Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

2 Answers 2

up vote 2 down vote accepted

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 $\frac1{\sqrt 3}$.

share|improve this answer

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 $(-1,1)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.