Approximations for have always fascinated me. When I was younger I always interpreted as some arbitrary constant relating the diameter of a circle to its perimeter, but it's really so much more than that. Circles appear so frequently in math that is everywhere. In this (potential) series, I will go over and prove various formulas for .
In this first one, we will go over a very famous one: , and Euler's proof (with more rigor) of the fact in . Let's get started.
The Weierstrass Factorization Theorem says that entire functions (if you do not know what that is, don't sweat it) with complex roots can be written as , where is a non-zero constant. For polynomials, this may not seem very remarkable, but this allows us to do things with functions like cosine and the exponential function. This may seem unrelated to our problem, but it is not.
Let us examine . This function has roots at for . Thus, by this theorem we can write
We can then group terms to get
Now we are getting closer to our original goal. We have an infinite sequence of squares, but we want those squares to be in the denominator. So let's divide by to get . Note that after we do this, will be different, so for clarity let's call it some other constant .
The next step is to determine . Recall that . Thus, at , the equation becomes . So hey, is ! How convenient. So our function becomes
Now let's quickly consider the Taylor/MacLaurin series for centered at . Most of you took Calc II and probably know this, but just to remind you, this is . So, the function that we found ourselves above should be equal to this function. It will be significantly difficult to compare the whole function, so let's just compare the coefficients -- they should be equal. The coefficient of it in this function is .
In our above function, the term will be determined by "picking" one in one of the binomials and "picking" in the rest of them. Thus, the coefficient term is equal to . And now we're home free:
And we get our desired equality!
Regarding the approximation itself, it certainly isn't the fastest one out there (we will investigate that one, by Ramanujan, in the last installation of this series, since it is a monster), but it is not the slowest either. When determining how quick an algorithm will yield convergence, it is good to look at the relative size of each term. If each term is only a little bit smaller than the last, then that usually means it converges slowly.
That's it for now. I hope you enjoyed reading this and I hope this was educational to you. Next time, we will investigate the Wallis Product
That one will be shorter as we've already done all the work in this one!