Problem 43003. Simpsons's rule (but not Homer Simpson)

Created by John D'Errico

Solve Later

{"relationships":[{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/document","targetMode":"","relationshipId":"rId1","target":"/matlab/document.xml"},{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/output","targetMode":"","relationshipId":"rId2","target":"/matlab/output.xml"}],"parts":[{"partUri":"/matlab/document.xml","relationship":[],"contentType":"application/vnd.mathworks.matlab.code.document+xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<w:document xmlns:w=\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\"><w:body><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>I wonder what Homer Simpson would have thought of Simpson's rule? Somehow I doubt his thoughts would have included the phrase Newton-Cotes, or even numerical integration.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>In this problem, I want you to use</w:t></w:r><w:r><w:t> </w:t></w:r><w:hyperlink w:docLocation=\"https://en.wikipedia.org/wiki/Simpson%27s_rule\"><w:r><w:t>Simpson's rule</w:t></w:r></w:hyperlink><w:r><w:t> to integrate a function, provided as a list of points in a vector. The points must be equally spaced in x, so all that need be provided is the spacing between points in x. As well, I'll be nice and always give you an odd number of points, since that is important for Simpson's rule to work smoothly.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>As a test case, we should know that the integral of cos(x) over the interval [0,pi/2] is 1, at least analytically that is true. Your Simpson's rule code, for only 7 points spanning that interval will do pretty well:</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[deltax = pi/12;\nFx = cos(linspace(0,pi/2,7));\nsimpsInt(Fx,deltax)\nans =\n 1.00002631217059]]></w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Thus, Fx is a vector of supplied function values, and deltax is the step size taken on the x-axis.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Homer would be proud.</w:t></w:r></w:p></w:body></w:document>"},{"partUri":"/matlab/output.xml","contentType":"text/xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"no\" ?><embeddedOutputs><metaData><evaluationState>manual</evaluationState><layoutState>code</layoutState><outputStatus>ready</outputStatus></metaData><outputArray type=\"array\"/><regionArray type=\"array\"/></embeddedOutputs>"}]}

I wonder what Homer Simpson would have thought of Simpson's rule? Somehow I doubt his thoughts would have included the phrase Newton-Cotes, or even numerical integration.In this problem, I want you to use <a href = "https://en.wikipedia.org/wiki/Simpson%27s_rule">Simpson's rule</a> to integrate a function, provided as a list of points in a vector. The points must be equally spaced in x, so all that need be provided is the spacing between points in x. As well, I'll be nice and always give you an odd number of points, since that is important for Simpson's rule to work smoothly.As a test case, we should know that the integral of cos(x) over the interval [0,pi/2] is 1, at least analytically that is true. Your Simpson's rule code, for only 7 points spanning that interval will do pretty well:<pre class="language-matlab">deltax = pi/12;
Fx = cos(linspace(0,pi/2,7));
simpsInt(Fx,deltax)
ans =
 1.00002631217059
</pre>Thus, Fx is a vector of supplied function values, and deltax is the step size taken on the x-axis.Homer would be proud.

Solve

Solution Stats

84 Solutions
24 Solvers

Last Solution submitted on Dec 27, 2020

Last 200 Solutions

Problem Comments

8 Comments

8 Comments

Show 5 older comments

John D'Errico on 1 Oct 2016

If this problem gains some interest, I'll start trying to post additional problems. My idea is to build a new problem set, for general numerical methods like this.

Peng Liu on 1 Oct 2016

Great problem along with a very clear description. Looking forward to your addtional problems in numerical analysis.

John D'Errico on 2 Oct 2016

I've asked Ned how to create a new problem group first. (I don't think if I have that privilege.) Then I plan to establish a few new problem groups for problems like this. Perhaps a numerical methods group, a computational geometry group, others? I hope there will also be some existing problems I can then include in addition to those I will write.

Peng Liu on 2 Oct 2016

Great idea to create a new problem group! Numerical analysis is definitely good problem group to start with. There are already some numerical integration problems which I have solved on Cody (such as a general integration problem: https://www.mathworks.com/matlabcentral/cody/problems/1031-composite-trapezoidal-rule-for-numeric-integration, and a problem dealing with trapezoidal rule: https://www.mathworks.com/matlabcentral/cody/problems/1031-composite-trapezoidal-rule-for-numeric-integration). I guess these problems can be included in a numerical analysis group.

Peng Liu on 2 Oct 2016

Sorry for the wrong link. A general numerical integration problem: https://www.mathworks.com/matlabcentral/cody/problems/1197-numerical-integration, and a trapezoidal rule problem: https://www.mathworks.com/matlabcentral/cody/problems/1031-composite-trapezoidal-rule-for-numeric-integration. I realize that Cody does not allow users to modify/delete their own comments, which is very inconvenient.

John D'Errico on 2 Oct 2016

Thanks for the problem links. I'll forward that idea about editing comments to the site developers.

James on 3 Oct 2016

Nice problem, but #7 in the Test Suite needs an assert in it.

Giovanni Mottola on 4 Oct 2016

There's something I don't understand. The first function I wrote, according to Simpson's rule, passed all tests except #6; here the absolute error was still very small (~1.53e-15) but greater than the tolerance, so my first solution failed the test set. Thus I had to add a small correction factor of 1e-15. Now the function passes all tests, but I can't understand the error in my implementation.

Solution Comments

Solution 992732

1 Comment

1 Comment

Jiawei Gong on 26 Apr 2020

I used the same formula and failed test 6. I think it's because of the machine error eps. Is that why you had 1e-15 at the end?

Problem Recent Solvers24

Problem Tags

integration numerical methods

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Problem 43003. Simpsons's rule (but not Homer Simpson)

Solution Stats

Last 200 Solutions

Problem Comments

Solution Comments

Problem Recent Solvers24

Suggested Problems

More from this Author4

Problem Tags

Community Treasure Hunt

Problem 43003. Simpsons's rule (but not Homer Simpson)

Solution Stats

Last 200 Solutions

Problem Comments

Solution Comments

Problem Recent Solvers24

Suggested Problems

More from this Author4

Problem Tags

Community Treasure Hunt

Players