Thread: Question 1

Find the range of the function,
y=sin(sin(sin(sin(sin x))))

Originally Posted by ThePerfectHacker

The rules is that you post thy answers in white.
Failure to follow these rules will result in excommunication.

Find the range of the function,
y=sin(sin(sin(sin(sin x))))

Just a question about the rules. Do you just want the answer, or do you want a proof for these as well?

-Dan

Originally Posted by topsquark

Just a question about the rules. Do you just want the answer, or do you want a proof for these as well?

-Dan

Post anything that helps obtain a solution.


This is mine 28th post!!!

Originally Posted by ThePerfectHacker

The rules is that you post thy answers in white.
Failure to follow these rules will result in excommunication.

Find the range of the function,
y=sin(sin(sin(sin(sin x))))

Radians or degrees?
----
EDIT (Responce from ThePerfectHacker):
In radians

The graph of the sine curve is progressively flattening out each sin( that you add. One may hypothesize that the curve is approaching zero.

The range is clearly between y = -1 .. 1, although I am not sure of the exact values.

Originally Posted by Glaysher

Radians or degrees?
----
EDIT (Responce from ThePerfectHacker):
In radians

At least at one level it is irrelevant what angle measure is
used.

RonL

Originally Posted by CaptainBlack

At least at one level it is irrelevant what angle measure is
used.

RonL

There is a big difference between radians and degrees to my thinking.

Answer: Range(Y)= [-0.6275, 0.6275]

How so? Why does it matter when determining the range. Perhaps you are thinking of the domain, although in this situation that is irrelevant, too.

Originally Posted by Random333

There is a big difference between radians and degrees to my thinking.

Answer: Range(Y)= [-0.6275, 0.6275]

Supose that in this case the range were expresible in the form:

[-tan(tan(7)), +tan(tan(7)]

while this may numerically be a different range depending on whether we are
in degrees or radians, the most elegant form of the answer does not care
two hoots about which angle mode we use.

A more interesting question about this problem might be what is the domain?
There are two natural assumptions one could make for the domain, and they
are R and C. Each of these assumptions gives rise to a different answer,
and each is equally plausible as a guess at the setters intention. If I were
solving this problem I would solve both of these possibilities.

RonL

Originally Posted by CaptainBlack

Supose that in this case the range were expresible in the form:

[-tan(tan(7)), +tan(tan(7)]

while this may numerically be a different range depending on whether we are
in degrees or radians, the most elegant form of the answer does not care
two hoots about which angle mode we use.

A more interesting question about this problem might be what is the domain?
There are two natural assumptions one could make for the domain, and they
are R and C. Each of these assumptions gives rise to a different answer,
and each is equally plausible as a guess at the setters intention. If I were
solving this problem I would solve both of these possibilities.

RonL

Then is answer simply:

[-sin(sin(sin(sin(1)))), sin(sin(sin(sin(1))))?

Because that seems too simple to be right. I was trying to think of a way of writing it exactly in terms of pi

EDIT: Correction - Typed too many sines!

Originally Posted by Glaysher

Then is answer simply:

[-sin(sin(sin(sin(sin(1))))), sin(sin(sin(sin(sin(1)))))?

Because that seems too simple to be right. I was trying to think of a way of writing it exactly in terms of pi

Or [sin(sin(sin(sin(1)))), sin(sin(sin(1)))]?

sin[0,2pi] = [-1,1],
sin[-1,1] = [sin(1),1],
sin[sin(1),1] = [sin(sin(1)),sin(1)],
sin[sin(sin(1)),sin(1)] = [sin(sin(sin(1))),sin(sin(1))],
sin[sin(sin(sin(1))),sin(sin(1))] = [sin(sin(sin(sin(1)))),sin(sin(sin(1)))].

Originally Posted by JakeD

Or [sin(sin(sin(sin(1)))), sin(sin(sin(1)))] ?

sin[0, 2pi] = [-1,1],
sin[-1,1] = [sin(1),1],

for theta in the range [-pi/2, pi/2] sin is monotonic increasing so:

sin([-1,1]) = [sin(-1), sin(1)] = [-sin(1), sin(1)]


RonL

Originally Posted by CaptainBlack

for theta in the range [-pi/2, pi/2] sin is monotonic increasing so:

sin([-1,1]) = [sin(-1), sin(1)] = [-sin(1), sin(1)]


RonL

Did you check mine after I corrected it as I had originally typed too many sines?

Originally Posted by Glaysher

Did you check mine after I corrected it as I had originally typed too many sines?

Yes I did, and it agrees with what I think the solution for the real
domain should be. However I'm not the setter of this problem, nor have I
discussed it with him so I don't know what he thinks a correct solution would
look like

RonL

(who thinks the quoting in this thread is too complex for white
text - you would not believe how messy some of my posts
have been while sorting out the quoted and reply text)

Originally Posted by Glaysher

Then is answer simply:

[-sin(sin(sin(sin(1)))), sin(sin(sin(sin(1))))?

I agree also, that's how I came up with my answer of:

Originally Posted by Random333

Answer: Range(Y)= [-0.6275, 0.6275]