Am I cheating in this case to evaluate $\pi$?

Question

Since $\lim_{x \to 0}$$\sin x \over x$$=1$，here let $x=$$\pi\over n$ , then we have $\lim_{{\pi\over n} \to 0}$$\sin {\pi\over n} \over {\pi\over n}$$=1$ , which implies $\pi=$$\lim_{n \to\infty}$$\ n*sin {\pi\over n}$ Here is the link of the animated graph of this method: Approximation of Pi

The idea is to evaluate $\pi$ by assigning a very large number to $n$, and then calculate the corresponding value of the function $f(n)=n*sin {\pi\over n}$ using a computer program: the bigger $n$ is, the closer to $\pi$ it will be.

So, here comes the question which makes me very confused: Am I cheating in this case by using the $\pi$ already existing in the function $f(n)=n*sin {\pi\over n}$ to generate the accurate value of $\pi$?

I'm not sure if I can do this. But the $\pi$ in the function actually acts as a notation of angle, which can be replaced with $180^\circ$, while the other $\pi$ here is a real number. They are different concepts So I guess it is not cheating.

Answer 1

Yes, this is cheating; your computer will surely be using the value of $\pi$ in the computation, which defeats the purpose - and this fact is somewhat unavoidable, given that $\sin(x)$ is likely computed by Taylor series - and plugging $\pi x$ into such a method certain requires knowing $\pi$ (as opposed to if $\sin$ was calculated geometrically, where $\pi$ might be a more natural unit - as has been referenced by comments, an interesting limit is the first equation in this question).

In spirit, if you had an equation of the form: $$\pi=f(\pi)$$ for some function $f$, then you could might be able to solve the above numerically. For instance, we could find $\pi$ by noting that $\pi=\sin(\pi)+\pi$ and, as the derivative of $\sin(x)+x$ is $0$ at $x=\pi$, we could perform a method called fixed-point iteration to find $\pi$ by choosing some starting point $x_1$ (for instance $3$) and setting $x_2=\sin(x_1)+x_1$ and so on. This converges to $\pi$, despite the definition looking initially circular.

However, we can prove the following fact: $$x=\lim_{n\rightarrow\infty}n\sin\left(\frac{x}n\right)$$ which means your equation works equally well for any $x$ that is not $\pi$ - so we cannot use it to figure out that value of $\pi$ as it does not uniquely specify it.

Answer 2

Since $\sin(\frac\pi{2^n})$ can be evaluated exactly (example), this isn't necessarily circular.

From Wikipedia:

The thing on the right is equal to $2^{k+1}\sin(\frac\pi{2^{k+1}})$, which is your equation with $n=2^{k+1}$.