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







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$endgroup$












  • $begingroup$
    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





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    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

    2287




    2287












    • $begingroup$
      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


















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