Project

The first method to approximate pi was developed by Archimedes. This method "finds an approximation of pi by determining the length of the perimeter of a polygon inscribed within a circle (which is less than the circumference of the circle) and the perimeter of a polygon circumscribed outside a circle (which is greater than the circumference). The value of pi lies between those two lengths" (Groleau). Archimedes began with using a hexagon for his method and then doubled the sides to a 12-sided polygon, a 24-sided polygon, and eventually a 48 and 96-sided polygons. Archimedes' work with this regular polygon method allowed him to approximate pi accurately to the thousandths place.

Another common method to approximate pi is known as the Monte Carlo approach. This approach uses probability to make an estimation of pi. The method works like so: Suppose that there is a circle with radius r inscribed within a square. Thus, the square has side lengths of exactly 2r, which is also the diameter of the circle. Suppose we are going to throw darts at the square. If the dart hits the circle, it's considered a success. If the dart hits inside the square, but not inside the circle, it's considered a miss. What's the probability of having success? It's the area of the circle divided by the area of the square. The area of the circle is \(\pi r^2\), and the area of the square is \(2r\cdot 2r=4r^2\). Thus, the probability of having success is \(\frac{\pi r^2}{4r^2}=\frac{\pi}{4}\) and pi can be approximated using a simulation since \(\pi \approx 4\frac{\text{the number of successes in the sample}}{\text{the total number of trials in the sample}}\) This approximation will become more accurate as the total number of trials approaches infinity (Lehana).

A third way to approximate pi is to use calculus or a Riemann sum. Consider a circle centered at the origin (0,0) with radius 1. Since the radius is 1, the area of the circle is pi. Consider the part of the circle that is in the first quadrant of the graph. The y-intercept of the curve is at (0,1) and the x-intercept is at (1,0). If we find the area under this curve, we will have found \(\frac{1}{4} \pi\), since the area of the circle is pi and one-fourth of the circle is in the first quadrant. In order to find the area under the curve, we can integrate or use a Riemann sum. To integrate, we know that the area of this unit circle is given by the equation \(x^2+y^2=1\). Since we are only interested in the curve located in the first quadrant we can rewrite the equation as \(y=\sqrt{1-x^2}\) and integrate from zero to one (Oliver). We can also use Riemann sums to approximate the area under this curve. This is done by partitioning the interval from x=0 and x=1 into z equal segments. These z equal segments become the bases for rectangles. We create rectangles that extend up to the curve. If we are doing a right Riemann sum, we create our rectangles of such a height that their top right corners touch the curve. If we are performing a left Riemann sum, we create our rectangles such that their top left corners touch the curve. We then find the area of each of these rectangles and sum together each of their areas to find the area under the curve. As z grows closer to infinity, our approximation for the area under the curve becomes more and more accurate. After we have found the area under the curve, we then multiply that area by four to find our approximation for pi, since the area under the curve in the first quadrant is \(\frac{\pi}{4}\) (Khan Academy).

The final way of approximating pi that we will discuss is the use of infinite series. In essence a Riemman sum can be considered an infinite sum. Since this idea of infinite sums works well to approximate pi, several famous mathematicians have dedicated themselves to finding series that approximate pi even more quickly. Oliver describes the first infinite series that was used to approximate pi, "From trigonometry, we know \(tan(\frac{\pi}{4}) = 1\). We can now use the inverse tangent function, \(arctan(x)\), to calculate \(\arctan(1) = \frac{\pi}{4}\). And luckily, we have a simple and easy formula for \(\arctan(x)\)." The formula for \(\arctan(x)\) is given by \(\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}\). Thus, to find \(\arctan(1)\), we can set x=1 and use the formula. Since \(\arctan(1) = \frac{\pi}{4}\), we can then multiply the formula that we found by 4 and we have an approximation of pi. This method is also known as the Gregory-Leibniz Series or the Madhava-Gregory series. Other mathematicians have continued to make series that approximate pi even more efficiently. One of these mathematicians was Srinivasa Ramanujan, and his approximation is now the basis for many computer algorithms approximating pi and is given by \(\frac{1}{\pi}=\frac{\sqrt{8}}{99^2}\sum_{n=0}^{\infty}\frac{(4n!)}{(4^nn!)^4} \frac{1103+26390n}{99^{4n}}\). This method of approximation is extremely powerful- with only one iteration we can approximate pi correctly to the 14th digit (Oliver).