# What is the range of the function y=sqrt(1-cosxsqrt(1-cosx(sqrt(1-cosx ......oo ?

Apr 9, 2018

I need double-check.

#### Explanation:

Apr 9, 2018

$\left[\frac{- 1 + \sqrt{5}}{2} , \frac{1 + \sqrt{5}}{2}\right]$

#### Explanation:

Given:

$y = \sqrt{1 - \cos x \sqrt{1 - \cos x \sqrt{1 - \cos x \sqrt{\ldots}}}}$

write $t$ for $\cos x$ to get:

$y = \sqrt{1 - t \sqrt{1 - t \sqrt{1 - t \sqrt{\ldots}}}}$

Square both sides to get:

${y}^{2} = 1 - t \sqrt{1 - t \sqrt{1 - t \sqrt{\ldots}}} = 1 - t y$

Add $t y - 1$ to both sides to get:

${y}^{2} + t y - 1 = 0$

This quadratic in $y$ has roots given by the quadratic formula:

$y = \frac{- t \pm \sqrt{{t}^{2} + 4}}{2}$

Note that we need to choose the $+$ sign of $\pm$, since the principal square root defining $y$ is non-negative.

So:

$y = \frac{- t + \sqrt{{t}^{2} + 4}}{2}$

Then:

$\frac{\mathrm{dy}}{\mathrm{dt}} = - \frac{1}{2} + \frac{t}{2 \sqrt{{t}^{2} + 4}}$

This is $0$ when:

$\frac{t}{\sqrt{{t}^{2} + 4}} = 1$

That is:

$t = \sqrt{{t}^{2} + 4}$

Squaring both sides:

${t}^{2} = {t}^{2} + 4$

So the derivative is never $0$, always negative.

So the maximum and minimum values of $y$ are attained when $t = \pm 1$, being the range of $t = \cos x$.

When $t = - 1$:

$y = \frac{1 + \sqrt{5}}{2}$

When $t = 1$

$y = \frac{- 1 + \sqrt{5}}{2}$

So the range of $y$ is:

$\left[\frac{- 1 + \sqrt{5}}{2} , \frac{1 + \sqrt{5}}{2}\right]$

graph{(y-(-(cos x)+sqrt((cos x)^2+4))/2) = 0 [-15, 15, -0.63, 1.87]}

Apr 9, 2018

See below.

#### Explanation:

We have

${y}_{\min} = \sqrt{1 - {y}_{\min}}$
${y}_{\max} = \sqrt{1 + {y}_{\max}}$

Here

${y}_{\min}$ is associated to the value $\cos x = 1$ and
${y}_{\max}$ is associated to $\cos x = - 1$

Now

${y}_{\min} = \frac{1}{2} \left(- 1 \pm \sqrt{5}\right)$ and
${y}_{\max} = \frac{1}{2} \left(1 \pm \sqrt{5}\right)$

then the feasible limits are

$\frac{1}{2} \left(- 1 + \sqrt{5}\right) \le y \le \frac{1}{2} \left(1 + \sqrt{5}\right)$

NOTE

With $y = \sqrt{1 + \alpha y}$

we have that $y$ is an increasing function of $\alpha$