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October 10th, 2006, 07:46 AM
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| | Question 1 Find the range of the function,
y=sin(sin(sin(sin(sin x))))
Last edited by ThePerfectHacker; December 5th, 2006 at 01:47 PM.
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October 10th, 2006, 11:03 AM
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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
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October 10th, 2006, 02:15 PM
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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 28   th post!!! | 
October 11th, 2006, 12:52 AM
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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
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Last edited by ThePerfectHacker; October 11th, 2006 at 05:15 AM.
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October 13th, 2006, 06:17 PM
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| | 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. | 
October 14th, 2006, 12:04 AM
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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
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October 14th, 2006, 06:52 PM
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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] | 
October 14th, 2006, 07:36 PM
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| | 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. | 
October 14th, 2006, 11:13 PM
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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
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Last edited by CaptainBlack; October 14th, 2006 at 11:47 PM.
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October 15th, 2006, 12:07 AM
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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!
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Last edited by Glaysher; October 15th, 2006 at 12:51 AM.
Reason: Correction
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October 15th, 2006, 12:16 AM
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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)))].
Last edited by JakeD; October 15th, 2006 at 12:32 AM.
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October 15th, 2006, 02:48 AM
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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
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October 15th, 2006, 03:19 AM
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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?
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October 15th, 2006, 03:42 AM
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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)
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October 15th, 2006, 04:29 AM
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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: Quote:
Originally Posted by Random333 Answer: Range(Y)= [-0.6275, 0.6275] | | | Thread Tools | | | | Display Modes | Linear Mode |
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