Recently, I came across an addendum to a problem that appears in many calculus texts, an addendum I had never explored. It intrigued me, and I hope it will capture your attention too.

The problem is that of girding the equator of the earth with a belt, then extending by one unit (here, taken as the foot) the radius of the circle so formed. The question is by how much does the circumference of the belt increase. This problem usually appears in the section of the calculus text dealing with linear approximations by the differential. It turns out that the circumference of the enlarged band is 2*Pi ft greater than the original band.

(An alternate version of this has the circumference of the band increased by one foot, with the radius then being increased by 0.16 ft.)

The addendum to the problem then asked how high would the enlarged band be over the surface of the earth if it were lifted at one point and drawn as tight as possible around the equator. At first, I didn't know what to think. Would the height be some surprisingly large number? And how would one go about calculating this height.

It turns out that the enlarged and lifted band would be some 616.67 feet above the surface of the earth! This is significantly larger than the increase in the diameter of the original band. So, the result is a surprise, at least to me.

This is the kind of amusement that retirement affords. I heartily recommend both the amusement and the retirement. The supporting calculations can be found in the attached worksheet: Girding.mw

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  governed by the logistic equation with a constant rate of harvesting satisfies the initial value problem . This model is typically analyzed by setting the derivative equal to zero and finding the two equilibrium solutions . A sketch of solutions  for different values of a suggests that the larger equilibrium is stable; the smaller, unstable.

When a is less that the unstable equilibrium,  becomes zero at a time , and the population becomes extinct. If  is not interpreted as pertaining to a population, its graph exists beyond , 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 , the time at which , depend on a and h?

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

To answer the first question, the explicit solution, which has the form , exhibits its vertical asymptote when . Solving this equation for  gives the time at which the vertical asymptote is located, a function that is as difficult to graph as . Again the remedy is to solve for, and graph, , with  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 22^nd 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 uu^T 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 u, H can be constructed and the reflection y calculated. Starting with x and y, u 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