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Demo on the Trapezoidal Rule and Simpson's Rule

Contents

The Trapezoidal RuleSimpson's Rule

The Trapezoidal Rule

The idea of the trapezoidal rule is to approximate a general curve by trapezoids, like this. Weillustrate with the problem of integrating sin(x) from 0 to pi. Of course, the true value of theintegral is 2.

ezplot('sin(x)', [0, pi]), hold on approx = zeros(1,7); %initialize vector of results for  j = 1:7

n = 2^j;x = pi*(0:1/n:1);plot(x, sin(x), 'r')weights = [1, 2*ones(1,n‐1), 1];approx(j) = pi/(2*n)*sin(x)*weights';

end disp('Using Trapezoidal Rule')disp(' n Approximation')for  j = 1:7

disp(['n = ', num2str(2^j, '%d'), ' ', num2str(approx(j), '%1.10f')])end 

Using Trapezoidal Rulen Approximation

n = 2 1.5707963268

n = 4 1.8961188979n = 8 1.9742316019n = 16 1.9935703438n = 32 1.9983933610n = 64 1.9995983886n = 128 1.9998996002

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Simpson's Rule

The idea of Simpson's rule is to approximate a general curve by arcs of parabolas, like this.We illustrate with the problem of integrating sin(x) from 0 to pi. Of course, the true value of 

the integral is 2.

figure, ezplot('sin(x)', [0, pi]), hold on x = 0:.01:1;plot(pi*x, 1 ‐ 4*(x‐.5).^2, 'r')approx = zeros(1, 7); %initialize vector of results for  j = 1:7

n = 2^j;x = pi*(0:1/n:1);

weights = [1,2*ones(1,n‐1)+2*mod([1:n‐1],2),1] ;approx(j) = pi/(3*n)*sin(x)*weights';

end 

disp('Using Simpson Rule')disp(' n Approximation')for  j = 1:7

disp(['n = ', num2str(2^j, '%d'), ' ', num2str(approx(j), '%1.10f')])end 

Using Simpson Rulen Approximation

n = 2 2.0943951024n = 4 2.0045597550n = 8 2.0002691699n = 16 2.0000165910n = 32 2.0000010334n = 64 2.0000000645n = 128 2.0000000040

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 Note that for smooth functions, the error in Simpson's rule is roughly proportional to 1/n^4,so that it doesn't require a large value of n to get reasonable accuracy. However, for very

 jagged functions, the trapezoidal rule can be more accurate. Here is a program to compute theSimpson's rule approximation to an integral, along with some examples.

type Simpson 

 function Q = Simpson(fun, a, b, n) %SIMPSON Numerically evaluate integral, using Simpson's rule. % syntax: Q = Simpson(fun, a, b, n) % FUN should be a vectorized function defined on the interval % from a to b. n should be a positive even integer. % example: Simpson(@sin, 0, pi,4) returns the approximation % 2.0046 to the integral 2 of the sine function from 0 to pi.

if mod(n,2)error('n must be even')

end;weights = [1,2*ones(1,n‐1)+2*mod([1:n‐1],2),1] ;

 x = a:(b‐a)/n:b;Q = (b‐a)/(3*n)*fun(x)*weights';return

Simpson(@(x) x.^3, 0, 1, 2)

ans =

0.2500

The above example shows that Simpson's rule is perfectly accurate for cubic polynomialfunctions. On the other hand, the square root function has a derivative that blows up at x= 0,so in this case the results won't be as good, even with a much bigger value of n:

Simpson(@sqrt, 0, 1, 2)

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ans =

0.6381

Simpson(@sqrt, 0, 1, 16)

ans =

0.6654

The true value of the ingral of sqrt(x) from 0 to 1 is 2/3, so the error as a function of n lookslike:

for  j = 1:10disp(['n = ', num2str(2^j, '%d'), ' ', 'error =', ... 

num2str(2/3 ‐ Simpson(@sqrt, 0, 1, 2^j), '%1.10f')])end 

n = 2 error =0.0285954792n = 4 error =0.0101404019

n = 8 error =0.0035873866n = 16 error =0.0012684780n = 32 error =0.0004484839n = 64 error =0.0001585636n = 128 error =0.0000560607n = 256 error =0.0000198205n = 512 error =0.0000070076n = 1024 error =0.0000024776

One can see from this example that doubling n still decreases the error by more than a factor of 2, but by less than a factor of 16, more like a factor of 3.

 Publ is hed wit h MATLAB® 7 .0.1