i need help with this question..I'm not so sure how to go about the arguments. Any help would be appreciated.
Consider a regular $n$-sided polygon tangential to and enclosing the unit circle to approximate $π$.
(a) Use geometrical arguments to show that the half-perimeter of the polygon $C_n = n \tan(\frac πn) ≥ π$.
(b) Calculate $C_4$ and $C_8$, not approximating irrationals by decimals.
(c) Use Richardson extrapolation with $h = \frac 1{n^2}$ to improve upon your results.
This is hard to write in text, since it's a very visual problem - if you have trouble following along, I suggest you grab a piece of paper and draw a picture of what I'm trying to describe. If there's a specific point you get stuck, I'll be happy to try and help.
The key fact about the circle is that every point on its perimeter is equally far from it's centre. We can use that fact to try and prove the formula. In particular, to find the side length of one of the sides of our polygon, we can construct a triangle where one of the corners is at the centre of the circle, and another is at a point where the polygon is tangent to the circle. Since this is a unit circle, the side length will have a length of 1.
The second thing we'll need is to realise the triangle we're constructing is right angled. This is because the side leading from the centre of the circle to it's edge is perpendicular to the tangent line of a circle.
Finally, we can think of the angle size of the point at the centre of the circle. The triangle we've constructed covers half of one side of our n-sided polygon, and there's nothing special about the half-side we constructed that distinguishes it from any other half-side. Therefore, each side of our polygon would take up an angle of 2π/n with respect to the centre of our circle, and a half side is half that - π/n.
So, we have a right-angled triangle, where one side is from the (and has a length of one). We also know the angle between the two sides. From there, we can calculate the length of the half side as tan(π/n). This is because the Opposite side of the triangle is the length we want to find, and the adjacent side has a length of one as the radius of our circle.
As there are 2n half-sides, the total length will be 2n tan(π/n). We want the half side, leading to the equation being n tan(π/n).