# $x_{n+1}=\frac{2x_n^3}{3x_n^2-1}$ Find all $a$ such that $\{x_n\}$ has finite limit

Let $\{x_n\}$ be a real sequence defined by: $$x_1=a \\ x_{n+1}=\frac{2x_n^3}{3x_n^2-1}$$

For all $n=1,2,3...$ and $a$ is a real number. Find all $a$ such that $\{x_n\}$ has finite limit when $n\to +\infty$ and find the finite limit in that cases.

I'm stuck when $a\in(-1,1)$ and $a<-1$ Please help me. Thanks

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Do you already have the three possible limit points? – Berci Nov 26 '12 at 18:43

If the sequence should converge to $x$, then we must have $$\tag1x(3x^2-1)=2x^3$$ because the given recursion implies $x_{n+1}(3x_n^2-1)=2x_n^3$. Equation $(1)$ can be transformed to $x^3-x=0$, i.e. $x\in \{-1,0,1\}$.

First note that if $a<0$ then we obtain the same $x_2=\frac{-2a^3}{3a^2-1}$ as we would with $-a$ instead. Therefore these cases are symmetric to the positive case: $a$ produces a converging sequence iff $-a$ does.

Note that $$\tag2\frac{x_{n+1}}{x_n}=\frac{2x_n^2}{3x_n^2-1}.$$ The right hand side is $<1$ if $x_n>1$. Also note that $$\tag3x_{n+1}-1=\frac{2x_n^3-3x_n^2+1}{3x_n^2-1}=\frac{(2x+1)(x-1)^2}{3x_n^2-1}$$ and the right hand side is positive if $x>1$. Thus by induction $a>1$ implies that all $x_n$ are $>1$ and the sequence is strictly decreasing. It must therefore converge to a limit $\ge 1$, hence to $+1$.

The case $a=1$ is trivial as it leads to the constant sequence $x_n=1$.

If $\frac1{\sqrt 3}<a<1$, then we read from $(3)$ that $x_2>1$, hence from then on, the sequence is again decreasing and converges $\to 1$.

If $a=\frac1{\sqrt 3}$, then the sequence is not defined because $x_2$ is not defined.

If $0<|x_n|<\frac1{\sqrt 5}$, then $2x_n^2<1-3x_n^2$ implies via $(2)$ that $\left\vert\frac{x_{n+1}}{x_n}\right\vert$ is strictly decreasing, hence $x_n\to 0$ if $0<a<\frac1{\sqrt 5}$.

If $x_n^2=\frac15$, then $(2)$ shows that $x_{n+1}=-x_n$. Hence $a=\frac1{\sqrt 5}$ leads to the nonconverging sequence $x_n=(-1)^{n-1}a$.

If $\frac15<x_n^2<\frac13$, then $0<1-3x_n^2<2x_n^2$, hence $(2)$ implies $|x_{n+1}|>|x_n|$. Therfore, with $\frac1{\sqrt 5}<a<\frac1{\sqrt 3}$ two things can happen:

• The sequence $x_n^2$ remains bounded by $\frac 13$. Then it is increasing and hence converges to some $b$ with $b=\frac{4b^3}{(3b-1)^2}$, which implies $b=0$ or $b=1$ or $b=\frac15$. Since the limit $b$ must also be $>\frac15$ and $\le \frac13$, this cannot happen.
• For some $n$, we have $x_n^2>\frac 13$. From then on, the sequence converges as seen above.
• For some $n$, we have $x_n^2=\frac 13$. Then $x_{n+1}$ is not defined.

The third case does happen for all members of a sequence beginning $$\tag 4 0.46560062143367758\ldots,\\0.4472135954999579\ldots,\\ 0.450201477782475\ldots,\\0.44770950581291\ldots$$

Summary: The only cases where the sequence does not converge are given by $a=\pm\frac1{\sqrt 5}$, $a=\pm\frac1{\sqrt 3}$ and when $|a|$ occurs in the sequence (4).

Remark: The sequence (4) is obtained by letting $a_0=\frac1{\sqrt 3}$ and then $a_{n+1}$ the unique solution of $2a_{n+1}^3-a_n(3a_{n+1}^2-1)=0$ between $\frac1{\sqrt 5}$ and $\frac1{\sqrt 3}$.

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HINT: Let $$f(x)=\frac{2x^3}{3x^2-1}-x\;,$$ and show first that any limit of a sequence $\langle x_n:n\in\Bbb Z^+\rangle$ must be a zero of $f$. Then show that $f\,'(x)<0$ for all $x\in(-1,1)$ for which the function is defined. Note that $$f(-1)=f(0)=f(1)=0$$ and that $f$ has vertical asymptotes at $x=\pm\frac1{\sqrt3}$. Conclude that $f(x)<0$ for $-1<x<-\frac1{\sqrt3}$ and $0<x<\frac1{\sqrt3}$, and $f(x)>0$ for $-\frac1{\sqrt3}<x<0$ and $\frac1{\sqrt3}<x<1$. Use this information to determine the limit of $\langle x_n:n\in\Bbb Z^+\rangle$ for values of $x_1$ in $(-1,1)$.

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