Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We have an inductively defined sequence $x_n=x_{n-1}+2y_{n-1}$ and $y_n=x_{n-1}+y_{n-1}$ where $x_n^2-2y_n^2=\pm 1$, where $x_0=1$ and $y_0=0$.

I need to prove that the sequence $\left(\frac{x_n}{y_n}\right)_{n=1}^\infty$ converges to $\sqrt2$.

Now I can see that $\left(\frac{x_n^2}{y_n^2}\right)_{n=1}^\infty$ converges to $2$, so that may be easier to prove.

It follows now that in order to prove this, $\forall\epsilon >0, \exists n>N$ s.t. $|\frac{x_n^2}{y_n^2}-2|<\epsilon$.

Can anybody point me in the right direction for solving this?

share|cite|improve this question
up vote 2 down vote accepted

That is not an equivalent statement. For example, $\{ (-1)^n \}$ doesn't converge, but the sequence of squares of each term converges to 1.

You will need to make an argument about the positive or negative nature of each term too. Once you've done that

Hint: $\frac {x_n ^2}{y_n^2} - 2 = \pm \frac {1}{y_n^2}$, and $y_n \rightarrow \pm \infty$, depending on your initial conditions.

share|cite|improve this answer
Hmm I see, but wouldn't this be an equivalent statement based on the fact that $x_n$ and $y_n$ are always positive? – Duncan Forster Jan 16 '13 at 23:36
@DuncanForster Considering that you didn't set any initial conditions, we could have $x_1 = -1, y_1=-1$, in which case we'd have $y_n \rightarrow -\infty$. In fact, it's possible for the terms to be oscillating ... The initial conditions are important. – Calvin Lin Jan 17 '13 at 2:43
Sorry I should have specified, $x_0=1$ and $y_0=0$. – Duncan Forster Jan 17 '13 at 3:42
@DuncanForster You should edit that into the original equation then. If you look at rob john's solution, you will find that there are initial conditions that lead to a different convergence. – Calvin Lin Jan 17 '13 at 16:20

The recursion can be written as $$ \begin{bmatrix}x_n\\y_n\end{bmatrix}=\begin{bmatrix}1&2\\1&1\end{bmatrix}\begin{bmatrix}x_{n-1}\\y_{n-1}\end{bmatrix} $$ The matrix has two eigenvalues: the roots of $\lambda^2-2\lambda-1=0$; that is, $\lambda=1\pm\sqrt{2}$.

The eigenvector with the eigenvalue of $1-\sqrt{2}$ is $$ \begin{bmatrix}1&2\\1&1\end{bmatrix}\color{#C00000}{\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}}=\begin{bmatrix}-2+\sqrt{2}\\-1+\sqrt{2}\end{bmatrix}=(1-\sqrt{2})\color{#C00000}{\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}} $$ The eigenvector with the eigenvalue of $1+\sqrt{2}$ is $$ \begin{bmatrix}1&2\\1&1\end{bmatrix}\color{#C00000}{\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}}=\begin{bmatrix}2+\sqrt{2}\\1+\sqrt{2}\end{bmatrix}=(1+\sqrt{2})\color{#C00000}{\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}} $$ Therefore, $$ \begin{align} \begin{bmatrix}x_n\\y_n\end{bmatrix} &=\begin{bmatrix}x_0\\y_0\end{bmatrix}\begin{bmatrix}1&2\\1&1\end{bmatrix}^n\\ &=\left(a\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}+b\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}\right)\begin{bmatrix}1&2\\1&1\end{bmatrix}^n\\ &=a(1+\sqrt2)^n\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}+b(1-\sqrt2)^n\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}\\ &=a(1+\sqrt2)^n\left(\color{#C00000}{\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}+\frac{b}{a}(2\sqrt{2}-3)^n\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}}\right) \end{align} $$ Assuming $a\ne0$, what is the limit of $x_n/y_n$ in red?

share|cite|improve this answer

Your Answer


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.