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Non linear recurrence

Limit for a Recurrence RelationSolve non-homogeneous linear recurrenceFind the linear recurrence of degree at most 2 for the following sequenceSolving a Linear Recurrence RelationSolve the Recurrence relation : $a_n= (a_n-1)^3a_n-2$ , $a_0=1, a_1=3$Recurrence proof problemSolve the recurrence $a_n=3a_n/3+2$ given $a_0=1$ and $n$ is a power of $3$Check solution of recurrence relation $a_0 = 3$, $a_1 = 7$, $a_n = 3a_n-1 - 2a_n-2$ for $n geq 2$Non-homogenous linear recurrence relation reasonable TRIAL solution?Linear Nonhomogeneous Recurrence Relations

0

$begingroup$

Let

$a_n+1=3a_n^2(1-a_n)$

Where $a_0$ is any complex number

For $f(x)=3x^2(1-x)$ I found fixed point $0,frac12(1pm fracisqrt3)$

Setting $a_0$ equal to one fixed point we get $a_n$ constant for all $n$.

Is it possible to find a formula for $a_n$ ?

recurrence-relations

share|cite|improve this question

edited Mar 13 at 20:12

Kamoulox

asked Mar 13 at 19:09

KamouloxKamoulox

185

$endgroup$

add a comment | 

0

$begingroup$

Let

$a_n+1=3a_n^2(1-a_n)$

Where $a_0$ is any complex number

For $f(x)=3x^2(1-x)$ I found fixed point $0,frac12(1pm fracisqrt3)$

Setting $a_0$ equal to one fixed point we get $a_n$ constant for all $n$.

Is it possible to find a formula for $a_n$ ?

recurrence-relations

share|cite|improve this question

edited Mar 13 at 20:12

Kamoulox

asked Mar 13 at 19:09

KamouloxKamoulox

185

$endgroup$

add a comment | 

$begingroup$

Let

$a_n+1=3a_n^2(1-a_n)$

Where $a_0$ is any complex number

For $f(x)=3x^2(1-x)$ I found fixed point $0,frac12(1pm fracisqrt3)$

Setting $a_0$ equal to one fixed point we get $a_n$ constant for all $n$.

Is it possible to find a formula for $a_n$ ?

recurrence-relations

share|cite|improve this question

edited Mar 13 at 20:12

Kamoulox

asked Mar 13 at 19:09

KamouloxKamoulox

185

$endgroup$

share|cite|improve this question

edited Mar 13 at 20:12

Kamoulox

asked Mar 13 at 19:09

KamouloxKamoulox

185

share|cite|improve this question

edited Mar 13 at 20:12

Kamoulox

asked Mar 13 at 19:09

KamouloxKamoulox

185

asked Mar 13 at 19:09

KamouloxKamoulox

185

asked Mar 13 at 19:09

KamouloxKamoulox

185

1 Answer
1

active

oldest

votes

1

$begingroup$

There is a general technique for this situation. Let
$, f(x) := 3x^2(1-x).,$ Now at the fixed point $,x=0,$ we have
$, f(x) = 3x^2 + O(x^3).,$ We want a function $,g(x),$ such
that $, g(x^2) = f(g(x)).,$ Given an initial ansatz of
$, g(x) = O(x),$ we use the previous equation to solve for the power series coefficients one at a time with the result
$$ g(x) = frac13 x + frac118 x^2 + frac11216 x^3 +
frac7324 x^4 + frac38910368 x^5 + O(x^6). $$
Now if $,a_0 = g(z),$ then $, a_n = g(z^2^n),$ which is quadratic convergence to the fixed point $0.$ It is unlikely that there is a closed form for $,g(x).,$ A similar result holds for the other fixed points.

share|cite|improve this answer

edited Mar 13 at 20:29

answered Mar 13 at 20:07

SomosSomos

14.6k11336

$endgroup$

add a comment | 

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1

$begingroup$

There is a general technique for this situation. Let
$, f(x) := 3x^2(1-x).,$ Now at the fixed point $,x=0,$ we have
$, f(x) = 3x^2 + O(x^3).,$ We want a function $,g(x),$ such
that $, g(x^2) = f(g(x)).,$ Given an initial ansatz of
$, g(x) = O(x),$ we use the previous equation to solve for the power series coefficients one at a time with the result
$$ g(x) = frac13 x + frac118 x^2 + frac11216 x^3 +
frac7324 x^4 + frac38910368 x^5 + O(x^6). $$
Now if $,a_0 = g(z),$ then $, a_n = g(z^2^n),$ which is quadratic convergence to the fixed point $0.$ It is unlikely that there is a closed form for $,g(x).,$ A similar result holds for the other fixed points.

share|cite|improve this answer

edited Mar 13 at 20:29

answered Mar 13 at 20:07

SomosSomos

14.6k11336

$endgroup$

add a comment | 

1

$begingroup$

There is a general technique for this situation. Let
$, f(x) := 3x^2(1-x).,$ Now at the fixed point $,x=0,$ we have
$, f(x) = 3x^2 + O(x^3).,$ We want a function $,g(x),$ such
that $, g(x^2) = f(g(x)).,$ Given an initial ansatz of
$, g(x) = O(x),$ we use the previous equation to solve for the power series coefficients one at a time with the result
$$ g(x) = frac13 x + frac118 x^2 + frac11216 x^3 +
frac7324 x^4 + frac38910368 x^5 + O(x^6). $$
Now if $,a_0 = g(z),$ then $, a_n = g(z^2^n),$ which is quadratic convergence to the fixed point $0.$ It is unlikely that there is a closed form for $,g(x).,$ A similar result holds for the other fixed points.

share|cite|improve this answer

edited Mar 13 at 20:29

answered Mar 13 at 20:07

SomosSomos

14.6k11336

$endgroup$

add a comment | 

$begingroup$

There is a general technique for this situation. Let
$, f(x) := 3x^2(1-x).,$ Now at the fixed point $,x=0,$ we have
$, f(x) = 3x^2 + O(x^3).,$ We want a function $,g(x),$ such
that $, g(x^2) = f(g(x)).,$ Given an initial ansatz of
$, g(x) = O(x),$ we use the previous equation to solve for the power series coefficients one at a time with the result
$$ g(x) = frac13 x + frac118 x^2 + frac11216 x^3 +
frac7324 x^4 + frac38910368 x^5 + O(x^6). $$
Now if $,a_0 = g(z),$ then $, a_n = g(z^2^n),$ which is quadratic convergence to the fixed point $0.$ It is unlikely that there is a closed form for $,g(x).,$ A similar result holds for the other fixed points.

share|cite|improve this answer

edited Mar 13 at 20:29

answered Mar 13 at 20:07

SomosSomos

14.6k11336

$endgroup$

share|cite|improve this answer

edited Mar 13 at 20:29

answered Mar 13 at 20:07

SomosSomos

14.6k11336

answered Mar 13 at 20:07

SomosSomos

14.6k11336

answered Mar 13 at 20:07

SomosSomos

14.6k11336