,then y can reach any number in [-1,1]

may i have a proof?

i just be very curious why it is ture and it's not my exam ,if the proof is too long ,please give me a e-mail ,thanks

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- April 21st 2010, 08:19 AM #1

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## A question about y=sin(n)

,then y can reach any number in [-1,1]

may i have a proof?

i just be very curious why it is ture and it's not my exam ,if the proof is too long ,please give me a e-mail ,thanks

- April 21st 2010, 08:52 AM #2

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First one should specify what "reach" means: there are (many) real number such that for all (just because is uncountable).

However, it is true that any number can be approximated by terms of the sequence with arbitrarily good accuracy: there exists a sequence such that . In other words, the set is dense in .

I think the best way to prove this is to prove a (very slightly) stronger statement about complex numbers: the sequence is dense in the complex unit circle. The result follows by projection (if is close to then is close to , and we take ).

The density of in the unit circle comes from a pigeon-hole principle and from the irrationality of .

The pigeon-hole principle argument is equivalent to the application given in this wikipedia page (paragraph beginning with "A notable problem...") where . I guess you can find other references with the keywords I gave you.

You can find plenty of references about the irrationality of on the web, or just admit it.

- April 21st 2010, 10:24 AM #3

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first,thanks to your idea, the idea is on the point ,because that cycle is necessary . And it's not complete, i have learned the complex analysis,the thinking from the friend above is good ,but i want a complete proof, the question refer to much kownledge in math ,esepcailly number theory and complex analysis.only the knowledge on may be not enough

could some give a proof? thanks to the friend above and your idea.