416












  • $begingroup$
    Do you mean $$x_n=4x_{n-1}+5,x_0=1$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 15 '18 at 15:48










  • $begingroup$
    Arithmetico–geometric sequence
    $endgroup$
    – Jean-Claude Arbaut
    Dec 15 '18 at 15:54




















  • $begingroup$
    Do you mean $$x_n=4x_{n-1}+5,x_0=1$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 15 '18 at 15:48










  • $begingroup$
    Arithmetico–geometric sequence
    $endgroup$
    – Jean-Claude Arbaut
    Dec 15 '18 at 15:54


















$begingroup$
Do you mean $$x_n=4x_{n-1}+5,x_0=1$$?
$endgroup$
– Dr. Sonnhard Graubner
Dec 15 '18 at 15:48




$begingroup$
Do you mean $$x_n=4x_{n-1}+5,x_0=1$$?
$endgroup$
– Dr. Sonnhard Graubner
Dec 15 '18 at 15:48












$begingroup$
Arithmetico–geometric sequence
$endgroup$
– Jean-Claude Arbaut
Dec 15 '18 at 15:54






$begingroup$
Arithmetico–geometric sequence
$endgroup$
– Jean-Claude Arbaut
Dec 15 '18 at 15:54












3 Answers
3






active

oldest

votes


















2












$begingroup$

If you calculate the first few terms explicitly, you will find that the $n$th term is the sum of an exponential and a geometric series. For example,



$$
X_3 = 4^3 + 5(4^2+4+1).
$$
So in general,
$$
X_n = 4^n + 5sum_{k=0}^{n-1}4^k = 4^n +5frac{1-4^n}{1-4},
$$



which should simplify to the answer you gave.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay, that makes sense. But it looks like a different solution then from what is written in my textbook?
    $endgroup$
    – ponikoli
    Dec 15 '18 at 16:18



















0












$begingroup$

Let $x_m=y_m+aimplies y_0=x_0-a=1-a$



$$5=x_n-4x_{n-1}=y_n+a-4(y_{m-1}+a)=y_n-4y_{n-1}-3a$$



Set $-3a=5iff a=?$ so that $y_n=4y_{n-1}=cdots=4^ry_{n-r},0le rle n$



$r=nimplies y_n=4^ny_0=4^n(1-a)$






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    This is a difference equation and it can be solved using Z-Transform. Take Z-transform of both sides of the equation and then use the initial condition. Ultimately, you will get Z-transform of X and then take its inverse z-transform to get the solution.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The loooong road.
      $endgroup$
      – Did
      Dec 17 '18 at 0:31











    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%2f3041632%2fsolving-recurrence-x-n-4x-n-15%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    If you calculate the first few terms explicitly, you will find that the $n$th term is the sum of an exponential and a geometric series. For example,



    $$
    X_3 = 4^3 + 5(4^2+4+1).
    $$
    So in general,
    $$
    X_n = 4^n + 5sum_{k=0}^{n-1}4^k = 4^n +5frac{1-4^n}{1-4},
    $$



    which should simplify to the answer you gave.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Okay, that makes sense. But it looks like a different solution then from what is written in my textbook?
      $endgroup$
      – ponikoli
      Dec 15 '18 at 16:18
















    2












    $begingroup$

    If you calculate the first few terms explicitly, you will find that the $n$th term is the sum of an exponential and a geometric series. For example,



    $$
    X_3 = 4^3 + 5(4^2+4+1).
    $$
    So in general,
    $$
    X_n = 4^n + 5sum_{k=0}^{n-1}4^k = 4^n +5frac{1-4^n}{1-4},
    $$



    which should simplify to the answer you gave.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Okay, that makes sense. But it looks like a different solution then from what is written in my textbook?
      $endgroup$
      – ponikoli
      Dec 15 '18 at 16:18














    2












    2








    2





    $begingroup$

    If you calculate the first few terms explicitly, you will find that the $n$th term is the sum of an exponential and a geometric series. For example,



    $$
    X_3 = 4^3 + 5(4^2+4+1).
    $$
    So in general,
    $$
    X_n = 4^n + 5sum_{k=0}^{n-1}4^k = 4^n +5frac{1-4^n}{1-4},
    $$



    which should simplify to the answer you gave.






    share|cite|improve this answer









    $endgroup$



    If you calculate the first few terms explicitly, you will find that the $n$th term is the sum of an exponential and a geometric series. For example,



    $$
    X_3 = 4^3 + 5(4^2+4+1).
    $$
    So in general,
    $$
    X_n = 4^n + 5sum_{k=0}^{n-1}4^k = 4^n +5frac{1-4^n}{1-4},
    $$



    which should simplify to the answer you gave.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 15 '18 at 16:04









    MatthiasMatthias

    3287




    3287












    • $begingroup$
      Okay, that makes sense. But it looks like a different solution then from what is written in my textbook?
      $endgroup$
      – ponikoli
      Dec 15 '18 at 16:18


















    • $begingroup$
      Okay, that makes sense. But it looks like a different solution then from what is written in my textbook?
      $endgroup$
      – ponikoli
      Dec 15 '18 at 16:18
















    $begingroup$
    Okay, that makes sense. But it looks like a different solution then from what is written in my textbook?
    $endgroup$
    – ponikoli
    Dec 15 '18 at 16:18




    $begingroup$
    Okay, that makes sense. But it looks like a different solution then from what is written in my textbook?
    $endgroup$
    – ponikoli
    Dec 15 '18 at 16:18











    0












    $begingroup$

    Let $x_m=y_m+aimplies y_0=x_0-a=1-a$



    $$5=x_n-4x_{n-1}=y_n+a-4(y_{m-1}+a)=y_n-4y_{n-1}-3a$$



    Set $-3a=5iff a=?$ so that $y_n=4y_{n-1}=cdots=4^ry_{n-r},0le rle n$



    $r=nimplies y_n=4^ny_0=4^n(1-a)$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Let $x_m=y_m+aimplies y_0=x_0-a=1-a$



      $$5=x_n-4x_{n-1}=y_n+a-4(y_{m-1}+a)=y_n-4y_{n-1}-3a$$



      Set $-3a=5iff a=?$ so that $y_n=4y_{n-1}=cdots=4^ry_{n-r},0le rle n$



      $r=nimplies y_n=4^ny_0=4^n(1-a)$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $x_m=y_m+aimplies y_0=x_0-a=1-a$



        $$5=x_n-4x_{n-1}=y_n+a-4(y_{m-1}+a)=y_n-4y_{n-1}-3a$$



        Set $-3a=5iff a=?$ so that $y_n=4y_{n-1}=cdots=4^ry_{n-r},0le rle n$



        $r=nimplies y_n=4^ny_0=4^n(1-a)$






        share|cite|improve this answer









        $endgroup$



        Let $x_m=y_m+aimplies y_0=x_0-a=1-a$



        $$5=x_n-4x_{n-1}=y_n+a-4(y_{m-1}+a)=y_n-4y_{n-1}-3a$$



        Set $-3a=5iff a=?$ so that $y_n=4y_{n-1}=cdots=4^ry_{n-r},0le rle n$



        $r=nimplies y_n=4^ny_0=4^n(1-a)$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 15 '18 at 16:14









        lab bhattacharjeelab bhattacharjee

        226k15157275




        226k15157275























            -1












            $begingroup$

            This is a difference equation and it can be solved using Z-Transform. Take Z-transform of both sides of the equation and then use the initial condition. Ultimately, you will get Z-transform of X and then take its inverse z-transform to get the solution.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              The loooong road.
              $endgroup$
              – Did
              Dec 17 '18 at 0:31
















            -1












            $begingroup$

            This is a difference equation and it can be solved using Z-Transform. Take Z-transform of both sides of the equation and then use the initial condition. Ultimately, you will get Z-transform of X and then take its inverse z-transform to get the solution.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              The loooong road.
              $endgroup$
              – Did
              Dec 17 '18 at 0:31














            -1












            -1








            -1





            $begingroup$

            This is a difference equation and it can be solved using Z-Transform. Take Z-transform of both sides of the equation and then use the initial condition. Ultimately, you will get Z-transform of X and then take its inverse z-transform to get the solution.






            share|cite|improve this answer









            $endgroup$



            This is a difference equation and it can be solved using Z-Transform. Take Z-transform of both sides of the equation and then use the initial condition. Ultimately, you will get Z-transform of X and then take its inverse z-transform to get the solution.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 15 '18 at 15:53









            Asit SrivastavaAsit Srivastava

            257




            257












            • $begingroup$
              The loooong road.
              $endgroup$
              – Did
              Dec 17 '18 at 0:31


















            • $begingroup$
              The loooong road.
              $endgroup$
              – Did
              Dec 17 '18 at 0:31
















            $begingroup$
            The loooong road.
            $endgroup$
            – Did
            Dec 17 '18 at 0:31




            $begingroup$
            The loooong road.
            $endgroup$
            – Did
            Dec 17 '18 at 0:31


















            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%2f3041632%2fsolving-recurrence-x-n-4x-n-15%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

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            Puebla de Zaragoza

            Image
            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
            Read more

            Musa

            Image
            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
            Read more