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

3 Answers
Apr 9, 2018

I need double-check.

Explanation:


my notebook...my notebook...

Apr 9, 2018

[(-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]}

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 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