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

3 Answers
Apr 9, 2018

Answer:

I need double-check.

Explanation:


my notebook...

Apr 9, 2018

Answer:

#[(-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

Answer:

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#