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12 years, 49 days

Dr. Robert J. Lopez, Emeritus Professor of Mathematics at the Rose-Hulman Institute of Technology in Terre Haute, Indiana, USA, is an award winning educator in mathematics and is the author of several books including Advanced Engineering Mathematics (Addison-Wesley 2001). For over two decades, Dr. Lopez has also been a visionary figure in the introduction of Maplesoft technology into undergraduate education. Dr. Lopez earned his Ph.D. in mathematics from Purdue University, his MS from the University of Missouri - Rolla, and his BA from Marist College. He has held academic appointments at Rose-Hulman (1985-2003), Memorial University of Newfoundland (1973-1985), and the University of Nebraska - Lincoln (1970-1973). His publication and research history includes manuscripts and papers in a variety of pure and applied mathematics topics. He has received numerous awards for outstanding scholarship and teaching.

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These are Posts that have been published by rlopez

At 3:00 PM EST on Thursday, December 15, Maplesoft hosted a momentous hour in my life, my "retirement party" ending my career at Maplesoft. It was a day I had planned some four years ago when I dropped to a lighter schedule, and a day my wife has been awaiting for six years.

Jim Cooper, CEO at Maplesoft, presented a very brief sketch of some milestones in my life, including my high school graduation in 1958, BA in 1963, MS in 1966, PhD in 1970, jobs at the University of Nebraska-Lincoln, Memorial University of Newfoundland, and the Rose-Hulman Institute of Technology. There was a picture of me taken from my high school graduation yearbook. There was a cake. There were kind words about my contributions to Maple, including "Clickable Calculus," the term and its meaning.

I was handed the microphone - I knew what I wanted to say. My wife was present in the gathering. I pointed to her and said that all the congratulations should go to her who had waited so patiently for my retirement for six years. I thanked Maplesoft and all its employees for nearly 14 of the best years of my life, for I have thoroughly enjoyed my return to Canada and my work (more like play) at Maplesoft. 

It's been a great opportunity to be part of the Maple experience, and now it's time for new ones. There'll be more woodworking in my basement woodshop where I make mostly noise and sawdust, some extra travel, more exercise and fresh air, long-delayed household projects, and whatever else my mate of 49 years asks.

But the best part of all is that I'll still have a connection to Maplesoft - I'll continue doing two webinars a month, will maintain and update much of the content I've created for Maple while at Maplesoft, and contribute additional content of relevance to the Maple community. 

A population p(t) governed by the logistic equation with a constant rate of harvesting satisfies the initial value problem diff(p(t), t) = (2/5)*p(t)*(1-(1/100)*p(t))-h, p(0) = a. This model is typically analyzed by setting the derivative equal to zero and finding the two equilibrium solutions p = 50+`&+-`(5*sqrt(100-10*h)). A sketch of solutions p(t) for different values of a suggests that the larger equilibrium is stable; the smaller, unstable.


When a is less that the unstable equilibrium, p(t) becomes zero at a time t[e], and the population becomes extinct. If p(t) is not interpreted as pertaining to a population, its graph exists beyond t[e], and actually has a vertical asymptote between the two branches of its graph.


In the worksheet "Logistic Model with Harvesting", two questions are investigated, namely,


  1. How does the location of this vertical asymptote depend on on a and h?
  2. How does the extinction time t[e], the time at which p(t) = 0, depend on a and h?

To answer the second question, an explicit solution p = p(a, h, t), readily provided by Maple, is set equal to zero and solved for t[e] = t[e](a, h). It turns out to be difficult both to graph the surface t[e](a, h) and to obtain a contour map of the level sets of this function. Instead, we solve for a = a(t[e], h) and obtain a graph of a(h) with t[e] as a slider-controlled parameter.


To answer the first question, the explicit solution, which has the form alpha*tan(phi(a, h, t))*beta(h)+50, exhibits its vertical asymptote when phi(a, h, t) = -(1/2)*Pi. Solving this equation for t[a] = t[a](a, h) gives the time at which the vertical asymptote is located, a function that is as difficult to graph as t[e]. Again the remedy is to solve for, and graph, a = a(h), with t[a] as a slider-controlled parameter.


Download the worksheet: Logistic_with_Harvesting.mw

The Saturday edition of our local newspaper (Waterloo Region Record) carries, as part of The PUZZLE Corner column, a weekly puzzle "STICKELERS" by Terry Stickels. Back on December 13, 2014, the puzzle was:

What number comes next?

1   4   18   96   600   4320   ?

The solution given was the number 35280, obtained by setting k = 1 in the general term k⋅k!.

On September 5, 2015, the column contained the puzzle:

What number comes next?

2  3  3  5  10  13  39  43  172  177  ?

The proposed solution was the number 885, obtained as a10 from the recursion

where a0 =2.

As a youngster, one of my uncles delighted in teasing me with a similar question for the sequence 36, 9, 50, 55, 62, 71, 79, 18, 20. Ignoring the fact that there is a missing entry between 9 and 50, the next member of the sequence is "Bay Parkway," which is what 22nd Avenue is actually called in the Brooklyn neighborhood of my youth. These are subway stops on what was then called the West End line of the subway that went out to Stillwell Avenue in Coney Island.

Armed with the skepticism inspired by this provincial chestnut, I looked at both of these puzzles with the attitude that the "next number" could be anything I chose it to be. After all, a sequence is a mapping from the (nonnegative) integers to the reals, and unless the mapping is completely specified, the function values are not well defined.

Indeed, for the first puzzle, the polynomial f(x) interpolating the points

(0, 1), (1, 4), (2, 18), (3, 93), (4, 600), (5, 4320)

reproduces the first six members of the given sequence, and gives 18593 (not 35280) for f(7). In other words, the pattern k⋅k! is not a unique representation of the sequence, given just the first six members. The worksheet NextNumber derives the interpolating polynomial f and establishes that f(n) is an integer for every nonzero integer n.

Likewise, for the second puzzle, the polynomial g(x) interpolating the points

(1, 2) ,(2, 3) ,(3, 3) ,(4, 5) ,(5, 10) ,(6, 13) ,(7, 39) ,(8, 43), (9, 172) ,(10, 177)

reproduces the first ten members of the given sequence, and gives -7331(not 885) for g(11). Once again, the pattern proposed as the "solution" is not unique, given that the worksheet NextNumber contains both g(x) and a proof that for integer n, all values of g(n) are integers.

The upshot of these observations is that without some guarantee of uniqueness, questions like "what is the next number" are meaningless. It would be far better to pose such challenges with the words "Find a pattern for the given members of the following sequence" and warn that the function capturing that pattern might not be unique.

I leave it to the interested reader to prove or disprove the following conjecture: Interpolate the first n terms of either sequence. The interpolating polynomial p will reproduce these n terms, but for k>n, p(k) will differ from the corresponding member of the sequence determined by the stated patterns. (Results of limited numerical experiments are consistent with the truth of this conjecture.)

Attached: NextNumber.mw

In a recent blog post, I found a single rotation that was equivalent to a sequence of Givens rotations, the underlying message being that teaching, learning, and doing mathematics is more effective and efficient when implemented with a tool like Maple. This post has the same message, but the medium is now the Householder reflection.

Given the vector x = , the Householder matrix H = I - 2 uuT reflects x to y = Hx, where I is the appropriate identity matrix, u = (x - y) / ||x - y|| is a unit normal for the plane (or hyperplane) across which x is reflected, and y necessarily has the same norm as x. The matrix H is orthogonal but its determinant is -1, making it a reflection instead of a rotation.

Starting with x and uH can be constructed and the reflection y calculated. Starting with x and yu and H can be determined. But what does any of this look like? Besides, when the Householder matrix is introduced as a tool for upper triangularizing a matrix, or for putting it into upper Hessenberg form, a recipe such as the one stated in Table 1 is the starting point.

In other words, the recipe in Table 1 reflects x to a vector y in which all entries below the kth are zero. Again, can any of this be visualized and rendered more concrete? (The chair who hired me into my first job averred that there are students who can learn from the general to the particular. Maybe some of my classmates in graduate school could, but in 40 years of teaching, I've never met one such student. Could that be because all things are known through the eyes of the beholder?)

In the attached worksheet, Householder matrices that reflect x = <5, -2, 1> to vectors y along the coordinate axes are constructed. These vectors and the reflecting planes are drawn, along with the appropriate normals u. In addition, the recipe in Table 1 is implemented, and the recipe itself examined. If you look at the worksheet, I believe you will agree that without Maple, the explorations shown would have been exceedingly difficult to carry out by hand.

Attached: RHR.mw

This post describes how Maple was used to investigate the Givens rotation matrix, and to answer a simple question about its behavior. The "Givens" part is the medium, but the message is that it really is better to teach, learn, and do mathematics with a tool like Maple.

The question: If Givens rotations are used to take the vector Y = <5, -2, 1> to Y2 = , about what axis and through what angle will a single rotation accomplish the same thing?

The Givens matrix G21 takes Y to the vector Y1 =, and the Givens matrix G31 takes Y1 to Y2. Graphing the vectors Y, Y1, and Y2 reveals that Y1 lies in the xz-plane and that Y2 is parallel to the x-axis. (These geometrical observations should have been obvious, but the typical usage of the Givens technique to "zero-out" elements in a vector or matrix obscured this, at least for me.)

The matrix G = G31 G21 rotates Y directly to Y2; is the axis of rotation the vector W = Y x Y2, and is the angle of rotation the angle  between Y and Y2? To test these hypotheses, I used the RotationMatrix command in the Student LinearAlgebra package to build the corresponding rotation matrix R. But R did not agree with G. I had either the axis or the angle (actually both) incorrect.

The individual Givens rotation matrices are orthogonal, so G, their product is also orthogonal. It will have 1 as its single real eigenvalue, and the corresponding eigenvector V is actually the direction of the axis of the rotation. The vector W is a multiple of <0, 1, 2> but V = <a, b, 1>, where . Clearly, W  V.

The rotation matrix that rotates about the axis V through the angle  isn't the matrix G either. The correct angle of rotation about V turns out to be

the angle between the projections of Y and Y2 onto the plane orthogonal to V. That came as a great surprise, one that required a significant adjustment of my intuition about spatial rotations. So again, the message is that teaching, learning, and doing mathematics is so much more effective and efficient when done with a tool like Maple.

A discussion of the Givens rotation, and a summary of the actual computations described above are available in the attached worksheet, What Gives with Givens.mw.

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