# Real Sequence/Examples/Root (2 + Root x(n))

## Example of Real Sequence

Let $\sequence {x_n}$ denote the real sequence defined as:

$x_n = \begin {cases} \sqrt 2 : n = 1 \\ \sqrt {2 + \sqrt {x_{n - 1} } } & : n > 1 \end {cases}$

Then $\sequence {x_n}$ converges to a root of $x^4 - 4 x^2 - x + 4 = 0$ between $\sqrt 3$ and $2$.

## Proof

$x^4 - 4 x^2 - x + 4 = 0$

Because $x_1 = \sqrt 2$, we have that:

$x_2 > \sqrt 2 = x_1$

Suppose that:

$x_n > x_n - 1$

for some $n \ge 2$.

Then:

 $\ds x_{n + 1}$ $=$ $\ds \sqrt {2 + \sqrt {x_n} }$ $\ds$ $>$ $\ds \sqrt {2 + \sqrt {x_{n - 1} } }$ $\ds$ $=$ $\ds x_n$

Hence $\sequence {x_n}$ is seen to be strictly increasing.

Next we note that if $x_n > 1$:

 $\ds x_{n + 1}$ $=$ $\ds \sqrt {2 + \sqrt {x_n} }$ $\ds$ $>$ $\ds \sqrt {2 + \sqrt 1}$ $\ds$ $=$ $\ds \sqrt 3$

As $x_n$ is strictly increasing, it follows that $x_n > \sqrt 3$ for all $n \ge 1$.

Similarly, we note that if $x_n < 2$:

 $\ds x_{n + 1}$ $=$ $\ds \sqrt {2 + \sqrt {x_n} }$ $\ds$ $<$ $\ds \sqrt {2 + \sqrt 2}$ $\ds$ $=$ $\ds \sqrt {2 + 2}$ $\ds$ $=$ $\ds 2$

So $\sequence {x_n} \to k$ where $\sqrt 3 < k \le 2$.

By definition of $x_n$:

 $\ds k$ $=$ $\ds \sqrt {2 + \sqrt k}$ $\ds \leadsto \ \$ $\ds k^2$ $=$ $\ds 2 + \sqrt k$ $\ds \leadsto \ \$ $\ds \paren {k^2 - 2}^2$ $=$ $\ds x$ $\ds \leadsto \ \$ $\ds k^4 - 4 k^2 + 4$ $=$ $\ds k$ $\ds \leadsto \ \$ $\ds k^4 - 4 k^2 - k + 4$ $=$ $\ds 0$

By investigating the shape of the graph, it is seen that this is the only root of $x^4 - 4 x^2 - x + 4 = 0$ strictly greater than $1$.

The other root, as is seen by inspection, is in fact $1$.

$\blacksquare$