Find a formula for the sequence $a_n$ if $a_1=1$ and $a_n+1=fraca_n^2+3a_n+1$ for $nge 1$

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It is easy to know that the sequence converges and the limit of it is $3$. However, to find a formula for this sequence is not easy. I hope someone could help me to solve this. Thanks!




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    Could you add some actual attempt besides it is not easy? Additionally, why do you need an explicit formula for $a_n$? Once you prove that $xmapsto fracx^2+3x+1$ is a contraction over a suitable space, the limit has to be $3$ by the Banach fixed point theorem.
    – Jack D'Aurizio♦
    11 hours ago










  • You may also check that the given iteration is Newton's method applied to $(x-3)^4 e^x$.
    – Jack D'Aurizio♦
    11 hours ago











  • Actually, all my attempts were in vain. What makes the problem difficult is the recursive formula only has one eigenvalue. For example, if we change the coefficient of $a_n^2$ to 2，we can find two eigenvalues a, b. Then we can show that $fraca_n+1^2-aa_n+1^2-b=(fraca_n^2-aa_n^2-b)^2$ holds. That was all I knew. Anyway, the Newton’s method you mentioned is interesting and I’ll look at it. @Jack D’Aurizio
    – Zhenyuan Lu
    11 hours ago











  • @JackD'Aurizio How did you come up with $(x-3)^4exp(x)$?
    – Batominovski
    11 hours ago






  • 1




    @Batominovski: by solving the DE $$ x-fracf(x)f'(x)=fracx^2+3x+1$$
    – Jack D'Aurizio♦
    11 hours ago















up vote
0
down vote

favorite
2












It is easy to know that the sequence converges and the limit of it is $3$. However, to find a formula for this sequence is not easy. I hope someone could help me to solve this. Thanks!




sequences-and-series




share|cite|improve this question

















  • 2




    Could you add some actual attempt besides it is not easy? Additionally, why do you need an explicit formula for $a_n$? Once you prove that $xmapsto fracx^2+3x+1$ is a contraction over a suitable space, the limit has to be $3$ by the Banach fixed point theorem.
    – Jack D'Aurizio♦
    11 hours ago










  • You may also check that the given iteration is Newton's method applied to $(x-3)^4 e^x$.
    – Jack D'Aurizio♦
    11 hours ago











  • Actually, all my attempts were in vain. What makes the problem difficult is the recursive formula only has one eigenvalue. For example, if we change the coefficient of $a_n^2$ to 2，we can find two eigenvalues a, b. Then we can show that $fraca_n+1^2-aa_n+1^2-b=(fraca_n^2-aa_n^2-b)^2$ holds. That was all I knew. Anyway, the Newton’s method you mentioned is interesting and I’ll look at it. @Jack D’Aurizio
    – Zhenyuan Lu
    11 hours ago











  • @JackD'Aurizio How did you come up with $(x-3)^4exp(x)$?
    – Batominovski
    11 hours ago






  • 1




    @Batominovski: by solving the DE $$ x-fracf(x)f'(x)=fracx^2+3x+1$$
    – Jack D'Aurizio♦
    11 hours ago













up vote
0
down vote

favorite
2









up vote
0
down vote

favorite
2






2





It is easy to know that the sequence converges and the limit of it is $3$. However, to find a formula for this sequence is not easy. I hope someone could help me to solve this. Thanks!




sequences-and-series




share|cite|improve this question













It is easy to know that the sequence converges and the limit of it is $3$. However, to find a formula for this sequence is not easy. I hope someone could help me to solve this. Thanks!





sequences-and-series





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 11 hours ago









J.G.

12.8k11323




12.8k11323









asked 12 hours ago









Zhenyuan Lu

843




843







  • 2




    Could you add some actual attempt besides it is not easy? Additionally, why do you need an explicit formula for $a_n$? Once you prove that $xmapsto fracx^2+3x+1$ is a contraction over a suitable space, the limit has to be $3$ by the Banach fixed point theorem.
    – Jack D'Aurizio♦
    11 hours ago










  • You may also check that the given iteration is Newton's method applied to $(x-3)^4 e^x$.
    – Jack D'Aurizio♦
    11 hours ago











  • Actually, all my attempts were in vain. What makes the problem difficult is the recursive formula only has one eigenvalue. For example, if we change the coefficient of $a_n^2$ to 2，we can find two eigenvalues a, b. Then we can show that $fraca_n+1^2-aa_n+1^2-b=(fraca_n^2-aa_n^2-b)^2$ holds. That was all I knew. Anyway, the Newton’s method you mentioned is interesting and I’ll look at it. @Jack D’Aurizio
    – Zhenyuan Lu
    11 hours ago











  • @JackD'Aurizio How did you come up with $(x-3)^4exp(x)$?
    – Batominovski
    11 hours ago






  • 1




    @Batominovski: by solving the DE $$ x-fracf(x)f'(x)=fracx^2+3x+1$$
    – Jack D'Aurizio♦
    11 hours ago













  • 2




    Could you add some actual attempt besides it is not easy? Additionally, why do you need an explicit formula for $a_n$? Once you prove that $xmapsto fracx^2+3x+1$ is a contraction over a suitable space, the limit has to be $3$ by the Banach fixed point theorem.
    – Jack D'Aurizio♦
    11 hours ago










  • You may also check that the given iteration is Newton's method applied to $(x-3)^4 e^x$.
    – Jack D'Aurizio♦
    11 hours ago











  • Actually, all my attempts were in vain. What makes the problem difficult is the recursive formula only has one eigenvalue. For example, if we change the coefficient of $a_n^2$ to 2，we can find two eigenvalues a, b. Then we can show that $fraca_n+1^2-aa_n+1^2-b=(fraca_n^2-aa_n^2-b)^2$ holds. That was all I knew. Anyway, the Newton’s method you mentioned is interesting and I’ll look at it. @Jack D’Aurizio
    – Zhenyuan Lu
    11 hours ago











  • @JackD'Aurizio How did you come up with $(x-3)^4exp(x)$?
    – Batominovski
    11 hours ago






  • 1




    @Batominovski: by solving the DE $$ x-fracf(x)f'(x)=fracx^2+3x+1$$
    – Jack D'Aurizio♦
    11 hours ago








2




2




Could you add some actual attempt besides it is not easy? Additionally, why do you need an explicit formula for $a_n$? Once you prove that $xmapsto fracx^2+3x+1$ is a contraction over a suitable space, the limit has to be $3$ by the Banach fixed point theorem.
– Jack D'Aurizio♦
11 hours ago




Could you add some actual attempt besides it is not easy? Additionally, why do you need an explicit formula for $a_n$? Once you prove that $xmapsto fracx^2+3x+1$ is a contraction over a suitable space, the limit has to be $3$ by the Banach fixed point theorem.
– Jack D'Aurizio♦
11 hours ago












You may also check that the given iteration is Newton's method applied to $(x-3)^4 e^x$.
– Jack D'Aurizio♦
11 hours ago





You may also check that the given iteration is Newton's method applied to $(x-3)^4 e^x$.
– Jack D'Aurizio♦
11 hours ago













Actually, all my attempts were in vain. What makes the problem difficult is the recursive formula only has one eigenvalue. For example, if we change the coefficient of $a_n^2$ to 2，we can find two eigenvalues a, b. Then we can show that $fraca_n+1^2-aa_n+1^2-b=(fraca_n^2-aa_n^2-b)^2$ holds. That was all I knew. Anyway, the Newton’s method you mentioned is interesting and I’ll look at it. @Jack D’Aurizio
– Zhenyuan Lu
11 hours ago





Actually, all my attempts were in vain. What makes the problem difficult is the recursive formula only has one eigenvalue. For example, if we change the coefficient of $a_n^2$ to 2，we can find two eigenvalues a, b. Then we can show that $fraca_n+1^2-aa_n+1^2-b=(fraca_n^2-aa_n^2-b)^2$ holds. That was all I knew. Anyway, the Newton’s method you mentioned is interesting and I’ll look at it. @Jack D’Aurizio
– Zhenyuan Lu
11 hours ago













@JackD'Aurizio How did you come up with $(x-3)^4exp(x)$?
– Batominovski
11 hours ago




@JackD'Aurizio How did you come up with $(x-3)^4exp(x)$?
– Batominovski
11 hours ago




1




1




@Batominovski: by solving the DE $$ x-fracf(x)f'(x)=fracx^2+3x+1$$
– Jack D'Aurizio♦
11 hours ago





@Batominovski: by solving the DE $$ x-fracf(x)f'(x)=fracx^2+3x+1$$
– Jack D'Aurizio♦
11 hours ago











2 Answers
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Note that
$$a_n+1-a_n=frac3-a_na_n+1text for every n=1,2,3,ldots,.$$
If $-1<a_n<3$, then we see that $a_n+1>a_n$. Furthermore, from
$$a_n+1-3=fraca_na_n+1,left(a_n-3right)text for every n=1,2,3,ldots,,$$
we deduce that, if $0<a_n<3$, then $0<a_n+1<3$ as well. Since $a_1=1in(0,3)$, we conclude that the sequence $left(a_nright)_ninmathbbZ_>0$ is an increasing sequence bounded above by $3$. The rest is easy.






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    up vote
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    down vote













    Define the function $ f(x) := (x^3 + 3)/(x + 1). $ Your iteration becomes
    $ a_n+1 = f(a_n). $ Define $ g(x) := (8x + 3)/(2x + 1) $ and
    $ t(x) := (3/4 x + 2 x^2)/(1 + 3 x + x^2). $ The algebraic identity
    $ g(t(x)) = f(g(x)) $ shows that $ f(x) $ is conjugate to $ t(x). $
    Define by power series the function $ u(x) := 3(x + 4x^2 + 128x^3/7 + 19520x^4/259 + 15516416x^5/45325 + O(x^6)). $ The equation
    $ t(u(x)) = u(3/4 x) $ shows that
    $ t(x) $ is conjugate to $ xmapsto 3/4x. $
    Putting it all together we get that
    $ f^(n)(g(u(x))) = g(u((3/4)^n x)) $ where $ f^(n) $ indicate $n$th composition of the function $f$. Now let $ x_1 approx -0.162 $ such that
    $ g(u(x_1) = a_1 = 1. $ Then $ a_n = g(u((3/4)^n x_1)). $ This is a formula for the general term of your sequence. I don't think there is a closed form for $ u(x). $






    share|cite|improve this answer





















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      2 Answers
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      Note that
      $$a_n+1-a_n=frac3-a_na_n+1text for every n=1,2,3,ldots,.$$
      If $-1<a_n<3$, then we see that $a_n+1>a_n$. Furthermore, from
      $$a_n+1-3=fraca_na_n+1,left(a_n-3right)text for every n=1,2,3,ldots,,$$
      we deduce that, if $0<a_n<3$, then $0<a_n+1<3$ as well. Since $a_1=1in(0,3)$, we conclude that the sequence $left(a_nright)_ninmathbbZ_>0$ is an increasing sequence bounded above by $3$. The rest is easy.






      share|cite|improve this answer

























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        down vote













        Note that
        $$a_n+1-a_n=frac3-a_na_n+1text for every n=1,2,3,ldots,.$$
        If $-1<a_n<3$, then we see that $a_n+1>a_n$. Furthermore, from
        $$a_n+1-3=fraca_na_n+1,left(a_n-3right)text for every n=1,2,3,ldots,,$$
        we deduce that, if $0<a_n<3$, then $0<a_n+1<3$ as well. Since $a_1=1in(0,3)$, we conclude that the sequence $left(a_nright)_ninmathbbZ_>0$ is an increasing sequence bounded above by $3$. The rest is easy.






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          Note that
          $$a_n+1-a_n=frac3-a_na_n+1text for every n=1,2,3,ldots,.$$
          If $-1<a_n<3$, then we see that $a_n+1>a_n$. Furthermore, from
          $$a_n+1-3=fraca_na_n+1,left(a_n-3right)text for every n=1,2,3,ldots,,$$
          we deduce that, if $0<a_n<3$, then $0<a_n+1<3$ as well. Since $a_1=1in(0,3)$, we conclude that the sequence $left(a_nright)_ninmathbbZ_>0$ is an increasing sequence bounded above by $3$. The rest is easy.






          share|cite|improve this answer













          Note that
          $$a_n+1-a_n=frac3-a_na_n+1text for every n=1,2,3,ldots,.$$
          If $-1<a_n<3$, then we see that $a_n+1>a_n$. Furthermore, from
          $$a_n+1-3=fraca_na_n+1,left(a_n-3right)text for every n=1,2,3,ldots,,$$
          we deduce that, if $0<a_n<3$, then $0<a_n+1<3$ as well. Since $a_1=1in(0,3)$, we conclude that the sequence $left(a_nright)_ninmathbbZ_>0$ is an increasing sequence bounded above by $3$. The rest is easy.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered 11 hours ago









          Batominovski

          22.2k22675




          22.2k22675




















              up vote
              0
              down vote













              Define the function $ f(x) := (x^3 + 3)/(x + 1). $ Your iteration becomes
              $ a_n+1 = f(a_n). $ Define $ g(x) := (8x + 3)/(2x + 1) $ and
              $ t(x) := (3/4 x + 2 x^2)/(1 + 3 x + x^2). $ The algebraic identity
              $ g(t(x)) = f(g(x)) $ shows that $ f(x) $ is conjugate to $ t(x). $
              Define by power series the function $ u(x) := 3(x + 4x^2 + 128x^3/7 + 19520x^4/259 + 15516416x^5/45325 + O(x^6)). $ The equation
              $ t(u(x)) = u(3/4 x) $ shows that
              $ t(x) $ is conjugate to $ xmapsto 3/4x. $
              Putting it all together we get that
              $ f^(n)(g(u(x))) = g(u((3/4)^n x)) $ where $ f^(n) $ indicate $n$th composition of the function $f$. Now let $ x_1 approx -0.162 $ such that
              $ g(u(x_1) = a_1 = 1. $ Then $ a_n = g(u((3/4)^n x_1)). $ This is a formula for the general term of your sequence. I don't think there is a closed form for $ u(x). $






              share|cite|improve this answer

























                up vote
                0
                down vote













                Define the function $ f(x) := (x^3 + 3)/(x + 1). $ Your iteration becomes
                $ a_n+1 = f(a_n). $ Define $ g(x) := (8x + 3)/(2x + 1) $ and
                $ t(x) := (3/4 x + 2 x^2)/(1 + 3 x + x^2). $ The algebraic identity
                $ g(t(x)) = f(g(x)) $ shows that $ f(x) $ is conjugate to $ t(x). $
                Define by power series the function $ u(x) := 3(x + 4x^2 + 128x^3/7 + 19520x^4/259 + 15516416x^5/45325 + O(x^6)). $ The equation
                $ t(u(x)) = u(3/4 x) $ shows that
                $ t(x) $ is conjugate to $ xmapsto 3/4x. $
                Putting it all together we get that
                $ f^(n)(g(u(x))) = g(u((3/4)^n x)) $ where $ f^(n) $ indicate $n$th composition of the function $f$. Now let $ x_1 approx -0.162 $ such that
                $ g(u(x_1) = a_1 = 1. $ Then $ a_n = g(u((3/4)^n x_1)). $ This is a formula for the general term of your sequence. I don't think there is a closed form for $ u(x). $






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Define the function $ f(x) := (x^3 + 3)/(x + 1). $ Your iteration becomes
                  $ a_n+1 = f(a_n). $ Define $ g(x) := (8x + 3)/(2x + 1) $ and
                  $ t(x) := (3/4 x + 2 x^2)/(1 + 3 x + x^2). $ The algebraic identity
                  $ g(t(x)) = f(g(x)) $ shows that $ f(x) $ is conjugate to $ t(x). $
                  Define by power series the function $ u(x) := 3(x + 4x^2 + 128x^3/7 + 19520x^4/259 + 15516416x^5/45325 + O(x^6)). $ The equation
                  $ t(u(x)) = u(3/4 x) $ shows that
                  $ t(x) $ is conjugate to $ xmapsto 3/4x. $
                  Putting it all together we get that
                  $ f^(n)(g(u(x))) = g(u((3/4)^n x)) $ where $ f^(n) $ indicate $n$th composition of the function $f$. Now let $ x_1 approx -0.162 $ such that
                  $ g(u(x_1) = a_1 = 1. $ Then $ a_n = g(u((3/4)^n x_1)). $ This is a formula for the general term of your sequence. I don't think there is a closed form for $ u(x). $






                  share|cite|improve this answer













                  Define the function $ f(x) := (x^3 + 3)/(x + 1). $ Your iteration becomes
                  $ a_n+1 = f(a_n). $ Define $ g(x) := (8x + 3)/(2x + 1) $ and
                  $ t(x) := (3/4 x + 2 x^2)/(1 + 3 x + x^2). $ The algebraic identity
                  $ g(t(x)) = f(g(x)) $ shows that $ f(x) $ is conjugate to $ t(x). $
                  Define by power series the function $ u(x) := 3(x + 4x^2 + 128x^3/7 + 19520x^4/259 + 15516416x^5/45325 + O(x^6)). $ The equation
                  $ t(u(x)) = u(3/4 x) $ shows that
                  $ t(x) $ is conjugate to $ xmapsto 3/4x. $
                  Putting it all together we get that
                  $ f^(n)(g(u(x))) = g(u((3/4)^n x)) $ where $ f^(n) $ indicate $n$th composition of the function $f$. Now let $ x_1 approx -0.162 $ such that
                  $ g(u(x_1) = a_1 = 1. $ Then $ a_n = g(u((3/4)^n x_1)). $ This is a formula for the general term of your sequence. I don't think there is a closed form for $ u(x). $







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered 9 hours ago









                  Somos

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