Why \pi? I assume this is the question on everyone’s mind. (Whether you’re a Math lover or not)

The simple answer would be that we all love pie, now don’t we?

Before I begin discussing any technicalities, I’d like to acknowledge that it is possible for some of us to find the concepts easy whilst others might struggle with them. This is the reason why I’m choosing to speak in a very simple and understandable manner. (I’m baby proofing my post!)

Firstly, let us have a look at the Maclaurin series of \arctan(x)

Aside: A Maclaurin series is a polynomial which approximates a function around the point x = 0. The level of accuracy decreases as you move further away from x = 0. The only way to get an exact answer and not an approximate is to let the sum go to infinity.

Below is a table of the first few derivatives of \arctan(x). We will now use this to determine what the Maclaurin polynomial should look like.

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It is now quite clear that the polynomial should look something like this:

png1

The next step is quite clever. Since \arctan(1) = \frac{\pi}{4} (Some of you may know this as 45 degrees); we can let x = 1 to arrive at an expression for \frac{\pi}{4}. It is now evident that if we multiply the series by 4 we will arrive at an expression for \pi. This is shown as follows:

png2

Therefore,

 

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Now let us try out the result we’ve just arrived at and see just why it is the least preferred method of approximating \pi. It is quite easy to see the pattern in the terms of the above sum. So let us see how our sum converges towards to the value of \pi as we increase the number of terms. These calculations can be done using WolframAlpha if your calculator has insufficient memory.

The sum we will be using is as follows:

png4

When the number of terms = 10, the sum equals 3.232315809505594

When the number of terms = 100, the sum equals 3.1514934010709914

This looks horrible, doesn’t it? Intuitively, we all know that \pi = 3.14… So what’s going wrong? A good answer would be that this series converges towards \pi at an extremely slow rate so even though it doesn’t seem to be accurate right now; it’ll most definitely be more accurate if we add more terms to the sum. Before we do this, let us try and understand why it converges so slowly. One reason is the fact that our sum provides accurate approximations around the point x = 0, but here we are approximating around the point x = 1. Secondly, we multiplied the sum for \frac{\pi}{4} by 4 which added to the inaccuracy when approximating \pi itself. Of course, there are a few other reasons as well.

Why don’t we sum ten million terms and see what we arrive at? This gives us a value of 3.1415927535897814. It is clear now that this value is equal to \pi up to 7 decimal places.

I hope that the above paragraph proves why this is the least preferred method of approximating \pi. The reason I chose to discuss it is because it is relatively simple to understand in comparison to other sums which do in fact arrive at much better approximations of \pi in a much shorter time span. Lastly, I think we can all agree that this method is quite beautiful in its own way.

 

 

 

 

 

 

 

 

How clear is this post?