Power Serieshard
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The power series representation of arctan(x)=n=0(1)nx2n+12n+1\arctan(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} converges for x[1,1]x \in [-1, 1]. Using Machin's formula π=12arctan(1/2)4arctan(1/5)\pi = 12\arctan(1/2) - 4\arctan(1/5) and the power series, which truncation gives the best estimate?