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 up 4 - 4 x up 2 - x + 4 is 0.png

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:

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

Next we note that if $x_n > 1$:

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

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

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

By definition of $x_n$:

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 $1$.