You might have noticed something slightly strange happening when we made our approximation for \(\pi\) using the Maclaurin polynomial for \(\arctan x\). We were slightly sneaky in that we performed a definite integral, but we didn’t seem to have any constant of integration.

The sneaky line was this one:

\[\arctan x\approx\int \sum_{i=0}^n(-1)^i x^{2i} dx= x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}+...=\sum_{i=0}^n \frac{(-1)^{i}x^{2i+1}}{2i+1}\]

Where we have written the \(\arctan x\) function as an integral and not written the constant of integration.

The point is that we should really be saying:

\[\int \frac{1}{1+x^2}dx=\arctan x+c\]

and so there should be a constant in the expression on the left. Then, when we perform the integration we are left with another constant (let’s call the original one above \(c_1\) and the second one \(c_2\). So what we really should have written was:

\[\arctan x+c_1\approx\int \sum_{i=0}^n(-1)^i x^{2i} dx= x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}+...+c_2=\sum_{i=0}^n \frac{(-1)^{i}x^{2i+1}}{2i+1}+c_2\]

Then we can write:

\[\arctan x\approx\sum_{i=0}^n \frac{(-1)^{i}x^{2i+1}}{2i+1}+c_2-c_1\]

We can then fix our constants of integration by knowing that \(\arctan 0=0\) and thus \(c_2=c_1\) and thus we don’t actually have any constants to worry about.

In the case that I went through in class however of \(\int e^{-x^2}dx\) it is certainly true that you do need an integration constant and thus:

\[\int e^{-x^2} dx\approx\sum_{i=0}^n \frac{(-1)^i)}{i!}\int x^{2i}dx=\sum_{i=0}^n \frac{(-1)^ix^{2i+1})}{(2i+1)i!}+c\]

 
I also mentioned Ramanujan’s remarkable formula for calculating \(pi\) which gives roughly 8 decimal places per term calculated in the series. This contrasts with the above formula which needs five billion terms for 10 digits of accuracy!

Ramanujan’s formula is:

\[\frac{1}{\pi}=\frac{2\sqrt{2}}{9801}\sum_{k=0}^\infty (4k)!\frac{(1103+26390k)}{(k!)^4396^{4k}}\]

The mathematics behind this is beautiful, and deep, and not something of which I am an expert, but even from an outsider’s point of view, it can still be seen as something remarkable. There is a nice review paper here which discusses some of the maths behind this result: