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

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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 $1$.

$\blacksquare$


Sources