Visualising the Convergence of Pi, Calculated by the Leibniz Method
This post is one of a series of articles adapted from my university dissertation on data visualisation — see the rest of the series.
Pi (π), the ratio of a circle's circumference to its diameter, is a mathematical phenomenon. It is found in a massive variety of fields such as statistics, geometry, complex numbers, calculus and physics. It is irrational, which means it cannot be expressed as the quotient of two whole numbers (and therefore its decimal digits continue infinitely). Two fraction approximations are 22/7, which is slightly larger than the exact value of π, and 355/113, which is somewhat more accurate but still slightly larger than the true value.
There are a number of methods to generate the stream of decimal digits in π. With a modern standard computer it is fairly simple to write a program to calculate the digits. A formula I explored was discovered by the German mathematician Gottfried Leibniz, whereby the value of a quarter of π could be defined by the infinite series:
π/4 = 1 − 1/3 + 1/5 − 1/7 + 1/9 − ...
This can also be written as an infinite series using summation notation:
∑ (n=0 to ∞) [(−1)ⁿ / (2n + 1)] = π/4
Which can be formulated by proving that:
π/4 = arctan(1) = ∫₀¹ 1/(1 + x²) dx
And so I wrote a MATLAB program which ran iterations, giving a more accurate approximation for π on each iteration. Note that to find π we can multiply both sides by 4.
p = 0;
i = 0;
while 1 == 1 % Infinite loop
p = p + (4*((-1)^i))/(2*i + 1);
fprintf('%7i Iterations: Pi is approximately equal to %1.16f...\n',i+1,p)
i = i + 1;
end
This program printed the approximation generated after each iteration:
1 Iterations: Pi is approximately equal to 4.0000000000000000...
2 Iterations: Pi is approximately equal to 2.6666666666666670...
3 Iterations: Pi is approximately equal to 3.4666666666666668...
4 Iterations: Pi is approximately equal to 2.8952380952380956...
5 Iterations: Pi is approximately equal to 3.3396825396825403...
6 Iterations: Pi is approximately equal to 2.9760461760461765...
7 Iterations: Pi is approximately equal to 3.2837384837384844...
8 Iterations: Pi is approximately equal to 3.0170718170718178...
9 Iterations: Pi is approximately equal to 3.2523659347188767...
10 Iterations: Pi is approximately equal to 3.0418396189294032...
The first few iterations are very inaccurate. It takes 628 iterations to converge to 3.14 (2 decimal places). An interesting visualisation is simply the printed lines of the approximations, as they are printed to the screen in extremely quick succession, so you can literally see each digit in turn flick between two values and then settle at the correct digit:
611 Iterations: Pi is approximately equal to 3.1432293137049125...
612 Iterations: Pi is approximately equal to 3.1399586677523370...
613 Iterations: Pi is approximately equal to 3.1432239738747860...
614 Iterations: Pi is approximately equal to 3.1399639901747047...
615 Iterations: Pi is approximately equal to 3.1432186687751931...
616 Iterations: Pi is approximately equal to 3.1399692780359567...
617 Iterations: Pi is approximately equal to 3.1432133980683981...
618 Iterations: Pi is approximately equal to 3.1399745316716370...
619 Iterations: Pi is approximately equal to 3.1432081614210308...
620 Iterations: Pi is approximately equal to 3.1399797514129597...
621 Iterations: Pi is approximately equal to 3.1432029585040153...
622 Iterations: Pi is approximately equal to 3.1399849375868794...
623 Iterations: Pi is approximately equal to 3.1431977889925018...
624 Iterations: Pi is approximately equal to 3.1399900905161586...
625 Iterations: Pi is approximately equal to 3.1431926525657983...
626 Iterations: Pi is approximately equal to 3.1399952105194355...
627 Iterations: Pi is approximately equal to 3.1431875489073047...
628 Iterations: Pi is approximately equal to 3.1400002979112887...
629 Iterations: Pi is approximately equal to 3.1431824777044470...
630 Iterations: Pi is approximately equal to 3.1400053530023024...
Notice how the second decimal place flicks between 3 and 4 before finally settling on 4. As you can see, the value starts at 4 (too high), then subtracts 4/3 to land at 2.667 (too low), then adds 4/5 to land on 3.467 (closer but still too high), then subtracts 4/7 to land on 2.895 (closer but still too low), and so on, constantly adding too much on, then taking too much off, always getting closer. This can be visualised by plotting a line going from iteration to approximation:
p = 0;
S = [];
for i = 0:20
p = p + (4*((-1)^i))/(2*i + 1);
S = [S p];
plot(S)
pause(0.1)
end
Which generates the plot:
This gives a very good visualisation of how the Leibniz formula works, by adding on an amount, taking off a smaller amount, then adding on a smaller amount, and taking off a smaller amount, getting ever closer to the true value.
I also noticed that a better approximation could be made by taking an average after every 2 iterations:
(too high + too low) / 2 = about right
This "about right" value was always much closer, because we know that the value for π has to lie somewhere in the middle of the last two approximations, due to the nature of the formulation method. And the "about right" average will get much more accurate much quicker. I revised the program:
p = 0;
S = [];
i = 0;
while 1 == 1
p = p + (4*((-1)^i))/(2*i + 1);
if rem(i,2) == 0
p1 = p;
S = [S p1];
try
plot(length(S)-10:length(S),S(length(S)-10:end))
axis([length(S)-10 length(S) 2 4])
pause(0.1)
end
else
p2 = p;
S = [S p2];
try
plot(length(S)-10:length(S),S(length(S)-10:end))
axis([length(S)-10 length(S) 2 4])
pause(0.1)
end
end
i = i + 1;
end
The iterations were now coming out as:
4.000000000000000
2.666666666666667
3.466666666666667
2.895238095238096
3.339682539682540
2.976046176046177
3.283738483738484
3.017071817071818
3.252365934718877
3.041839618929403
3.232315809405594
Which are obviously a lot more accurate more quickly. After 50 iterations the plot starts to lock in on a much smaller margin between the upper and lower limit, and almost looks like a straight line after 250 iterations:
