Question on solutions of linear recurrences












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  1. Consider a linear recurrence of order $k$ with constant coefficients:


$$ a_n = lambda_1a_{n-1}+dots+lambda_ka_{n-k} + f(n), nge k.$$



How do I show that all solutions $a_n$ of this linear recurrence can be written as $a_n = a_n ^{(h)} + a_n^{(p)}$, with $a_n ^{(h)}$ an arbitrary solution of the corresponding homogeneous linear recurrence, and $a_n ^{(p)}$ a particular solution of the given linear recurrence.




  1. Consider this homogeneous linear recurrence of order $2$: $a_n = lambda_1 a_{n-1}+lambda_2 a_{n-2}, , nge 2$.


Suppose that the characteristic polynomial of this recurrence has one unique solution $r = lambda_1 / 2$. This means that $a_n = (frac{lambda_1}{2})^n $ is a solution of the given linear recurrence. How do I prove that $nr^n$ is also a solution of the homogeneous linear recurrence?



Thanks!






discrete-mathematics recurrence-relations







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    $begingroup$



    1. Consider a linear recurrence of order $k$ with constant coefficients:


    $$ a_n = lambda_1a_{n-1}+dots+lambda_ka_{n-k} + f(n), nge k.$$



    How do I show that all solutions $a_n$ of this linear recurrence can be written as $a_n = a_n ^{(h)} + a_n^{(p)}$, with $a_n ^{(h)}$ an arbitrary solution of the corresponding homogeneous linear recurrence, and $a_n ^{(p)}$ a particular solution of the given linear recurrence.




    1. Consider this homogeneous linear recurrence of order $2$: $a_n = lambda_1 a_{n-1}+lambda_2 a_{n-2}, , nge 2$.


    Suppose that the characteristic polynomial of this recurrence has one unique solution $r = lambda_1 / 2$. This means that $a_n = (frac{lambda_1}{2})^n $ is a solution of the given linear recurrence. How do I prove that $nr^n$ is also a solution of the homogeneous linear recurrence?



    Thanks!






    discrete-mathematics recurrence-relations







    share|cite|improve this question









    $endgroup$















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      0








      0


      0



      $begingroup$



      1. Consider a linear recurrence of order $k$ with constant coefficients:


      $$ a_n = lambda_1a_{n-1}+dots+lambda_ka_{n-k} + f(n), nge k.$$



      How do I show that all solutions $a_n$ of this linear recurrence can be written as $a_n = a_n ^{(h)} + a_n^{(p)}$, with $a_n ^{(h)}$ an arbitrary solution of the corresponding homogeneous linear recurrence, and $a_n ^{(p)}$ a particular solution of the given linear recurrence.




      1. Consider this homogeneous linear recurrence of order $2$: $a_n = lambda_1 a_{n-1}+lambda_2 a_{n-2}, , nge 2$.


      Suppose that the characteristic polynomial of this recurrence has one unique solution $r = lambda_1 / 2$. This means that $a_n = (frac{lambda_1}{2})^n $ is a solution of the given linear recurrence. How do I prove that $nr^n$ is also a solution of the homogeneous linear recurrence?



      Thanks!






      discrete-mathematics recurrence-relations







      share|cite|improve this question









      $endgroup$





      1. Consider a linear recurrence of order $k$ with constant coefficients:


      $$ a_n = lambda_1a_{n-1}+dots+lambda_ka_{n-k} + f(n), nge k.$$



      How do I show that all solutions $a_n$ of this linear recurrence can be written as $a_n = a_n ^{(h)} + a_n^{(p)}$, with $a_n ^{(h)}$ an arbitrary solution of the corresponding homogeneous linear recurrence, and $a_n ^{(p)}$ a particular solution of the given linear recurrence.




      1. Consider this homogeneous linear recurrence of order $2$: $a_n = lambda_1 a_{n-1}+lambda_2 a_{n-2}, , nge 2$.


      Suppose that the characteristic polynomial of this recurrence has one unique solution $r = lambda_1 / 2$. This means that $a_n = (frac{lambda_1}{2})^n $ is a solution of the given linear recurrence. How do I prove that $nr^n$ is also a solution of the homogeneous linear recurrence?



      Thanks!






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      discrete-mathematics recurrence-relations






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      asked Jan 26 at 14:30









      ZacharyZachary

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          1. Hint. Take two arbitrary solutions of your nonhomogeneous equation. What did you get as their difference? A solution of the homogeneous part of the equation. Now fix one particular solution of the nonhomogeneous equation and reverse the process.


          2. We have one root $r$ of multiplicity 2 of the characteristic polynomial. That means $x^2-lambda_1 x-lambda_2 = (x-r)^2 = x^2-2r x+r^2$. As you already noted, $lambda_1 = 2r$ and from above we also have $lambda_2 = -r^2$. Using this, just plug $a_k = kr^k$ for $k=n,n-1,n-2$, into your equation and you get easily the result.







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            1. Hint. Take two arbitrary solutions of your nonhomogeneous equation. What did you get as their difference? A solution of the homogeneous part of the equation. Now fix one particular solution of the nonhomogeneous equation and reverse the process.


            2. We have one root $r$ of multiplicity 2 of the characteristic polynomial. That means $x^2-lambda_1 x-lambda_2 = (x-r)^2 = x^2-2r x+r^2$. As you already noted, $lambda_1 = 2r$ and from above we also have $lambda_2 = -r^2$. Using this, just plug $a_k = kr^k$ for $k=n,n-1,n-2$, into your equation and you get easily the result.







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              $begingroup$


              1. Hint. Take two arbitrary solutions of your nonhomogeneous equation. What did you get as their difference? A solution of the homogeneous part of the equation. Now fix one particular solution of the nonhomogeneous equation and reverse the process.


              2. We have one root $r$ of multiplicity 2 of the characteristic polynomial. That means $x^2-lambda_1 x-lambda_2 = (x-r)^2 = x^2-2r x+r^2$. As you already noted, $lambda_1 = 2r$ and from above we also have $lambda_2 = -r^2$. Using this, just plug $a_k = kr^k$ for $k=n,n-1,n-2$, into your equation and you get easily the result.







              share|cite|improve this answer









              $endgroup$
















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                $begingroup$


                1. Hint. Take two arbitrary solutions of your nonhomogeneous equation. What did you get as their difference? A solution of the homogeneous part of the equation. Now fix one particular solution of the nonhomogeneous equation and reverse the process.


                2. We have one root $r$ of multiplicity 2 of the characteristic polynomial. That means $x^2-lambda_1 x-lambda_2 = (x-r)^2 = x^2-2r x+r^2$. As you already noted, $lambda_1 = 2r$ and from above we also have $lambda_2 = -r^2$. Using this, just plug $a_k = kr^k$ for $k=n,n-1,n-2$, into your equation and you get easily the result.







                share|cite|improve this answer









                $endgroup$




                1. Hint. Take two arbitrary solutions of your nonhomogeneous equation. What did you get as their difference? A solution of the homogeneous part of the equation. Now fix one particular solution of the nonhomogeneous equation and reverse the process.


                2. We have one root $r$ of multiplicity 2 of the characteristic polynomial. That means $x^2-lambda_1 x-lambda_2 = (x-r)^2 = x^2-2r x+r^2$. As you already noted, $lambda_1 = 2r$ and from above we also have $lambda_2 = -r^2$. Using this, just plug $a_k = kr^k$ for $k=n,n-1,n-2$, into your equation and you get easily the result.








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                answered Jan 26 at 15:18









                Roman HricRoman Hric

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