## Teaching Three-Dimensional Geometry, I

I have recently had a rather unusual opportunity.  I’ve talked a bit over the last few months about my consulting work producing online videos for a flipped classroom; I’ve been working busily on the Geometry unit.

Now the last section of this unit is on three-dimensional geometry, and I’ve been given pretty free reign as to what to cover in this 20-lectures series of 5-7 minute videos.  And given my interest in polyhedra (which I could focus on exclusively with no shortage of things to discuss!), I felt I had a good start.

But the challenge was also to cover some traditional topics (cones, cylinders, spheres, etc.) — as well as more advanced topics — while not using mathematics beyond what I’ve used in the first several sections of the Geometry unit.

There is, of course, no “correct” answer to this problem.  But I thought I’d share how I’d approach this series of lectures, since geometry is such a passion of mine — and I know it is for many readers as well.  The process of reforming high school geometry courses is now well underway; I hope to contribute to this discussion with today’s post.

Where to start?  Cones and cylinders — a very traditional beginning.  But I thought I’d start with surface areas.  Now for cylinders, this is pretty straightforward.  It’s not much more difficult for cones, but the approach is less obvious than for cylinders.

Earlier in the unit, we derived the formula for the area of a sector of a circle, so finding the lateral surface area of a cone is a nice opportunity to revisit this topic.  And of course, finding the lateral surface area of a cylinder involves just finding the area of a rectangle.

Now what do both of these problems have in common?  Their solution implies that cones and cylinders are flat.  In other words, we reduce what is apparently a three-dimensional problem (the surface area of a three-dimensional object) to a two-dimensional problem.

This is in sharp contrast to finding the surface area of a sphere — you can’t flatten out a sphere.  In fact, the entire science of cartography has evolved specifically in response to this inability.

So this is a nice chance to introduce a little differential geometry!  And no, I don’t really intend to go into differential geometry in any detail — but why not take just a minute in a lecture involving spheres to comment on why the formulas for the surface areas of cones and cylinders are fairly easy to derive, and why — at this level — we’re just given the formula for the surface area of a sphere.

I try to mention such ideas as frequently as I can — pointing out contrasts and connections which go beyond the usual presentation.  Sure, it may be lost on many or most students, but it just may provide that small spark for another.

I think such comments also get at the idea that mathematics is not a series of problems with answers at the back of the book…on the face of it, there is no apparent reason for a student to think that finding the surface area of a cone would be simpler than finding the surface area of a sphere.  This discussion gets them thinking.

Next, I’m planning to discuss Archimedes’ inscription of a sphere in a cylinder (which involves the relative volumes).  This is a bit more straightforward, and it’s a nice way to bring in a little history.

I also plan to look at inscribing a sphere in a right circular cone whose slant height is the same as the diameter of the base, so that we can look at a two-dimensional cross-section to solve the problem.  In particular, this revisits the topic of incircles of triangles in a natural way — I find it more difficult to motivate why you’d want to find an incircle when looking at a strictly two-dimensional problem.

Now on to calculus!  Yes, calculus.  One great mystery for students is the presence of “1/3” in so many volume formulas.  There is always the glib response — the “3” is for “three” dimensions, like the “2” in “1/2 bh” is for “two” dimensions.

When deriving these formulas using integration, this is actually exactly a fairly solid explanation.  But for high school students who have yet to take calculus?

It is easy to approximate the volume of a right circular cone by stacking thin circular disks on top of each other.  If we let the disks get thinner and take more and more of them, we find the volume of the cone as limit of these approximations.  All you need is the sum

$\displaystyle\sum_{k=1}^n k^2=\dfrac{n(n+1)(2n+1)}6.$

I plan to prove that

$\displaystyle\sum_{k=1}^nk=\dfrac{n(n+1)}2,$

and then prove (or perhaps just suggest — I’m not sure yet) the formula for the sum of squares.

I think a fairly informal approach could be successful here.  But I do think such discussions are necessary — in calculus, I’ve routinely asked students why certain formulas they remember are true, and they struggle.  As a simple example, students can rarely tell me why the hypotenuse of a 30-60-90 triangle is twice as long as the shorter leg.

When teachers just give students formulas and ask them to plug numbers in to get answers to oversimplified word problems, of course there is a sense of mystery/confusion — where did these formulas come from?  I’m hoping that this discussion suggests that there is a lot more to mathematics than just a bunch of formulas to memorize.

As usual, I realize I have much more to say on this topic than I had originally supposed…I’ve only discussed up to the fifth lecture so far!  Since I have not had extensive experience teaching more traditional topics at the high school, it has been an interesting challenge to tackle the usual geometry topics in a way that grabs students’ attention.  It’s a challenge I enjoy, and of course I’ll have much more to say about it next week….

## Guest Blogger: Scott Kim, IV

Well, this is the last installment of Scott Kim’s blog post on transforming mathematics education!  These are all important issues, and when you think about them all at once, they seem insurmountable.  It takes each of us working one at a time in our local communities, as well as groups of us working together in broader communities, to effect a change.  What is crucial is that we not only discuss these issues, but we do something about them.  Those of us who participated in the discussion a month ago at the Bay Area Mathematical Artists Seminar are definitely interested in both discussing and doing.

Scott suggests we need to move past our differences and find constructive ways to act.  No, this isn’t easy.  But we need to do this to solve any problem, not just those surrounding mathematics education.  It’s time for some of us to start working on these issues, and many others of us to continue working.  We can’t just sit and watch, passively, any more.  It’s time to act.  What are you waiting for?

## Level 4. Resistance from SOCIETY (quarreling crew)

Sailing is a team sport. You can’t get where you want to go without a cooperative crew. Similarly, math education reform is a social issue. You can’t change how math is taught unless parents, teachers, administrators and policy makers are on board. Most adults cling to the way they were taught as if it were the only way to teach math, largely out of ignorance — they simply aren’t aware of other approaches.

Here are three ways society needs to change the way it thinks about math and math education in order for change to happen.

4a. Attitude. The United States has an attitude problem when it comes to math teachers. First, we underpay and under-respect teachers. And the situation is only getting worse as math graduates flock to lucrative high-tech jobs instead of the teaching profession. The book The Smartest Kids in the World and How They Got That Way describes how FInland turned their educational system around — they decided to pay teachers well, set high qualification standards, and give teachers considerable autonomy to teach however they think is best, with the remarkable result that student respect for teachers is extremely high.

Second, it is socially acceptable, even a badge of honor, to say that you were never good at math. You would never say the same thing about reading. Many people do not in fact read books, but no one would publicly brag that they were never good at reading. Our society supports the idea that parents should read to their kids at night, but perpetuates the idea that being no good at math is just fine.

Solution: respect teachers by paying them well, and value math literacy as much as we value reading literacy.

4b. Vision. The national conversation about math education in the United States is locked in a debate about whether we should teach the basics, or the concepts. As a result we see over the decades that the pendulum swings back and forth between No Child Left Behind and standardized testing on one extreme, and New Math and Common Core Math on the other extreme. As long as the pendulum keeps swinging, we will never settle on stable solution. The resolution, of course, is that we need both. In practice, schools that overemphasize rote math find that they must supplement with conceptual exercises, and schools that overemphasize conceptual understanding find that they must supplement with mechanical drill. We need both rote skills and conceptual understanding, just as kids learning to read need both the mechanical skills of grammar and vocabulary, and the conceptual skills of comprehension and argument construction.

Solution: We need a vision of math education that seamlessly integrates mechanical skills and conceptual understanding, in a way that works within the practical realities of teacher abilities and schoolday schedules. To form a vision, don’t just ask people what they want. A vision should go further than conventional wisdom. As Henry Ford is reported to have said (but probably didn’t), “If I had asked people what they wanted, they would have said faster horses.” Or as Steve Jobs did say, “It’s really hard to design products by focus groups. A lot of times, people don’t know what they want until you show it to them.”

4c. The will to act. As a child I grumbled about the educational system I found myself in. As a young adult I started attending math education conferences (regional meetings of the National Council of Teachers of Mathematics), and was astonished to find that all the thousands of teachers at the conference knew perfectly well what math education should look like — full of joyful constructive activities that challenged kids to play with ideas and think deeply. Yet they went back to their schools and largely continued business as usual. They knew what to do, but were unwilling or unable to act, except at a very small scale.

Solution: Yes, a journey of a thousand miles starts with a single step. And change is slow. But if we’re to get where we want to go, we need to think bigger. Assume that big long lasting change is possible, and in the long term, inevitable. As Margaret Mead said, “Never doubt that a small group of thoughtful, committed citizens can change the world; indeed, it’s the only thing that ever has.” I’m starting my small group. Others I know are starting theirs. What about you?

## Circle Geometry

Today, I thought I’d share a little more about things learned along the way with my curriculum consulting.  As I mentioned before, I’m creating a series of online lectures for the Geometry unit.  This past week, the section I was working on (and will still be working on into next week) is Circle Geometry.

As I also remarked earlier, I’m using the University of Chicago School Mathematics Project’s textbook on Geometry as a reference.  In this text are many theorems about the measure of the angle between two intersecting lines in terms of the measures of the intercepted arcs.

This image is certainly familiar.

The question I had to consider was how to organize all these results in a coherent 5–7-minute lecture.  It turns out that there was too much for just one lecture, so I did spread it out into two.  But I still needed a flow.

Although the results were not new to me, I had never taught this topic before.  My main experience teaching geometry at the high school level was designing and teaching a course on spherical trigonometry as it applies to studying polyhedra.  So this gave me an opportunity to stand back and just think about putting it all together.

I was happy with what I came up with — an approach which could be classified under “combinatorial geometry.”  I decided to pose the following question:

In other words, if you have two intersecting lines, and you draw a circle so that it intersects both lines, what configurations are possible?

Looked at in this way, there are just two considerations:  whether the intersection of the lines is outside, on, or inside the circle, and whether the lines are secant or tangent lines.

It’s not difficult to make the enumeration, so I’ll just give it briefly here.  There is only one configuration if the intersection of the lines lies inside the circle, since both lines must be secant lines.

When the intersection of the lines is on the circle, one of the lines may be tangent, although both cannot be since there is a unique tangent at any given point on a circle.  And when the intersection of the lines is outside the circle, zero, one, or two of the lines may be tangent to the circle.

This enumeration allows for a systematic approach.  If you’ve ever worked through find the angle measures, you know that starting with the arrangement in the upper right corner is the way to begin.  I won’t go through all the details, but I will just indicate that the following figure is all you need:

This simple case is analyzed by considering $\angle QOR$ as an angle exterior to $\Delta POQ.$  The analysis of all the other cases builds from this.

I decided to include a discussion not found in the UCSMP text — continuity.  Of course this is not a topic which can be rigorously discussed at say, the 10th-grade level.  But why not give students an intuition of the idea?

This series illustrates the case that the intersection of the lines is outside the circle, and one of the lines is tangent.  We look at this as the limiting case of a series of pairs of secant lines.

This argument depends upon the fact that the measurement of all arcs and angles varies continuously as $S$ moves around the circle.  While, as mentioned, this cannot be addressed rigorously, it is a very intuitive argument.  Moreover, there are many different software packages you could use to make an animation of this process, and display all the arc and angle measurements as $S$ moves around the circle.

There is no reason not to introduce this argument.  In my pair of lectures, I used more traditional geometrical arguments as well.  It doesn’t hurt students to be exposed to a wide range of proof ideas.

I summarize all of these results in the following graphic.

The measure of the angle indicated with the red dot is half the measure of the intercepted arc, or the sum/difference of the measures of the intercepts arcs, shown in red and blue.  An arc in blue indicates its measure is to be subtracted rather than added.  I was very happy with this graphic.  I think that if a student followed the lecture, they could state every result just by looking at it.

This also proved to be a great segue into looking at the power of a point.  I thought I’d begin with the figure in the upper left, proving the usual theorem using similar triangles.

And now for another continuity argument!

This is a nice way to see that the power of any point on the circle is 0.  It is also a nice contrast to the theorems about the angle between the intersecting lines:  when $PT$ and $RT$ eventually reach 0, you’re not able to conclude anything about a relationship between $QT$ and $ST.$

This means that there is no theorem relating lengths of segments for the two cases when the intersection of the lines lies on the circle.   I use the following graphic to indicate this, with two cases grayed out when the power of the point offers no conclusion.

Now all of these results are the usual ones found in high school geometry textbooks; nothing new here.  But for me, just having to step back and think about how to put them all together was a fun challenge.

Again, I am surprised at how much I’m learning even though I’m just putting together a few slides on elementary geometry.  The process of writing these lectures is an engaging one, and I hope the students who will eventually watch them will benefit from a perspective not found in more traditional textbooks.

## What Is…A Polygon?

A haven’t made a post in quite some time in my “What is a Geometry?” thread.  In working on my online lectures in the section on Polygons, I of course needed to define just what a polygon is.  This turned out to be a little more challenging than I had imagined.  I thought that the issues that arose would make this discussion an interesting continuation of the “What is a Geometry?” thread.

In general, I think that the Wikipedia does a good job with mathematics — but specifically, the definition of a polygon leaves quite a bit to be desired.  I’ll reproduce it here for you:

In elementary geometry, a polygon is a plane figure that is bounded by a finite chain of straight line segments closing in a loop to form a closed polygonal chain, or circuit.  These segments are called its edges or sides, and the points where two edges meet are the polygon’s vertices (singular: vertex) or corners.

OK, maybe not so bad of a start.  There are lots of examples given which fit this definition, but many which do not.  For example, this definition allows consecutive segments to lie on the same line, which is typically disallowed in most other definitions of polygons.

So maybe a clause may be added to the definition which does not allow this.  But then we encounter a polygon like this (I’m using screenshots from my lecture as illustrations):

My definition begins with a list of vertices — but the problem is still the same.  The vertex labeled “4” is on the edge joining the vertices labeled “1” and “2.”  Again, this is usually avoided.

And what about the following figure?

With the Wikipedia definition, a vertex can be an endpoint of more than two edges of a polygon.  Again, problematic.  There would be no way to distinguish this figure from a single polygon and two different triangles sharing a vertex.

Moreover, there is no condition saying that the straight line segments need to be distinct.  So the same segment might occur multiple times as an edge of a polygon.

None of these behaviors is illustrated anywhere on the Wikipedia page.  I’ve done some Wikipedia editing a while back, and would be interested in working on this page when I have more time to devote to such things.

So what is the fix?  I’m using the Geometry text of the University of Chicago School Mathematics Project as a reference, which is one of the most rigorous geometry texts around.  Here is their definition:

They remark that this is the definition used in 23 out of the 45 geometry text they surveyed.  And in fact, it is the rewrite of a definition in previous editions:  “A polygon is the union of segments in the same plane such that each segment intersects exactly two others, one at each of its endpoints.”  This definition was problematic, though, since by this definition, the following is actually a polygon!

Now this revised definition solves all of the problems above — but I couldn’t use it.  Why not?

One of the sections I’ll be writing lectures for is three-dimensional geometry — and (of course) I’ll be saying a lot about polyhedra in this section.  There are Platonic and Archimedean solids, as well as the Kepler-Poinsot polyhedra, like the small stellated dodecahedron shown below.

The faces of the small stellated dodecahedron are pentagrams, five meeting at each vertex.

But the UCSMP definition does not allow edges to cross.  Each edge meets exactly two others, at each of its endpoints.  So that means that an edge cannot cross another in its interior.

Now I just can’t talk about polyhedra without talking about nonconvex examples.  Sure, it is possible to talk about pentagrams with edges crossing as decagons without crossing edges.

But this would be the height of absurdity.  Besides the fact that none of the dozens of books and articles I’ve read on polyhedra in the past few decades ever do such a thing — and I’m sure none ever will.

So I had to go it alone.  I’ll share with you my definition — but I can’t say it’s the best.  The difficulty lies with being mathematically precise while still making the definition accessible to high school students.  Here it is:

A polygon is determined by a list of its vertices. Edges of the polygon connect adjacent vertices in the list, and there is also an edge connecting the last vertex in the list to the first one. All vertices in the list must be different. Finally, no three consecutive vertices of the polygon can lie on the same line, and no vertex can lie in the interior of another edge.

I don’t think this is too bad.  But there is still a subtle glitch, which I haven’t worked out yet, and which doesn’t necessarily need to be worked out at this level.  When I talk about triangles, for example, I allow cases where the sides have lengths 3, 4, and 7, for example.  But I qualify such a triangle by calling it a degenerate triangle.

Since a triangle is a polygon, a degenerate triangle should be a degenerate polygon, right?  The problem is that calling something a “degenerate polygon” gives the impression that it is actually some type of polygon.  But a degenerate triangle, by my definition, is not a polygon.  So when I use the term degenerate polygon, I’m not actually talking about a polygon….

So I’ll let you think this over.  I just wanted to share how surprised I was at how subtle the definition of something so “simple” could be.  An ordinary polygon.

If you find this sort of question intriguing, you might go online and research all the various definitions of polyhedron.  Convex polyhedra are easy to define (as are convex polygons), but when you get into the different types of behavior possible in the nonconvex cases, well, it becomes problematic.  In fact, no one, as far as I know, has ever come up with a satisfactory definition for “polyhedron.”  Might even do a blog post on that some day….

## Calculus VIII: Miscellaneous Problems, I

In this post, I’ll continue discussing problems I’ve been encountering in the calculus textbook I’m reading.  Some problems are involved enough to require an entire post devoted to them; others are interesting but relatively short.  Today, I’ll discuss four shorter problems.

The first problem is Exercise 26 on page 33:

If a cylindrical hole be drilled through a solid sphere, the axis of the cylinder passing through the center of the sphere, show that the volume of the portion of the sphere left is equal to the volume of a sphere whose diameter is the length of the hole.

This is not a difficult problem to solve; I’ll leave the simple integral to the reader.  This has always been a favorite volume problem of mine, and this is the earliest reference I’ve seen to it.

Perhaps it was a classic even back then — remember, the book was published in 1954.  The author usually attributes problems he uses to their sources, but this problem has no attribution.  I would be interested to know if anyone knows of an earlier reference to this problem.

The second problem is not from an exercise, but is discussed in Art. 37 on page 48.  It’s one of those “of course!” moments, leaving you to wonder why you never thought to try it yourself….

Why is the antiderivative of $y=x^{-1}$ the natural logarithm?  There are a few different ways this is usually shown, but here’s one I haven’t seen before:  consider the limit

$\displaystyle\lim_{n\to-1}\int_a^bx^n\,dx,\quad 0

It seems so obvious when you see it written down, but I’ve never thought to take this limit before.  You get

$\displaystyle\lim_{n\to-1}\dfrac{b^{n+1}-a^{n+1}}{n+1}.$

Now apply L’Hopital’s rule!  And there you have it:

$\displaystyle\lim_{n\to-1}\int_a^bx^n\,dx=\log b-\log a.$

I think that perhaps when writing $x^n,$ I’m so conditioned to thinking of $n$ as a constant that I never thought of turning it into the variable.  It’s a nice proof.

Next is Art. 68, which begins on page 82.  Again, you’ll agree that it seems pretty obvious after the discussion, but I’ve never seen this diagram drawn before.  This is likely because hyperbolic trigonometry is downplayed in today’s calculus curriculum.  You might recall the comment I made about a colleague once saying they didn’t teach hyperbolic trigonometry since it wasn’t on the AP exam.

So let’s look at the hyperbola $x^2-y^2=a^2.$  The goal of this exercise is to find a geometrical interpretation of the relationship

$\sec\theta=\cosh u,$

which is key to connecting circular and hyperbolic trigonometry by means of the gudermannian, as I have discussed earlier.

Draw the auxiliary circle $x^2+y^2=a^2,$ and consider the point

$P=(a\cosh u,a\sinh u).$

Now drop a perpendicular from $P$ on the x-axis to the point $N=(a\cosh u,0).$  Next, draw a tangent from $N$ to the auxiliary circle, meeting it at $T.$  Finally, join $T$ to the origin.

Since $NT$ is tangent to the circle, we know that $\Delta NTO$ is a right triangle.  Therefore $ON=a\sec\theta.$  But by construction, $ON=a\cosh u,$ and so

$\sec\theta=\cosh u.$

Yep, that’s all there is to it!  A geometrical illustration of the gudermannian function.  So very simple.  And incidentally, the author goes on to discuss the gudermannian function in the next section.

For the last example, I’ll need to skip ahead a little bit, since my next exploration is a bit too involved and may need an entire post.  As a teaser, I’ll just say that I learned a completely new way to derive Cardan’s formula for solving a cubic equation!  It involves calculus and quite a bit of algebra.  At some point, I’d like to dive in a little deeper and see if I can relate this new proof with the usual one — but again, that for another time.

So this last example (Art. 96 on page 109) is about differentiating

$y=e^{ax}\sin(bx).$

Of course this is just a simple application of the product rule:

$\dfrac{dy}{dx}=e^{ax}(a\sin(bx)+b\cos(bx)).$

But why stop here?  We can go further, using an idea very common when working with physics applications.  We seek to write

$a\sin(bx)+b\cos(bx)=c\sin(bx+\theta).$

Since

$c\sin(bx+\theta)=c\sin(bx)\cos(\theta)+c\cos(bx)\sin(\theta),$

this amounts to solving

$c\cos(\theta)=a,\quad c\sin(\theta)=b.$

This is straightforward:

$c=\sqrt{a^2+b^2},\quad\theta=\arctan(b/a).$

Thus,

$\dfrac{dy}{dx}=\sqrt{a^2+b^2}\,e^{ax}\sin(bx+\arctan(b/a)).$

This means that taking the derivative of $e^{ax}\sin(bx)$ amounts to multiplying the function by $\sqrt{a^2+b^2}$ and increasing the angle in the sine function by $\arctan(b/a).$  Therefore

$\dfrac{d^n}{dx^n}e^{ax}\sin(bx)=(a^2+b^2)^{n/2}e^{ax}\sin(bx+n\arctan(b/a)).$

I actually did the proof by induction to verify this.  It’s pretty cumbersome.

Note that this also implies that

$\displaystyle\int e^{ax}\sin(bx)\,dx=\dfrac{e^{ax}\sin(bx-\arctan(b/a))}{\sqrt{a^2+b^2}}.$

The same results hold with sine being replaced by cosine.  Such elegant results.

I hope you found these problems as interesting as I did!  There are so many calculus gems in this book.  I’ll continue to keep sharing….

## Calculus VII: Approximations

Although I’ll have a very busy summer with consulting, I’ve taken some time to start reading more again.  You know, those books which have been sitting on your shelves for years….

So I’ve started Volume I of A Treatise on the Integral Calculus by Joseph Edwards.

I include a picture of the cover page, since you can google it and download a copy online.  Between Volumes I and II, there’s about 1800 pages of integral calculus….

Since I’ll likely be working with a calculus curriculum later this year, I thought I’d look at some older books and see what calculus was like back in the day.  I’m continually surprised at how much there is to learn about elementary calculus, despite having taught it for over 25 years.

My approach will be a simple one — I’ll organize my posts by page number.  As I read through the books and solve interesting problems, I’ll share with you things I find novel and interesting.  The more I read books like these and think about calculus, the more I think most current textbooks simply are not up to the task of presenting calculus in any meaningful way.  Sigh.

This is not the time to be on my soapbox — this is the time for some fun!  So here is the first topic:  Weddle’s Rule, found on page 21.

Ever hear of it?  Bonus points if you have — but I never did.  It’s another approximation rule for integrals.  Here it is: given a function $f$ on the interval $[a,b],$ divide the interval into six equal subintervals with points $x_0, x_1,\ldots x_6$ and corresponding function values $y_0=f(x_0),\ldots,y_6=f(x_6).$  Then

$\displaystyle\int_a^bf(x)\,dx\approx \dfrac{b-a}{20}\left(y_1+5y_2+y_3+6y_4+y_5+5y_6+y_7\right).$

Yikes!  Where did that come from?  I’ll present my take on the idea, and offer a theory.  If there are any historians of mathematics out there, I’d be happy to hear if my theory is correct.

One reason most of us haven’t heard of Weddle’s Rule is that approximations aren’t as important as they were before calculators and computers.  So many exercises in this book involve approximation techniques.

So how would you come up with Weddle’s Rule?  I’ll share my (likely mythical) scenario with you.  It’s based on some notes I wrote up a while ago on Taylor series.  So before diving into Weddle’s Rule, I’ll show you how I’d derive Simpson’s Rule — the technique is the same, but the algebra is easier.  And by the way, if anyone has seen this technique before, please let me know!  I’m sure it must have been done before, but I’ve never been able to find a source illustrating it.

Let’s assume we want to approximate

$F(x)=\displaystyle\int_a^xf(t)\,dt$

by using three equally-spaced points on the interval $[a,x].$  In other words, we want to find weights $p,$ $q,$ and $r$ such that

$S(x)=\left(p f(a)+ q f\left(\dfrac{a+x}2\right)+rf(x)\right)(x-a)\approx F(x).$

How might we approach this?  We can create Taylor series for $F(x)$ and $S(x)$ about the point $a.$  The first is easy using the Fundamental Theorem of Calculus, assuming sufficient differentiability:

$F(x)=f(a)(x-a)+\dfrac{f'(a)}{2!}(x-a)^2+\dfrac{f''(a)}{3!}(x-a)^3+\cdots$

Now to construct the Taylor series of $S(x)$ about $x=a,$ we need to evaluate several derivatives at $a.$ This is not difficult to do by hand, but it is easy to do using Mathematica and a command such as

Doing so yields the following:

Now the problem becomes a simpler algebra problem — to force as many of the coefficients of the derivatives on the right-hand side to be $1$ as possible.  This will make the derivatives of $F$ and $S$ match, and the Taylor polynomials will be equal up to some order.

Solving the first three such equations,

yields, as we expect, $p=1/6,$ $q=2/3,$ and $r=1/6.$ Note that these values also imply that

$\dfrac12q+4r=1,$

but

$\dfrac5{16}q+5r=\dfrac{25}{24}.$

This implies that

$S(x)-F(x)=\dfrac1{24}\cdot\dfrac{(x-a)^5}{5!}+O((x-a)^6)$

on each subinterval, so that

$S(x)-F(x)=O((x-a)^5)$

on each subinterval, giving that Simpson’s rule is $O((x-a)^4).$

So how we apply these to derive Weddle’s rule?  We could try to find weights $w_1,\ldots w_7$ to create an approximation

$W(x)=\left(w_1 f(a)+w_2f\left(\dfrac{5a+x}6\right)+\cdots+w_7f(x)\right)(x-a).$

If we apply precisely the same procedure as we did with Simpson’s Rule, we get the following as the sequence of weights to create the best approximation:

$\dfrac{41}{840},\ \dfrac9{35},\ \dfrac9{280},\ \dfrac{34}{105},\ \dfrac9{280},\ \dfrac9{35},\ \dfrac{41}{480}.$

Not exactly easy to work with — remember, no calculators or computers.

So let’s make the approximation a little worse.  Recall how the weights were found — a system of seven equations in seven unknowns was solved, analogous to the three equations in three unknowns for Simpson’s rule.  Instead, we specify $w_1,$ and solve the first six equations in terms of $w_1.$  This gives us

Now all weights must be positive; this gives the constraint

$0.046\overline6\approx\dfrac7{150}

Let’s put $w_1=1/20,$ which is in the interval just described.  This gives the sequence of weights to be

$\dfrac1{20},\ \dfrac5{20},\ \dfrac1{20},\ \dfrac6{20},\ \dfrac1{20},\ \dfrac5{20},\ \dfrac1{20},$

where all fractions are written with the same denominator.  Now imagine factoring out the $1/2,$ and you notice that all divisions are by 10.  Can you see the advantage?  If you have a table of values for your functions, you just need to multiply function values by a single-digit number, and then move the decimal place over one.  An approximators dream!

So Weddle’s approximation is exact for fifth-degree polynomials, even though it is possible to use six subintervals to get weights which are exact for sixth-degree polynomials.  Yes, we lose an order of accuracy — but now our computations are much easier to carry out.

Was this Weddle’s thinking?  I can’t be sure; I wasn’t able to locate the original article online.  But it is a way for me to make sense out of Weddle’s rule.

I will admit that in a traditional calculus class, I don’t address approximations in this way.  There is a time crunch to get “everything” done — that is, everything the student is expected to know for the next course in the calculus sequence.

Should these concepts be taught?  I’ll make a brief observation:  in reading through the first 200 pages of this calculus book, it seems that all that has changed since 1954 is that content was pared down significantly, and more calculator exercises were added.

This is not the solution.  We need to rethink what students need to now know and how that material should be taught in light of emerging technology.  So let’s get started!

## Pythagorean Triples

I recently began writing some lectures for an online course — I’ll talk more about the nature of the course in next week’s post.  The broad topic is geometry, of course a favorite — and the specific topic for this unit is Triangles.

You can’t talk about triangles without talking about the Pythagorean Theorem.  Part of my job is also to compose problems for the lectures as well as for quizzes and exams, and to my surprise, I came up with a few interesting ones.  So I thought I’d share them with you.  I am always a fan of sharing mathematics as it happens!

The questions I wrote are based on the following parameterization of Pythagorean Triples:  Given positive integers $p$ and $q,$ then

$(q^2-p^2,2pq,q^2+p^2)$

is a Pythagorean Triple.  This parameterization generates all primitive Pythagorean Triples — that is, triples whose sides share no common factor.  But it is not possible to get $(9,12,15),$ for example, using this parameterization.  Of course $(9,12,15)$ is just three times the triple $(3,4,5);$ therefore, if you can generate all primitive Pythagorean Triples, you can take multiples of them to generate all Pythagorean Triples.

I thought of my first problem walking down the sidewalk going to lunch the other day.  The simplest Pythagorean Triple, $(3,4,5),$ has side lengths which are in arithmetic progression.  What other Pythagorean Triples have this property?

The simplest way to approach this is to parameterize such a triple by $(a,a+d,a+2d),$ where $a>0$ is the smallest integer in the arithmetic progression and $d>0$ is the common difference.  Since the triangle is a right triangle, we must have

$a^2+(a+d)^2=(a+2d)^2,$

which we may rewrite as

$a^2-2ad-3d^2=0.$

Now this factors:

$(a+d)(a-3d)=0,$

resulting in solutions $a=-d$ or $a=3d.$  We did assume that $d>0,$ so we eliminate the solution $a=-d.$  Note that this would generate the triple $(a,0,-a),$ and in fact $a^2+0^2=(-a)^2.$ But one side length is zero and another is negative, so no triangle is possible with these side lengths.

What about the solution $a=3d$?  Here, we get

$(a,a+d,a+2d)=(3d,4d,5d),$

which you can observe is just a multiple $d$ of the primitive Pythagorean Triple $(3,4,5).$

The conclusion?  The only Pythagorean Triples possible whose side lengths are in arithmetic progression are multiples of the $(3,4,5)$ right triangle.

I really didn’t know the answer would come out so nicely — but since the algebra involved was fairly straightforward, I thought I could include this as a non-routine example of an application of the Pythagorean Theorem at the high school level.

The previous problem was part of a lecture.  The next problem was written as a possible exam question for teachers; once I realized I had more than one interesting problem, I thought there would be enough for a blog post….

I was just looking for interesting patterns in Pythagorean Triples, and noticed that with the $(6,8,10)$ triangle, the area and perimeter were both $24.$  A coincidence?  Were there other triangles with this property?

Of course there had to be finitely many — as the side lengths get larger, the area gets larger faster than the perimeter, as the area is essentially a quadratic function, while the perimeter is essentially a linear function.  So how many others are there?  Make a mental note of your guess before reading further….

We begin by parameterizing by

$(k(q^2-p^2),2kpq,k(q^2+p^2));$

the factor of $k$ is necessary since the two-variable version generates all primitive Pythagorean Triples, but not necessarily all Pythagorean Triples.

Setting the perimeter and area equal to each other results in

$\dfrac12k(q^2-p^2)\cdot2kpq=k(q^2-p^2+2pq+q^2+p^2),$

Cancelling out factors of $k,$ $q,$ and $p+q$ results in

$kp(q-p)=2.$

This equation clearly has just three solutions, since one of the factors must be $2$ and the other two factors must be $1.$

None is particular difficult; let’s take them one at a time.  When $k=2,$ then $p=q-p=1,$ so that $q=2.$  Substituting back into the parameterization, we obtain the Pythagorean Triple $2(3,4,5),$ which is the triple $(6,8,10).$

When $p=2,$ then $k=q-p=1,$ so that $q=3.$ This generates a new Pythagorean Triple, $(5,12,13).$

Finally, when $q-p=2,$ then $k=p=1$ and $q=3,$ so that the Pythagorean Triple $(8,6,10)$ is generated.  Of course this is just a duplicate of the first solution.

Surprised that there was just one more solution?  I was!  It was such a nice, straightforward solution, that I couldn’t help but include it.

There was a third problem which I liked, but the algebra was a little too intense — there was a nice geometrical solution, but it required ideas learned later on in the course.  So here it is if you want a challenge:  suppose you are given two right triangles, and you know that their perimeters and areas are the same.  Prove that they are congruent.

I think you might enjoy solving this purely algebraically.  I did like it so much, though, that I included a simpler version in one of my lectures:  suppose you are given two right triangles, and you know that their hypotenuses are both of length $8$ and that their perimeters are equal.  Prove that the triangles are congruent.

To be honest, I never knew I’d find problem solving with the Pythagorean Theorem so interesting.  It’s nice to know that there is always more geometry to learn!  Even with something as apparently simple as the venerable Pythagorean Theorem….