An algebra problem by A Former Brilliant Member

Algebra Level 2

Solve for $x$ if

$\large x=\sqrt{1-\sqrt{1-\sqrt{1-…}}}$ .

Give your answer correct to three decimal places.

1 solution

A Former Brilliant Member
Jul 10, 2017

$\large x=\sqrt{1-\sqrt{1-\sqrt{1-…}}}$

Square both sides.

$\large x^2=\left (\sqrt{1-\sqrt{1-\sqrt{1-…}}}\right)^2$

$\large x^2=1-\sqrt{1-\sqrt{1-…}}$

but: $\large x=\sqrt{1-\sqrt{1-\sqrt{1-…}}}$

Therefore,

$\large x^2=1-x$

$\large x^2+x-1=0$

By using the quadratic formula to solve for $\large x$ , we get

$\large x=$ $\large \boxed{0.618}$

The value of $x$ cannot be negative (because of the definition of the square root function), so the solution is $0.618$ .

How do you know that the limit defining the recursive sequence

$a_n = \begin{cases} \sqrt{1}, n = 1 \\ a_{n + 1} = \sqrt{1 - a_n}, n \geq 2 \end{cases}$

converges?

Zach Abueg - 3 years, 11 months ago