There 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... .
This 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:
Beginning with the integral
and writing 1/(1 + t2) = 1 - t2 + t4 - t6 + ... and integrating term-by-term, we obtain the infinite series
tan-1(x) = x - x3/3 + x5/5 - x7/7 + ... ,
which converges for -1 <= x <= 1. Substituting x = 1, we obtain Leibniz's series
pi = 4(1 - 1/3 + 1/5 - 1/7 + 1/9 - ...) .
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
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, Sk(tan(pi/n)) denotes the approximation obtained by using the first k terms of the above series.
-- Steven Weintraub
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