Approximation of pi is "too precise" .

The function below should approximate pi adding about 2 digits of precision for increasing n. Why is piApprox2(3) exactly = pi and not just close to pi? Is their any built in knowledge Matlab uses, that I am not aware of?

When extracting 10 Mio digits of the result P (via Pc = char(vpa(P,10000000))) and looking at the last 10 characters of Pc, they are exactly the same as when looking at these digits in pi.

function P = piApprox2(n)
% approximates pi using n approximation steps of formula (7) from: 
% https://www.davidhbailey.com/dhbpapers/pi-formulas.pdf 
s1 = 0;
s2 = 0;
s3 = 0;
s4 = 0;
for k = 0:n
    s1 = s1 + (-1)^k / ((2*k+1)*57^(2*k+1));
    s2 = s2 + (-1)^k / ((2*k+1)*239^(2*k+1));
    s3 = s3 + (-1)^k / ((2*k+1)*682^(2*k+1));
    s4 = s4 + (-1)^k / ((2*k+1)*12943^(2*k+1)); 
end
P = 4 * ( 44*s1 + 7*s2 - 12*s3 + 24*s4);
end

[Walter Roberson]

vpa(P,10000000)

You should be using

vpa(sym(P,'f'),10000000)

because the default conversion of double to symbolic involves approximating.

[Thomas]

Thank you very much - that explains it!

Answer 1

[Image Analyst]

For what it's worth, when I tried a pi estimation series, it was so accurate that after 2 terms, MATLAB couldn't tell the difference from pi. So I had a similar problem.

% Computes Ramanujan's formula for pi.
% http://www.scientificamerican.com/article/equations-are-art-inside-a-mathematicians-brain/
% On the bottom of the heap of beautiful formulas, mathematicians consistently rated 
% Srinivasa Ramanujan's infinite series for estimating pi as the most ugly.

% Problem: convergence is so fast that double precision is not
% precise enough to show the convergence curve.

clc;    % Clear the command window.
close all;  % Close all figures.
clear;  % Erase all existing variables. Or clearvars if you want.
workspace;  % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 24;

% Compute the sum for a bunch of different number of terms
% in the series to see the shape of the error curve
% as it converges towards the true value of pi.
for kMax = 1 : 10 
	thisSum = 0;
	for k = 0 : kMax
		numerator = factorial(4 * k) * (1103 + 26390*k);
		denominator = factorial(k)^4 * 396^(4*k);
		thisTerm = numerator / denominator;
		thisSum = thisSum + thisTerm;
	end
	theConstant = 2 * sqrt(2) / 9801;
	oneOverPi = theConstant * thisSum;
	estimatedPi(kMax) = 1 / oneOverPi
	theError(kMax) = estimatedPi(kMax) - pi
end

% Plot the value of estimated pi.
subplot(2, 1, 1);
plot(estimatedPi, 'b*-', 'LineWidth', 2);
grid on;
title('Estimated \pi', 'FontSize', fontSize);
xlabel('Number of Terms in the Sum', 'FontSize', fontSize);

% Plot the value of the error (difference between true pi and estimated pi).
subplot(2, 1, 2);
plot(theError, 'b*-', 'LineWidth', 2);
grid on;
title('Error', 'FontSize', fontSize);
xlabel('Number of Terms in the Sum', 'FontSize', fontSize);

% Set up figure properties:
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
% Get rid of tool bar and pulldown menus that are along top of figure.
set(gcf, 'Toolbar', 'none', 'Menu', 'none');
% Give a name to the title bar.
set(gcf, 'Name', 'Estimation of pi', 'NumberTitle', 'Off') 

I don't have the symbolic toolbox so I guess I'm out of luck. Any way to see convergence without that toolbox?