Take a circle. Now, draw a hexagon inside of it. A regular hexagon is comprised of 6 equilateral triangles, and in this example we can say that they have a side length of one. That means the diameter of the circle is two, since the regular hexagon’s diameter is two as well. Since the total perimeter of the hexagon is 6, and the hexagon is inside the circle, and the area of a circle is πr^2, and r (which is half of a circle’s diameter, which is the distance from one side of the circle to the other), is one, that means this example circle is equal to π, since the equation would be π(1)^2, and 1^2 is just one, and π times one is just π. This means that π is the area of this circle, and so π must be more than the circumference (the perimeter of the circle) over the diameter, which is 6/2, which means π is more than 3. Now, draw a square outside of the circle. Since that square contains the circle, the square is larger than the circle. Since each side of the square is corresponding to the diameter of a circle, the total perimeter of the square is eight, which means since the area of the circle is less than the area of the square, π is less than 8/2, which means π is less than 4, which is true. π, or 3.14 is between three and four. Now, bisect the hexagon inside the circle to a regular dodecagon, and you can find that the total perimeter is equal to 6.212. The ratio of that dodecagon is 3.106, so π is more than 3.106. Do the same thing for the square, turning it into a larger dodecagon, and the perimeter is about 6.431. The ratio of that is about 3.2155 or so. At this point, the equations can become significantly more difficult, looking something like 6 √(2-√(3))< π<12(2-√(3)), since there are square roots, and square roots of square roots. For centuries, people calculated more and more of π by bisecting larger and larger n-gons, the most standoffish in particular being the Dutch “Ludolph van Ceulen”, who calculated the area and ratios of a shape with 2^62 sides, or 4,611,686,018,427,387,904 sides. All of that rewards you with 35 correct digits of π; 3.14159265358979323846264338327950288. Then, sir Isaac Newton came around. One day, while quarantined due to an outbreak of bubonic plague, he was doing some calculations along the lines of (1+x^2). Multiply it out, and you get 1+2x+x^2. You can do the same for (1+x)^3, for 1+3x+3x^2+x^3. Keep going up, and you get numbers that distinctly represent Paschal’s Triangle. Each layer of Paschal’s Triangle can be solved with the Binomial Theorem, (1+x)^n=1+nx+n(n-1)x2/2!+n(n-1)(n-2)x^3/3!, and so on forever. Newton discovered that not only can the Theorem be used with negative integers, but also fractions. You can use this to find the area of a circle, too. The area of a unit circle is 1, and a quarter of it is 1/4, but actually π/4 since 1= π in this scenario. Take (using Trigonometry integration for |,) 0|1(1-x^2)^1/2dx= 0|1[1-1/2x^2-1/8x^4-1/16x^6-…]dx, and you can replace 0|1(1-x^2)^1/2 with π/4. Then, integrate to get π=4[x-1/2x/3^3-1/8x/4^4-…]. Set x to 1, and the equation simplifies out to π=4[1-1/6-1/40-1/112-…]. This will cause a slow decline toward zero. The goal, however, is to get there as fast as possible. Newton changed 0|1 to 0|1/2, which is half of a quarter of a circle. You can break that small section up into a 30 degree triangle with an area of π/12 and a right triangle with an area of √3/8. If you substitute that into the end of π=4[x-1/2x/3^3-1/8x/4^4-…], to make π=4[x-1/2x/3^3-1/8x/4^4-…-√3/8], you get 3.14161, just two digits off. Keep going on with the fractions and exponents, and you can get as many digits of π as you can calculate.
