Calculating Pi using Elementary CalculusThere are many powerful and sophisticated methods known for the calculation of pi, and, using these methods, pi has been calculated to billions of decimal places. Here we present two methods for the calculation of pi which only use elementary calculus, but nevertheless are surprisingly effective. By way of comparison, note that pi = 3.14159265358979323846... . Method 1This method uses the fact that
Setting x = pi/n, we see that, for n large, pi is approximately equal to n·sin(pi/n). To use this, we must be able to compute sin(pi/n). Recall the trigonometric identities
valid for 0 <= x <= pi, where we take the positive square roots. Thus, if we know sin(pi/n) and cos(pi/n), we may compute sin(pi/2n) and cos(pi/2n), and then proceed recursively. Of course, we know sin(pi/6) = 1/2 and cos(pi/6) = sqrt(3)/2. We thus obtain the following table:
Method 2Beginning with the integral
and writing 1/(1 + t^{2}) = 1  t^{2} + t^{4}  t^{6} + ... and integrating termbyterm, we obtain the infinite series
which converges for 1 <= x <= 1. Substituting x = 1, we obtain Leibniz's series
However, because x = 1 is an endpoint of the interval of convergence, this series converges very slowly and so is impractical for calculation. (For example, the first five terms give the approximations 4., 2.666, 3.466, 2.895, 3.339 to pi.) On the other hand, we may substitute x = 1/sqrt(3) = tan^{1}(pi/6). Then, after some algebraic manipulation, we obtain the series
and the first five terms of this series give the approximations 3.46410, 3.07920, 3.15618, 3.13785, 3.14260 to pi, the last of which differs from pi by litle more than .001. Indeed, we can do even better. Substituting x = tan(pi/n) for n >= 4 (so that the series converges), we obtain the series
Now tan(x/2) = sqrt((1  cos(x))/(1 + cos(x))), so we may again compute tan(pi/n) recursively, starting with n = 6. We obtain the following approximations to pi, where in the last two columns of this table, S_{k}(tan(pi/n)) denotes the approximation obtained by using the first k terms of the above series.
 Steven Weintraub

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