Which irrational number represents the infinite simple continued fraction [0;7]?
-So from my current understanding [o;7] can be represented as the following:
$ = \frac{1}{7 + \frac{1}{7 + \frac{1}{7 +...}}}$
So, x would be $x = \frac{1}{7 + \frac{1}{7 + x}}$
$\frac{7(7+x)}{7+x} + \frac{1}{7+x}$
$\frac{49 + 7x}{7+x} + \frac{1}{7+x}$
$ \frac{50 + 7x}{7+x}$
From here we get:
$x(50 + 7x) = 7 + x$
$50x = 7x^2 = 7 + x$
$49x + 7x^2 = 7$
$ 49x = 7x^2 - 7 = 0$
$x = \frac{-49\pm \sqrt{2205}}{14}$
- UPDATED, the square root looked a bit weird to me wanted to confirm my answer or suggestions, any help is appreciated.
Let $x$ be the value of the continued fraction $[0;7]$. We have that $$ x = \frac{1}{7 + \frac{1}{7 + \frac{1}{7 +...}}}$$
Now we notice that $x$ appears in this continued fraction, so we can write it as: $$x = \frac{1}{7 + x}$$
This simplifies to $x^2+7x-1=0$, and solving for $x$ gives $x=\frac{-7}{2}\pm \frac{\sqrt{53}}{2}$