Please verify this for me...

[Dr. Zimski]

... and explain any implied wisdom (or possibly duhr-duhr-duuhrhrhr) regarding...

 

Pi = n[sin(180/n)], lim n approaches infinity

[ajb]

Try a series expansion of [math]Sin(\frac{\pi}{n})[/math] for large [math]n[/math] or equivalently small [math]\frac{\pi}{n}[/math].

 

See what you get to first order and the multiply by [math]n[/math].

 

I would call this an "engineering proof" as it is not exactly rigorous, but will probably be ok for your purposes.

 

(I assume you were not asking for an epsilon delta proof?)

[Shadow]

Consider the polygon in the attachment. It is a regular octagon. In this case, [math]\alpha = \frac{360^o}{8} = 45^o[/math]. In a regular n-gon, [math]\alpha=\frac{360^o}{n}[/math]. This means that [math]\frac{\alpha}{2}=\frac{360^o}{2n}=\frac{180^o}{n}[/math]. Thus, if we wanted to calculate [math]x[/math] on the picture, it would be [math]x = r \cdot sin(\frac{180^o}{n})[/math]. Now consider an n-gon with a radius of one. In such an n-gon, [math]x = sin(\frac{180^o}{n})[/math]. And since [math]y=2 \cdot x[/math], we can easily calculate the perimeter of an regular n-gon: [math] P=2 \cdot n \cdot sin(\frac{180^o}{n})[/math], where [math]P[/math] denotes the perimeter. Now, the bigger n is, the more the n-gon will resemble a circle. Thus, as n approaches infinity, the n-gon will become a circle. So, we can say that [math]\lim_{n\to\infty} 2 \cdot n \cdot r \cdot sin(\frac{180^o}{n}) = 2 \pi r[/math]. In other words, the perimeter of an n-gon with n approaching infinity is equal to the circumference of a circle. Since [math]r=1[/math] we can just ignore the r (even if it didn't equal one they would cancel out), and the twos cancel out, we're left with

[math]\lim_{n\to\infty}n \cdot sin(\frac{180^o}{n}) = \pi[/math]

 

And there you go. Hope this helps.

Edited July 29, 2009 by Shadow