How to find limit of a second-order recursion












0












$begingroup$


Suppose there is a sequence



$$u_n = au_{n-1} - a^2u_{n-1}^2 + bu_{n-2} - b^2u_{n-2}^2$$



with the boundary condition, $u_0, u_1$ both are positive and less than $1$.



How can I show that this sequence is convergent?



Actually, my goal is to show that $u_n$ converges to a value greater than $0$.



Possibly, when $a + b > 1$ then $limlimits_{nrightarrow infty} u_n neq 0$ as I experiment with a computer progam.






calculus recursion







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$endgroup$








  • 1




    $begingroup$
    Isn't this just $u_n=cu_{n-1}+du_{n-2}$ with $c=a-a^2,,d=b-b^2$? Study the roots of the auxiliary polynomial $lambda^2+(a^2-a)lambda+b^2-b$.
    $endgroup$
    – J.G.
    Dec 24 '18 at 8:42










  • $begingroup$
    I am sorry for this confusion. There is a typo here. There are squares of $u_{n-1}$ and $u_{n-2}$. Thanks.
    $endgroup$
    – user2843539
    Dec 24 '18 at 8:46


















0












$begingroup$


Suppose there is a sequence



$$u_n = au_{n-1} - a^2u_{n-1}^2 + bu_{n-2} - b^2u_{n-2}^2$$



with the boundary condition, $u_0, u_1$ both are positive and less than $1$.



How can I show that this sequence is convergent?



Actually, my goal is to show that $u_n$ converges to a value greater than $0$.



Possibly, when $a + b > 1$ then $limlimits_{nrightarrow infty} u_n neq 0$ as I experiment with a computer progam.






calculus recursion







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Isn't this just $u_n=cu_{n-1}+du_{n-2}$ with $c=a-a^2,,d=b-b^2$? Study the roots of the auxiliary polynomial $lambda^2+(a^2-a)lambda+b^2-b$.
    $endgroup$
    – J.G.
    Dec 24 '18 at 8:42










  • $begingroup$
    I am sorry for this confusion. There is a typo here. There are squares of $u_{n-1}$ and $u_{n-2}$. Thanks.
    $endgroup$
    – user2843539
    Dec 24 '18 at 8:46
















0












0








0


1



$begingroup$


Suppose there is a sequence



$$u_n = au_{n-1} - a^2u_{n-1}^2 + bu_{n-2} - b^2u_{n-2}^2$$



with the boundary condition, $u_0, u_1$ both are positive and less than $1$.



How can I show that this sequence is convergent?



Actually, my goal is to show that $u_n$ converges to a value greater than $0$.



Possibly, when $a + b > 1$ then $limlimits_{nrightarrow infty} u_n neq 0$ as I experiment with a computer progam.






calculus recursion







share|cite|improve this question











$endgroup$




Suppose there is a sequence



$$u_n = au_{n-1} - a^2u_{n-1}^2 + bu_{n-2} - b^2u_{n-2}^2$$



with the boundary condition, $u_0, u_1$ both are positive and less than $1$.



How can I show that this sequence is convergent?



Actually, my goal is to show that $u_n$ converges to a value greater than $0$.



Possibly, when $a + b > 1$ then $limlimits_{nrightarrow infty} u_n neq 0$ as I experiment with a computer progam.






calculus recursion




calculus recursion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 10:12









J.G.

32.6k23250




32.6k23250










asked Dec 24 '18 at 8:36









user2843539user2843539

64




64








  • 1




    $begingroup$
    Isn't this just $u_n=cu_{n-1}+du_{n-2}$ with $c=a-a^2,,d=b-b^2$? Study the roots of the auxiliary polynomial $lambda^2+(a^2-a)lambda+b^2-b$.
    $endgroup$
    – J.G.
    Dec 24 '18 at 8:42










  • $begingroup$
    I am sorry for this confusion. There is a typo here. There are squares of $u_{n-1}$ and $u_{n-2}$. Thanks.
    $endgroup$
    – user2843539
    Dec 24 '18 at 8:46
















  • 1




    $begingroup$
    Isn't this just $u_n=cu_{n-1}+du_{n-2}$ with $c=a-a^2,,d=b-b^2$? Study the roots of the auxiliary polynomial $lambda^2+(a^2-a)lambda+b^2-b$.
    $endgroup$
    – J.G.
    Dec 24 '18 at 8:42










  • $begingroup$
    I am sorry for this confusion. There is a typo here. There are squares of $u_{n-1}$ and $u_{n-2}$. Thanks.
    $endgroup$
    – user2843539
    Dec 24 '18 at 8:46










1




1




$begingroup$
Isn't this just $u_n=cu_{n-1}+du_{n-2}$ with $c=a-a^2,,d=b-b^2$? Study the roots of the auxiliary polynomial $lambda^2+(a^2-a)lambda+b^2-b$.
$endgroup$
– J.G.
Dec 24 '18 at 8:42




$begingroup$
Isn't this just $u_n=cu_{n-1}+du_{n-2}$ with $c=a-a^2,,d=b-b^2$? Study the roots of the auxiliary polynomial $lambda^2+(a^2-a)lambda+b^2-b$.
$endgroup$
– J.G.
Dec 24 '18 at 8:42












$begingroup$
I am sorry for this confusion. There is a typo here. There are squares of $u_{n-1}$ and $u_{n-2}$. Thanks.
$endgroup$
– user2843539
Dec 24 '18 at 8:46






$begingroup$
I am sorry for this confusion. There is a typo here. There are squares of $u_{n-1}$ and $u_{n-2}$. Thanks.
$endgroup$
– user2843539
Dec 24 '18 at 8:46












2 Answers
2






active

oldest

votes


















0












$begingroup$

This is a long comment, or possibly half an answer.



If the sequence has a limit $L$ then $$L=aL-a^2L^2+bL-b^2L^2tag{1},$$which can be rearranged viz. $0=L(a+b-1-(a^2+b^2)L)$. Therefore $L$ is nonexistent, $L=0$ or $$L=frac{a+b-1}{a^2+b^2}tag{2},$$the latter being the only case in which $u_n$ converges to a non-zero value. Indeed, in this case that value is positive iff $a+b>1$, as you've conjectured.



The hard part is checking whether the sequence has a limit. Define $$epsilon_n:=u_n-L=a(epsilon_{n-1}+L)-a^2(epsilon_{n-1}+L)^2+b(epsilon_{n-2}+L)-b^2(epsilon_{n-2}+L)^2-L.$$This simplifies for a root of (1) to $$epsilon_n=a(1-2aL)epsilon_{n-1}-a^2epsilon_{n-1}^2+b(1-2bL)epsilon_{n-2}-b^2epsilon_{n-2}^2tag{3}.$$The real question is whether (2) and (3) imply $epsilon_nto 0$. I suspect that can be proved from your boundary conditions (possibly with some further requirement on $a,,b$), but I'm not sure how. But if any proof strategy works, it'll be that one.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for a detailed suggestion. I will try to work with (3).
    $endgroup$
    – user2843539
    Dec 24 '18 at 14:00



















0












$begingroup$

If there is a limit u, then
$u = au - a^2u^2 + bu - b^2u^2.$



Requiring u to be nonzero gives
$1 = a - a^2u + b - b^2u.$



Whereupon $u = (a + b - 1)/(a^2 + b^2).$

Whence, if 1 < a + b and u is a nonzero limit, then 0 < u.






share|cite|improve this answer









$endgroup$














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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

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    active

    oldest

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    0












    $begingroup$

    This is a long comment, or possibly half an answer.



    If the sequence has a limit $L$ then $$L=aL-a^2L^2+bL-b^2L^2tag{1},$$which can be rearranged viz. $0=L(a+b-1-(a^2+b^2)L)$. Therefore $L$ is nonexistent, $L=0$ or $$L=frac{a+b-1}{a^2+b^2}tag{2},$$the latter being the only case in which $u_n$ converges to a non-zero value. Indeed, in this case that value is positive iff $a+b>1$, as you've conjectured.



    The hard part is checking whether the sequence has a limit. Define $$epsilon_n:=u_n-L=a(epsilon_{n-1}+L)-a^2(epsilon_{n-1}+L)^2+b(epsilon_{n-2}+L)-b^2(epsilon_{n-2}+L)^2-L.$$This simplifies for a root of (1) to $$epsilon_n=a(1-2aL)epsilon_{n-1}-a^2epsilon_{n-1}^2+b(1-2bL)epsilon_{n-2}-b^2epsilon_{n-2}^2tag{3}.$$The real question is whether (2) and (3) imply $epsilon_nto 0$. I suspect that can be proved from your boundary conditions (possibly with some further requirement on $a,,b$), but I'm not sure how. But if any proof strategy works, it'll be that one.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for a detailed suggestion. I will try to work with (3).
      $endgroup$
      – user2843539
      Dec 24 '18 at 14:00
















    0












    $begingroup$

    This is a long comment, or possibly half an answer.



    If the sequence has a limit $L$ then $$L=aL-a^2L^2+bL-b^2L^2tag{1},$$which can be rearranged viz. $0=L(a+b-1-(a^2+b^2)L)$. Therefore $L$ is nonexistent, $L=0$ or $$L=frac{a+b-1}{a^2+b^2}tag{2},$$the latter being the only case in which $u_n$ converges to a non-zero value. Indeed, in this case that value is positive iff $a+b>1$, as you've conjectured.



    The hard part is checking whether the sequence has a limit. Define $$epsilon_n:=u_n-L=a(epsilon_{n-1}+L)-a^2(epsilon_{n-1}+L)^2+b(epsilon_{n-2}+L)-b^2(epsilon_{n-2}+L)^2-L.$$This simplifies for a root of (1) to $$epsilon_n=a(1-2aL)epsilon_{n-1}-a^2epsilon_{n-1}^2+b(1-2bL)epsilon_{n-2}-b^2epsilon_{n-2}^2tag{3}.$$The real question is whether (2) and (3) imply $epsilon_nto 0$. I suspect that can be proved from your boundary conditions (possibly with some further requirement on $a,,b$), but I'm not sure how. But if any proof strategy works, it'll be that one.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for a detailed suggestion. I will try to work with (3).
      $endgroup$
      – user2843539
      Dec 24 '18 at 14:00














    0












    0








    0





    $begingroup$

    This is a long comment, or possibly half an answer.



    If the sequence has a limit $L$ then $$L=aL-a^2L^2+bL-b^2L^2tag{1},$$which can be rearranged viz. $0=L(a+b-1-(a^2+b^2)L)$. Therefore $L$ is nonexistent, $L=0$ or $$L=frac{a+b-1}{a^2+b^2}tag{2},$$the latter being the only case in which $u_n$ converges to a non-zero value. Indeed, in this case that value is positive iff $a+b>1$, as you've conjectured.



    The hard part is checking whether the sequence has a limit. Define $$epsilon_n:=u_n-L=a(epsilon_{n-1}+L)-a^2(epsilon_{n-1}+L)^2+b(epsilon_{n-2}+L)-b^2(epsilon_{n-2}+L)^2-L.$$This simplifies for a root of (1) to $$epsilon_n=a(1-2aL)epsilon_{n-1}-a^2epsilon_{n-1}^2+b(1-2bL)epsilon_{n-2}-b^2epsilon_{n-2}^2tag{3}.$$The real question is whether (2) and (3) imply $epsilon_nto 0$. I suspect that can be proved from your boundary conditions (possibly with some further requirement on $a,,b$), but I'm not sure how. But if any proof strategy works, it'll be that one.






    share|cite|improve this answer









    $endgroup$



    This is a long comment, or possibly half an answer.



    If the sequence has a limit $L$ then $$L=aL-a^2L^2+bL-b^2L^2tag{1},$$which can be rearranged viz. $0=L(a+b-1-(a^2+b^2)L)$. Therefore $L$ is nonexistent, $L=0$ or $$L=frac{a+b-1}{a^2+b^2}tag{2},$$the latter being the only case in which $u_n$ converges to a non-zero value. Indeed, in this case that value is positive iff $a+b>1$, as you've conjectured.



    The hard part is checking whether the sequence has a limit. Define $$epsilon_n:=u_n-L=a(epsilon_{n-1}+L)-a^2(epsilon_{n-1}+L)^2+b(epsilon_{n-2}+L)-b^2(epsilon_{n-2}+L)^2-L.$$This simplifies for a root of (1) to $$epsilon_n=a(1-2aL)epsilon_{n-1}-a^2epsilon_{n-1}^2+b(1-2bL)epsilon_{n-2}-b^2epsilon_{n-2}^2tag{3}.$$The real question is whether (2) and (3) imply $epsilon_nto 0$. I suspect that can be proved from your boundary conditions (possibly with some further requirement on $a,,b$), but I'm not sure how. But if any proof strategy works, it'll be that one.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 24 '18 at 10:11









    J.G.J.G.

    32.6k23250




    32.6k23250












    • $begingroup$
      Thank you for a detailed suggestion. I will try to work with (3).
      $endgroup$
      – user2843539
      Dec 24 '18 at 14:00


















    • $begingroup$
      Thank you for a detailed suggestion. I will try to work with (3).
      $endgroup$
      – user2843539
      Dec 24 '18 at 14:00
















    $begingroup$
    Thank you for a detailed suggestion. I will try to work with (3).
    $endgroup$
    – user2843539
    Dec 24 '18 at 14:00




    $begingroup$
    Thank you for a detailed suggestion. I will try to work with (3).
    $endgroup$
    – user2843539
    Dec 24 '18 at 14:00











    0












    $begingroup$

    If there is a limit u, then
    $u = au - a^2u^2 + bu - b^2u^2.$



    Requiring u to be nonzero gives
    $1 = a - a^2u + b - b^2u.$



    Whereupon $u = (a + b - 1)/(a^2 + b^2).$

    Whence, if 1 < a + b and u is a nonzero limit, then 0 < u.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      If there is a limit u, then
      $u = au - a^2u^2 + bu - b^2u^2.$



      Requiring u to be nonzero gives
      $1 = a - a^2u + b - b^2u.$



      Whereupon $u = (a + b - 1)/(a^2 + b^2).$

      Whence, if 1 < a + b and u is a nonzero limit, then 0 < u.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        If there is a limit u, then
        $u = au - a^2u^2 + bu - b^2u^2.$



        Requiring u to be nonzero gives
        $1 = a - a^2u + b - b^2u.$



        Whereupon $u = (a + b - 1)/(a^2 + b^2).$

        Whence, if 1 < a + b and u is a nonzero limit, then 0 < u.






        share|cite|improve this answer









        $endgroup$



        If there is a limit u, then
        $u = au - a^2u^2 + bu - b^2u^2.$



        Requiring u to be nonzero gives
        $1 = a - a^2u + b - b^2u.$



        Whereupon $u = (a + b - 1)/(a^2 + b^2).$

        Whence, if 1 < a + b and u is a nonzero limit, then 0 < u.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 24 '18 at 10:10









        William ElliotWilliam Elliot

        8,9562820




        8,9562820






























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