Calculating Supremum of sin n for Positive Integers

[vagabond]

I've got a question in analysis:
How to calculate the supremum of sin n for positive integers n?

I have tried hard but still cannot figure out it.
Thanks very much to answer my question in advance!

 

[Data]

This is a question with a fairly involved answer. The answer is 1 (as you might expect), but this is complicated by the fact that \sin n never actually takes the value 1 for any integer n. In fact, \sin n takes infinitely many values in every subinterval of \left[ -1, 1\right], but it still manages to "miss" most of them (this is in the same way that there are infinitely many rationals on every real subinterval, but the rationals still have measure 0). In other words, every x \in \left[ -1, \ 1 \right] is an accumulation point of \{\sin n\}_{n=0}^\infty.

Try approaching it this way: write n in the form

n = 2\pi k + r, \ 0 < r < 2 \pi, \ k \in \mathbb{Z}.

You should prove that this decomposition is unique, in the sense that if you also have n = 2\pi p + q, \ 0 < q < 2\pi, \ p \in \mathbb{Z} then q=r, \ p=k, and you should prove that you can always do this. Then, see if you can find a way to show that r can be made arbitrarily close to \pi / 2 by choosing n appropriately, and thus that \sin n can be made arbitrarily close to \sin (2\pi + \pi/2) = 1.

 

[M98Ranger]

Calculating the supremum of sin n for positive integers n can be a challenging task in analysis. One approach to solving this problem is to use the properties of the sine function and the definition of supremum.

Firstly, we know that the sine function is bounded between -1 and 1, with a period of 2π. This means that for any positive integer n, sin n will lie between -1 and 1. Therefore, the supremum of sin n cannot be larger than 1.

Next, we can use the definition of supremum, which states that the supremum of a set is the smallest upper bound of that set. In this case, the set is the set of all possible values of sin n for positive integers n. Since we have already established that sin n is bounded between -1 and 1, the supremum of this set must be 1.

Therefore, the supremum of sin n for positive integers n is 1. This can also be seen graphically by plotting the values of sin n for positive integers n on a graph. The graph will show that the function approaches 1 as n increases.

I hope this helps answer your question. Keep practicing and exploring different approaches in analysis, and you will continue to improve your understanding of mathematical concepts. Good luck!