Why is Hutton's formula best for calculating pi? We are asked to prove that $\frac{\pi}{4} = 2\arctan{\frac{1}{3}} + \arctan{\frac{1}{7}}$ (Hutton's formula) which I have managed to do. We are then told to consider why this formula is better to calculate the decimal digits of pi using the Maclaurin series for arctan(x) than $\frac{\pi}{4} = \arctan(1)$. I am assuming it has something to do with efficiency but can anyone explain it to me please? Thanks
 A: If you consider the series $\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} = \arctan(1)$, you can notice that this is a conditionally convergent series.  Heuristically you can think of this as a slowly converging series.  Notice that the individual terms decrease at a rate of $1/(2n+1)$
On the other hand the series $\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \frac{1}{3^{2n+1}} = \arctan(1/3)$ and the series $\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \frac{1}{7^{2n+1}} = \arctan(1/7)$ are both dominated by geometric series.  They would converge faster, since the individual terms exponentially go to zero.
A: To understand why this is true, you should try compute the truncated series
$$\sum_{n=0}^{N} \frac{(-1)^nx^{2n+1}}{2n+1}$$ (with $x=1$, $x=1/3$ and $x=1/7$) and see to which value of $N$ you have to do it to get the second decimal of $\pi$ right. 
A: The MacLaurin expansion of a function is basically the best approximation of the function by some polynomial around zero. 
Thus the difference between the function and its MacLaurin expansion is smaller when you are closer to zero, thats why one would get a more accurate answer using Hutton's formula instead of approximating $\arctan(1)$ with its MacLaurin series.
