Does anyone know why Machin's original formula for $\pi$ converges so much faster than Leibniz's formula for $\pi$?
Machin's original formula: $\pi=\sum_{n=0}^\infty \frac{16(-1)^n239^{2n+1}-4(-1)^n5^{2n+1}}{1195^{2n+1}(2n+1)}$
Leibniz's formula: $\pi=\sum_{n=0}^\infty \frac{4(-1)^n}{2n+1}$
I think it has something to do with the centre of convergences, but am not sure.
On
Machin's formula is superior because the terms tend much more quickly to $0$ than the terms of Leibniz's formula.
The arctan-function has the expansion $$\arctan(x)=x-1/3x^3+1/5x^5-+\cdots$$ and the idea of Machin's formula is to cleverly combine values of the arctan-function with small arguments.
By the way : Very fast to $\pi$ converges the following sequence
$$x_1=3$$
$$x_{n+1}=x_n+\sin(x_n)$$
On
When you write Machin's formula as $$ \pi=\sum_{n=0}^\infty \frac{16(-1)^n239^{2n+1}-4(-1)^n5^{2n+1}}{1195^{2n+1}(2n+1)} , $$ then you may well wonder where it came from. But Machin used $$ \frac{\pi}{4} = \arctan\frac{1}{5} - \arctan\frac{1}{239} $$ and then applied the Tayor series of $\arctan$ evaluated at these two points. Of course that Taylor series converges faster at $1/5$ than at $1$.
Plug: I write this for undergraduates or advanced high school students. (And 1987-era computers.) There is a discussion of the rate of convergence of the Machin series.
G. A. Edgar, "Pi: difficult or easy?" Math. Mag. 60 (1987) 141--150.
Both formulas are based on the Taylor development of the arc tangent,
$$\sum_n\frac{(-1)^nx^{2n+1}}{2n+1}.$$
The absolute ratio of successive terms is
$$\frac{2n-1}{2n+1}x^2.$$
So when $x=1$ (Leibnitz), the general term decreases desperately slowly, while in the Machin formula, it decreases by more than a factor $25$ on every new term, giving each time more than one exact digit.
There are much faster formulas (Machin isn't used anymore for the computation records), and even a formula that gives the $n^{th}$ digit without computing the previous (though in base $16$ only).