The power series expansion of $\tan^{-1}(x)$ is $$\tan^{-1}(x)=x-\frac 13 x^3+\frac 15 x^5-\frac 17 x^7+ \cdots .$$ Use the above series to determine a series for calculating $\pi$.
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Mhenni Benghorbal
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Kyle Matthew
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Your post is presently (close to) unreadable. For some basic information about writing math at this site see e.g. here, here, here and here. – Lord_Farin Apr 22 '13 at 11:17
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just subs $x=1$ in the series you are given. – Mhenni Benghorbal Apr 22 '13 at 11:34
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Why don't you go with this – ABC Apr 22 '13 at 11:49
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Your series is : $$\arctan(x)=x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+\cdots$$ (this is easy to obtain from $\ \displaystyle\arctan(x)=\int \frac {dx}{1+x^2}$ using Taylor series)
A simple result is : $$\arctan(1)=\frac{\pi}4=1-\frac 13+\frac 15-\frac 17+\cdots$$
so that multiplying by $4$ ... but the convergence is very slow.
Other formulas are more useful like (John Machin 1706) and many others : $$\frac {\pi}4= 4\,\arctan\left(\frac 15\right)-\arctan\left(\frac 1{239}\right)$$ (see Wikipedia for other methods)
Raymond Manzoni
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