I am trying to solve an exercise regarding the calculation of $\pi$ via power sums, but I am puzzled by the wording of a question.
a) Show that $(5+i)^4(239-i)$ is a real positive multiple of $1+i$
b) Show Machin's Formula:
$$\frac{\pi}{4}=4 \arctan \frac{1}{5}-\arctan \ \frac{1}{239}$$
c) Find the power series of arctan around 0. This can be done for
instance by integrating the power series of its derivative.
d) Use partial sums for arctan to calculate estimates for $\pi$. How does this method compare to calculating estimates via $\pi=4 \arctan(1)$?
I solved a)-c) but have gotten stuck at d). I know from c) that the power series of $\arctan$ at $0$ is given by
$$\arctan(x)=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}$$
But I am at a loss what to do next, as I do not see how to calculate $\pi$ in any other way than using
$$\pi=\arctan(1)=\sum_{n=0}^\infty (-1)^n \frac{1^{2n+1}}{2n+1}$$
Can somebody explain what is supposed to be done here or how this is question is solved?
Best Answer
I think it’s just asking you to substitute $\frac15$ and $\frac1{239}$ into the series to get approximations for $\pi$ and to note that the convergence is much better than for $\arctan1$ because of the additional factor of $\frac1{25}$ per step.