This is a Pi Approximation Leibnez Series (converges to max 16 decimals after the 5 quadrillionth nth term, so like never) Nilakantha Series: (converges to 16 decimals after the 170,000th nth term, so sometime later today :) Bailey-Borwein-Plouffe-Series: (converges to 16 decimals after like 12 terms, blink and you'll miss it)
Pi is a irrational number that can only be approximated. (There are infinite number of digits in pi) To approximate pi, mathematicians use an operation called series. *The faster a series converges the less terms of n are required to achieve a certain accuracy of approximation. Explanation of a Series: *Read Carefully* Series are a sum of the terms of a sequence. An example of a sequence is: 1, 2, 4, 8, 16. (This is a Geometric Sequence, as each term is being multiplied by 2) A Geometric sequence is modeled by a1(r)^(n-1) where: a1= The first term r = The common ratio n= nth term For our example sequence we can model it as 1(2)^(n-1) Where 2 is the ratio (multiply by 2) and 1 is the first term. Now for the series. The funky "E" is called a sum (short for summation), and the number on the bottom represents which term to start with, and the value on top represents the value to end with. Let's use a summation of our previous geometric sequence 1(2)^(n-1) from n=0 to n=4. We find the 0th term, the 1st term, the 2nd term, the 3rd term, and the 4th term and add them together. (1+2+4+8+16) to get 31 as our sum. Now for the infinite series, where n starts at some constant (usually 0 or 1) and goes to infinite.(∞) We can see that if we found the sum of 1+2+4+8+16+32+64... to infinite we would end up with infinite as our sum. This means that this series would diverge as the series doesn't approach a number. (infinite is not a number) So what series would converge? For a geometric series the ratio has to be less then 1 for convergence. For example [0,∞] E(1/2)^(n-1) Here the ratio is equal to 1/2 which is less then 1, so it converges. How? Let's find the first few terms. a(1)=1, a(2)=1/2, a(3)=1/4, a(4)=1/8... and so on. We can see that as n increases the terms becomes smaller and smaller (approaching 0) Eventually the terms will become of negligible value and the sum will reach some value. For Geometric Series we can find what value the sum converges at with this formula: a1/1-r, where a1= First term r= ratio. For the example 1(1/2)^(n-1) our first term is 1, and our ratio is 1/2. 1/(1-1/2) becomes 1/(1/2) or 2, which is where our geometric series converges! I hope this helps anyone to grasp an idea of what series are, and how they are used to approximate pi. **********************Happy PI day!************************ Using my radical knowledge of Calculus BC principles. (shoutout Mrs. S!) I abhor ratio test, but can't stop using it :|
