Question

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

Answer 1

I need double-check.

Explanation:

Answer 2

`[(-1+sqrt(5))/2, (1+sqrt(5))/2]`

Explanation:

Given:

`y = sqrt(1-cos xsqrt(1-cos xsqrt(1-cosxsqrt(...))))`

write `t` for `cos x` to get:

`y = sqrt(1-tsqrt(1-tsqrt(1-tsqrt(...))))`

Square both sides to get:

`y^2 = 1-tsqrt(1-tsqrt(1-tsqrt(...))) = 1-ty`

Add `ty-1` to both sides to get:

`y^2+ty-1 = 0`

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

`y = (-t+-sqrt(t^2+4))/2`

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

So:

`y = (-t+sqrt(t^2+4))/2`

Then:

`(dy)/(dt) = -1/2+t/(2sqrt(t^2+4))`

This is `0` when:

`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 = +-1`, being the range of `t = cos x`.

When `t = -1`:

`y = (1+sqrt(5))/2`

When `t = 1`

`y = (-1+sqrt(5))/2`

So the range of `y` is:

`[(-1+sqrt(5))/2, (1+sqrt(5))/2]`

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

Answer 3

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 `cosx = -1`

Now

`y_min = 1/2(-1pm sqrt5)` and
`y_max = 1/2(1 pm sqrt5)`

then the feasible limits are

`1/2(-1+sqrt5) le y le 1/2(1+sqrt5)`

NOTE

With `y = sqrt(1+alpha y)`

we have that `y` is an increasing function of `alpha`