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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!
sequences-and-series
edited 11 hours ago
J.G.
12.8k11323
asked 12 hours ago
Zhenyuan Lu
843
 |Â
show 2 more comments
up vote
0
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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!
sequences-and-series
edited 11 hours ago
J.G.
12.8k11323
asked 12 hours ago
Zhenyuan Lu
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
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
edited 11 hours ago
J.G.
12.8k11323
asked 12 hours ago
Zhenyuan Lu
843
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
edited 11 hours ago
J.G.
12.8k11323
asked 12 hours ago
Zhenyuan Lu
843
edited 11 hours ago
J.G.
12.8k11323
edited 11 hours ago
J.G.
12.8k11323
edited 11 hours ago
J.G.
12.8k11323
12.8k11323
asked 12 hours ago
Zhenyuan Lu
843
asked 12 hours ago
Zhenyuan Lu
843
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
 |Â
show 2 more comments
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
 |Â
show 2 more comments
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.
answered 11 hours ago
Batominovski
22.2k22675
add a comment |Â
up vote
0
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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). $
answered 9 hours ago
Somos
10.7k1830
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered 11 hours ago
Batominovski
22.2k22675
add a comment |Â
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.
answered 11 hours ago
Batominovski
22.2k22675
add a comment |Â
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.
answered 11 hours ago
Batominovski
22.2k22675
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.
answered 11 hours ago
Batominovski
22.2k22675
answered 11 hours ago
Batominovski
22.2k22675
answered 11 hours ago
Batominovski
22.2k22675
answered 11 hours ago
Batominovski
22.2k22675
22.2k22675
add a comment |Â
add a comment |Â
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). $
answered 9 hours ago
Somos
10.7k1830
add a comment |Â
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). $
answered 9 hours ago
Somos
10.7k1830
add a comment |Â
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). $
answered 9 hours ago
Somos
10.7k1830
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). $
answered 9 hours ago
Somos
10.7k1830
answered 9 hours ago
Somos
10.7k1830
answered 9 hours ago
Somos
10.7k1830
answered 9 hours ago
Somos
10.7k1830
10.7k1830
add a comment |Â
add a comment |Â
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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