Calculating Pi using Elementary Calculus Calculating Pi using Elementary Calculus
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... .
Method 1
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
sin(x/2) = sqrt((1-cos(x))/2), cos(x/2) = sqrt((1+cos(x))/2), 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:
| n | sin(pi/n) | cos(pi/n) | n·sin(pi/n) |
| 6 | 0.50000000 | 0.86602540 | 3.00000000 |
| 12 | 0.25881904 | 0.96592582 | 3.10582854 |
| 24 | 0.13052619 | 0.99144486 | 3.13262868 |
| 48 | 0.06540312 | 0.99785892 | 3.13935027 |
| 96 | 0.03271908 | 0.99946458 | 3.14103199 |
| 192 | 0.01636173 | 0.99986613 | 3.14145242 |
| 384 | 0.00818113 | 0.99996653 | 3.14155767 |
Method 2
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
pi = 2·sqrt(3)(1 - 1/3·3 + 1/5·32 - 1/7·33 + 1/9·34 - ...) 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
pi = n(tan(pi/n) - tan3(pi/n)/3 + tan5(pi/n)/5 - tan7(pi/n)/7 + tan9(pi/n)/9 - ...). 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.
| n | S5(tan(pi/n)) | S10(tan(pi/n)) |
| 6 | 3.14260474566308467280 | 3.14159051063808009964 |
| 12 | 3.14159317934909787183 | 3.14159265358927023954 |
| 24 | 3.14159265403260366883 | 3.14159265358979323810 |
-- Steven Weintraub
