# Calculating Supremum of sin n for Positive Integers

• vagabond
In summary, the question is asking how to calculate the supremum of sin n for positive integers n, which is 1. However, this is complicated by the fact that sin n never takes the value of 1 for any integer n. The solution involves writing n in a specific form and showing that r can be made arbitrarily close to pi/2, thus making sin n approach 1.
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!

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 $\sin (2\pi + \pi/2) = 1$.

Last edited:

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!

## 1. What is the definition of supremum?

The supremum, or least upper bound, of a set is the smallest number that is greater than or equal to all elements in the set.

## 2. How do you calculate the supremum of sin n for positive integers?

The supremum of sin n for positive integers can be calculated by taking the limit as n approaches infinity of sin n. This limit will be equal to 1, making 1 the supremum of sin n for positive integers.

## 3. Why is the supremum of sin n for positive integers equal to 1?

This is because as n approaches infinity, the sine function oscillates between -1 and 1, with 1 being the highest value it can reach. Therefore, 1 is the least upper bound for sin n for positive integers.

## 4. Can the supremum of sin n for positive integers be reached?

No, the supremum of sin n for positive integers is a limit that is approached but never reached. In other words, there is no positive integer n that will make sin n exactly equal to 1.

## 5. What is the significance of calculating the supremum of sin n for positive integers?

Calculating the supremum of sin n for positive integers is important in understanding the behavior of the sine function and its upper bounds. It also has applications in calculus and other mathematical fields.

• Topology and Analysis
Replies
2
Views
1K
• Calculus
Replies
8
Views
353
• Topology and Analysis
Replies
2
Views
1K
• Calculus
Replies
3
Views
1K
• Calculus
Replies
3
Views
2K
• Optics
Replies
3
Views
1K
• Calculus
Replies
8
Views
359
• Calculus
Replies
3
Views
2K
• Calculus
Replies
2
Views
8K
• Calculus
Replies
24
Views
3K