A MOOC on Numerical Methods Released

After the rigorous and comprehensive development and assessment of the NSF funded innovative open courseware on Numerical Methods since 2002, we are offering a FREE Massive Open Online Course (MOOC) in Numerical Methods – Part 1 of 2 at https://www.canvas.net/browse/usflorida/courses/numerical-methods

The MOOC is Part 1 of a two-part course in Numerical Methods.  The course covers the mathematical procedures of differentiation, nonlinear equations and simultaneous linear equations.  We had the MOOC on Udemy but we are migrating it to CANVAS in two stages. CANVAS has a broader appeal for free MOOCs, it has a user friendly interface, looks familiar for many students using CANVAS, and has the capability of online quizzes that are algorithmic.

Start your journey today whether you are learning numerical methods for the first time or just need a refresher.  Unlike other MOOCs, you have a lifetime access to the course and you can pace yourself. Ask questions within the course and we will keep the conversation going!


About: Numerical methods are techniques to approximate mathematical procedures (example of a mathematical procedure is an integral).  Approximations are needed because we either cannot solve the procedure analytically (example is the standard normal cumulative  distribution function) or because the analytical method is intractable (example is solving a set of a thousand simultaneous linear equations for a thousand unknowns).

Materials Included: Textbook Chapters, Video Lectures, Quizzes, Solutions to Quizzes

How Long to Complete: About 20 hours of lectures need to be watched and estimated  time to read textbook and do quizzes is 40 hours.  It is a typical 7-week semester length course.

Course Structure: For each section, you have video lectures, followed by a textbook chapter, a quiz and its complete solution, and automatically graded online quizzes.


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Rejecting roots of nonlinear equation for a physical problem?

To make the fulcrum of a bascule bridge, a long hollow steel shaft called the trunnion is shrink fit into a steel hub. The resulting steel trunnion-hub assembly is then shrink fit into the girder of the bridge. This is done by first immersing the trunnion in a cold medium such as dry-ice/alcohol mixture.  After the trunnion reaches the steady state temperature of the cold medium, the trunnion outer diameter contracts.  The trunnion is taken out of the medium and slid though the hole of the hub.   When the trunnion heats up, it expands and creates an interference fit with the hub.


In 1995, on one of the bridges in USA, this assembly procedure did not work as designed.  Before the trunnion could be inserted fully into the hub, the trunnion got stuck.  So a new trunnion and hub had to be ordered at a cost of $50,000.  Coupled with construction delays, the total loss was more than hundred thousand dollars.

Why did the trunnion get stuck?  This was because the trunnion had not contracted enough to slide through the hole.

Now the same designer is working on making the fulcrum for another bascule bridge.  Can you help him so that he does not make the same mistake?

For this new bridge, he needs to fit a hollow trunnion of outside diameter  in a hub of inner diameter .  His plan is to put the trunnion in dry ice/alcohol mixture (temperature of the fluid – dry ice/alcohol mixture is -108 degrees F) to contract the trunnion so that it can be slided through the hole of the hub.  To slide the trunnion without sticking, he has also specified a diametrical clearance of at least 0.01 inches  between the trunnion and the hub.   What temperature does he need to cool the trunnion to so that he gets the desired contraction?

For the solution of this problem click here

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End of semester grading VBA module

In spite of learning management systems making assigning letter grades simpler, these systems still leave much to be desired when asked to incorporate extra credit or curving of an assessment grade.  To overcome this drawback, I download the grades to a excel spreadsheet and calculate the overall score of each student.  Based on the overall score, one needs to then assign a letter grade for the transcripts.  To assign 13 different plus/minus grading letters manually can be a good candidate for making a mistake in a large size class.  So I use a VBA code (how to access VBA editor) to assign letter grades.

c1 = overall percentage score
fungrade = letter grade

Autar Kaw, End of semester grading VBA module, last retrieved at https://autarkaw.wordpress.com/2015/12/21/end-of-semester-grading-vba-module/, December 21, 2015.

Function fungrade (c1 As Integer) As String
Dim gra As String
Select Case c1
Case Is >= 98
gra = “A+”
Case Is >= 90
gra = “A”
Case Is >= 86
gra = “A-”
Case Is >= 83
gra = “B+”
Case Is >= 80
gra = “B”
Case Is >= 76
gra = “B-”
Case Is >= 73
gra = “C+”
Case Is >= 70
gra = “C”
Case Is >= 66
gra = “C-”
Case Is >= 63
gra = “D+”
Case Is >= 60
gra = “D”
Case Is >= 56
gra = “D-”
Case Is >= 0
gra = “F”
End Select
fungrade = gra
End Function

PS. You can modify the above given VBA code as needed.  If you want a less hard-coded version, you can modify by having two more inputs – 1) a vector of lowest limit of score for a particular letter grade, and 2) a corresponding vector of the same length of letter grades.

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A quadrature formula example

To relate how quadrature formulas are derived, see a simple example of how to do so

See the pdf file for solution.


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Example to show that a polynomial of order n or less that passes through (n+1) data points is unique.

Problem: Through three data pairs (0,0), (3,9) and (4,12), an interpolating polynomial of order 2 or less is found to be y=3x. Prove that there is no other polynomial of order 2 or less that passes through these three points.

See the pdf file for solution.


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Why multiply possible form of part of particular solution form by a power of the independent variable when solving an ordinary differential equation

When solving a fixed-constant linear ordinary differential equation where the part of the homogeneous solution is same form as part of a possible particular solution, why do we get the next independent solution in the form of x^n* possible form of part of particular solution?  Show this through an example.

See the pdf file


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Largest number that can be stored in a floating word of 7 bits

QUESTION: What is the largest base-10 positive number that can be stored using 7 bits, where the 1st bit is used for the sign of the number; the 2nd bit for sign of the exponent; 3 bits for mantissa, and the rest of the bits for the exponent?

ANSWER: Remember the base is 2.
1st bit will need to be zero as the number is positive.

2nd bit will need to be zero as that will make the exponent positive as 2^positive. number will give higher number than 2^negative number.

The mantissa bits will need to be 111 as you are looking for largest number and that will give the number to be 1.111 (the 1 before radix point is automatic) in base of 2 or 1*2^0+1*2^(-1)+1*2^(-2)+1*2^(-3)=1.875 in base of 10.

Now the exponent: it uses 2 bits. This will need to be 11 in base 2 and that is 3 in base 10. So the exponent part is 2^(+3)=8.

Largest number is +1.875*8=15

Now think what will give you the smallest positive number.


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