"By the middle of the 20th century, mathematics was undergoing a rapid phase of growth, stimulated by its widespread applications and powerful new methods."
As the title and opening sentence imply, this chapter is all about some very modern branches of math, some of which I've discussed in the blog before. Stewart begins by mentioning Douglas Adams, who declared that the meaning of life is 42:
"This incident is a parody of a famous statement in which Laplace summed up the mathematical view of determinism...."
I don't need to quote Laplace's entire statement here -- suffice it to say that if the universe were fully deterministic, it would be possible to predict the future using a single formula. But one of the first people to challenge Laplace's determinism was Henri Poincare, who tried to predict the future for just three bodies (or objects):
"This complexity is now seen as a classic example of chaos. His entry won King Oscar II's prize, even though it did not fully solve the problem posed."
I don't need to provide all the details here, since I've already blogged about Poincare "winning the Oscar" for our previous side-along reading. Instead, we move on to the Dutch engineer Balthazar van der Pol, who made an electronic circuit that beats just like a human heart:
"His work was given a solid mathematical basis during the Second World War by John Littlewood and Mary Cartwright, in a study that originated in the electronics of radar."
We move on to nonlinear dynamics. It was a weatherman, Edward Lorenz, who discovered just how chaotic the weather can be:
"He also discovered that if the same equations are solved using initial values of the variables that differ only very slightly from each other, then the differences become amplified until the new solution differs completely from the original one."
This became known as the butterfly effect or the Lorenz attractor. Meanwhile, David Ruelle and Floris Takens applied this idea to fluid flow:
"A common type of fluid flow, laminar flow, is smooth and regular, just what you would expect from a deterministic theory."
But turbulent flow, on the other hand, is chaotic and random. At this point, the author begins to describe some theoretical "monsters":
"During the early development of calculus, mathematicians had assumed that any continuously varying quantity must possess a well-defined rate of change almost everywhere."
But as Weierstrauss discovered -- and Sam Shah reminds us, there exist functions that are continuous everywhere but differentiable nowhere. And Benoit Mandlebrot discovered some even more monstrous curves, known as fractals:
"Today, scientists have absorbed fractals into their normal ways of thinking, just as their predecessors did at the end of the 19th century with those maverick mathematical monstrosities."
Again, one of our previous side-along reading books was Mandelbrot's work on fractals, and so these are already familiar to us.
Stewart now moves on to the other main theme of this chapter, complexity. He tells us that simple problems can have complex solutions:
"In the early 1980's George Cowan, formerly head of research at Los Alamos, decided that one way forward lay in the newly developed theories of nonlinear analysis."
We learn that he created the Santa Fe Institute, devoted to the study of complex systems. And of complexity theory, the author writes:
"And it uses nonlinear dynamics and other areas of mathematics to understand what the computers reveal. In one type of new mathematical model, known as a cellular automaton, such things as trees. birds and squirrels are incarnated as tiny coloured squares."
John von Neumann was the originator of cellular automata theory. A few weeks ago on the blog I mentioned John Conway -- his "Game of Life" is based on cellular automata." Of the work of von Neumann and Conway on cellular automata:
"This turns out to be an excellent way to work out which factors are important, and to uncover general insights into why complex systems do what they do. A complex system that defies analysis by traditional modelling techniques is the formation of river basins and deltas."
And the ultimate connection between cellular automata and biological life is profound:
"Complex systems support the view that on a lifeless planet with sufficiently complex chemistry, life is likely to arise spontaneously and to organize itself into ever more complex and sophisticated forms."
Let's think back to what we learned in this book about how mathematics was created:
"If it is obvious where to go next, anyone can do it. And so, over some four millennia, the elaborate, elegant structure that we call mathematics came into being."
On that note, Stewart concludes the book as follows:
"Mathematics has never been so active, it has never been so diverse, and it has never been so vital to our society. Welcome to the Golden Age of mathematics."
The sidebars in this final chapter are on Poincare's blunder (which we've read about before), a biography of Mary Lucy Cartwright, what nonlinear analysis did for them, and what nonlinear analysis does for us.
I hope you enjoyed our reading of Ian Stewart's book. I liked reading so much about so many different areas of mathematics, as opposed to just a single field like many of our previous books.
Speaking of books, I'm still thinking about yesterday's Algebra I lesson. I've seen graphing using intercepts taught as a lesson in many Algebra I texts, but I don't remember ever being taught this method as a young student.
Well, recall that on the same day that I bought Ian Stewart's book from the library, I also purchased a teacher's edition of the very text that I used as a young Algebra I student. (This book was published by McDougal Littell and written by Brown and Dolciani.) So I can look at the actual text to see whether I was ever taught graphing using intercepts back then.
For starters, linear functions don't even appear until Chapter 8 of Brown/Dolciani. Most texts today teach linear functions in the first semester, but my old text covers factoring first.
Here are the lessons in Chapter 8 of Brown/Dolciani:
8-1. Equations in Two Variables
8-2. Points, Lines, and Their Graphs
8-3. Slope of a Line
8-4. The Slope-Intercept Form of a Linear Equation
8-5. Determining the Equation of a Line
8-6. Functions Defined by Tables and Graphs.
8-7. Functions Defined by Equations
8-8. Linear and Quadratic Functions
8-9. Direct Variation
8-10. Inverse Variation
Graphing first appears in Lesson 8-2 of this text. After introducing the coordinate plane and its parts, the student learns about linear equations:
"Since two points determine a line, you need to find only two solutions of a linear equation in order to graph it. However, it is a good idea to find a third solution as a check. The easiest solutions to find are those where the line crosses the x-axis (y = 0) and the y-axis (x = 0)."
But the word "intercept" never appears in this lesson (much less "graphing using intercepts"). In the following lesson, slope is introduced, and so from this point on, it's expected that students will use slope to graph lines.
Lesson 8-4 is on slope-intercept form, and so the term "y-intercept" obviously appears here. On the other hand, "x-intercept" doesn't appear until the next lesson, when students are asked to find equations of lines given two parts (such as slope and x-intercept, or two points).
I also notice that in this lesson, the point-slope form is never taught. If the students are given a point and a slope, they are expected to plug these into y = mx + b to find b. While slopes of parallel lines appear in the main part of Lesson 8-4, slopes of perpendicular lines appear only in an "Extra," yet I do recall being tested on slopes of perpendicular lines that year.
(I know this because I remember trying to show up the teacher by finding the angle between the lines when they weren't perpendicular. I don't recall how I tried to do this, but I do know that she wrote "Angle measures wrong!" on my paper. This is also the chapter where I showed her up by using "delta" in the Slope Formula. Her response was "Delta is one of my favorite Greek letters!" Since then, I've heard that some Algebra I teachers now actually use deltas to teach slope.)
At the end of this chapter, there is a "Chapter Test." Of the six questions listed for Lesson 8-2, four are on graphing single points, and only two are on graphing lines. But the equations for the two lines are written in slope-intercept form. If I were a student taking that test (which I probably once was), I'd just graph them using the slope and y-intercept, not using both intercepts. Other questions that require students to graph lines (Lessons 8-4 and 8-8) ask the students to write the slope-intercept form.first.
All of this shows how Algebra I has changed in the quarter-century since I took the class. The standard form (to be graphed using intercepts) and point-slope form are taught more now, while back then only the slope-intercept form was emphasized.
OK, that's enough about Algebra I. Let's finally get to the U of Chicago Geometry text.
Lesson 6-3 of the U of Chicago text is called "Rotations." In the modern Third Edition, we must backtrack to Lesson 4-5 to learn about rotations.
This is what I wrote last year about today's lesson:
Yesterday, we discussed translations on the coordinate plane, and so now we move on to rotations. I point out that we learned how to perform translations of the form (x, y) -> (x + h, y + k) -- which turns out to be every translation in the plane.
But with rotations, we only perform a precious few of them. The only rotations that appear on the PARCC and other Common Core tests are those of magnitude 90, 180, or 270. Yet we've seen a few of these rotations centered at points other than the origin on the PARCC.
We'll begin with rotations that are centered at the origin, though. Just as we used the Two Reflections Theorem for Translations yesterday, today we'll use the Two Reflections Theorem for Rotations. So to perform the rotation of 180 degrees about the origin, we compose two reflections in mirrors that intersect at the origin, at an angle of half of 180, or 90 degrees. The obvious choices for mirrors are the x- and y-axes. We've already proved that the reflection image of (x, y) in the x-axis is (x, -y) and the reflection image of (x, y) in the y-axis is (-x, y). It doesn't matter in which order we compose these as reflections in perpendicular mirrors always commute. So we prove that the rotation image of (x, y) centered at the origin and of magnitude 180 degrees is (-x, -y).
Now our other common rotation magnitude is 90 degrees -- and this time, it will make a difference whether it's clockwise or counterclockwise. The angle between the mirrors will now have to be half of 90, or 45 degrees. There's one mirror to consider that will help us with a 45-degree angle -- the line whose equation is y = x.
We've hinted at several proofs involving reflection over the line y = x. Let's look at the quadrilateral whose vertices are (0, 0), (a, 0), (a, a), and (0, a). We can show that this figure is a kite.
So now we can apply the properties of a kite -- the Kite Symmetry Theorem. The diagonal of our kite running from (0, 0) to (a, a) bisects the angle between the x- and y-axes -- and since we know that the angle between the axes is 90 degrees, the diagonal must form a 45-degree angle with each axis. And reflecting across this symmetry diagonal must map the axes to each other and x = a to y = a.
Recall that at this point, we don't know the equations of lines, so we aren't yet certain that the graph of y = x is even a line (which we'd better figure out before trying to use it as a mirror). But we see that the value of a in the above proof is arbitrary -- it's true for every single real number a (although in case a is negative, we should probably say that the kite has sides of length |a|, not a). Therefore every single point of the graph of y = x lies on the bisector of the angle between the axes -- that is, the graph of y = x is exactly that line. And reflecting in that line maps x = a to y = a and vice versa -- that is, it switches x and y. Therefore the image of (a, b) must be (b, a).
Now that we know how to reflect in the line y = x, let's use it to perform a 90-degree rotation. It's probably easiest just to start with the reflection in y = x first, so (x, y) maps to (y, x). As for the second mirror, it depends on whether we want to go clockwise or counterclockwise. To go clockwise, the second mirror must be 45 degrees clockwise of the first mirror, y = x. That is the x-axis, and to reflect in it, we change the sign of the second coordinate. So (y, x) reflected in the second mirror is the point (y, -x), so mapping (x, y) to (y, -x) rotates points 90 degrees clockwise. To go counterclockwise, the second mirror must be 45 degrees counterclockwise of the first mirror, y = x. That is the y-axis, and to reflect in it, we change the sign of the first coordinate. So (y, x) reflected in the second mirror is the point (-y, x), so mapping (x, y) to (-y, x) rotates points 90 degrees counterclockwise.
Notice that some of the PARCC questions mention 270-degree rotations -- for example, there was a released question that mentions a 270-degree clockwise rotation. Of course, a 270-degree clockwise rotation is equivalent to a 90-degree counterclockwise rotation, so it maps (x, y) to (-y, x). If students forget this, they can still take half of 270 degrees to get 135 degrees clockwise, and they can see that 135 degrees clockwise from the line y = x is still the y-axis, just as it would have been if they'd gone 45 degrees counterclockwise instead.
Last year, I created a quick worksheet to help students perform any of the reflections and rotations mentioned in this post. (This was late in the year when we were covering PARCC questions, but now I'm giving this lesson much earlier.) It takes the coordinate plane labeled with the positive x-, negative x-, positive y-, and negative y-axes. Students can then perform the rotations on the axes to see what happens. For example, let's try our 270-degree clockwise rotation. After we rotate the paper 270 degrees clockwise, we see that where the x ought to be, we see -y instead, and where the y ought to be, it's +x. Thus the image must be (-y, x).
Okay, so we've taken care of all the rotations centered at the origin, But on the PARCC, there are questions with rotations centered at other points. These questions that I've seen direct the students to take a triangle ABC and rotate it around one of its vertices -- let's say C. Well that makes things a little easier, since then the rotation image of C is C itself. So then there are only two points that we need to find, A' and B'.
It's possible, in principle, to find formulas to determine the image of (x, y) under reflections in mirrors parallel to the axes and rotations centered at points other than the origin. We've seen, for example, that the point (x, y) reflected in the line x = a is (2a - x, y). An interesting question is, where exactly does the 2 in 2a - x come from?
To find out, we notice that if we were reflecting in the y-axis (which is parallel to x = a), then the point (x, y) is mapped to (-x, y). Now that extra 2a term looks just like a horizontal translation of exactly 2a units.
So somehow, our reflection in the line x = a appears to be the composite of a reflection in the y-axis and a horizontal translation. (This is not a glide reflection between the mirror is perpendicular to the direction of translation -- we found out last year that such a composite yields a simple reflection, not a glide reflection.)
Using symbols, let's call the composite transformation T. It is the composite of a y-axis reflection, which we'll call r_y, and a horizontal translation of 2a units, H_2a:
T = H_2a o r_y
But the horizontal translation is itself the composite of two reflections. The two mirrors here must be vertical mirrors spaced exactly half of 2a, or a units apart. We might as well let the two mirrors be the y-axis itself and the line x = a.
T = H_2a o r_y
= r_(x = a) o r_y o r_y
= r_(x = a) o I
= r_(x = a)
which is exactly what we wanted -- a reflection in the line x = a.
Likewise, we see that the reflection in the line y = b maps (x, y) to (x, 2b - y). The composite of both reflections is a 180-degree rotation about the point (a, b), which maps (x, y) to (2a - x, 2b - y) -- and that's also the composite of a 180-degree rotation about the origin and yet another translation.
Now 90-degree rotations about points other than the origin are even trickier, because now we'd have to reflect about mirrors with equations like y = x + b -- and we don't even know that's a line yet. The algebra involved in this reflection gets very messy.
Of course, if we try to visualize the rotation, another composite transformation jumps at us. To perform a 90-degree rotation (either clockwise or counterclockwise) about the point (a, b), it appears that we can first perform the translation that maps (a, b) to (0, 0), then perform the rotation centered at the origin, and finally translate (0, 0) back to (a, b).
This seems to work, but is there any reason why it should? Let's use symbols again -- in order to remember what the symbols stand for, we let "rot" stand for the rotation and "trans" stand for the translation mapping (0, 0) to (a, b). Then trans^-1 can stand for the inverse translation -- the one mapping (a, b) to (0, 0). This gives us:
T = trans o rot o trans^-1
This composite has a name in classes like linear algebra and above -- conjugation. That is, we are conjugating the rotation by the translation.
We now want to rewrite both the translation and origin-rotation with two mirrors each. And as usual, we want to choose the mirrors strategically so that some of the reflections cancel out. For the translation, we'll let k be the line joining the points (0, 0) and (a, b). Then l will be the line perpendicular to k passing through the origin, m will be the line perpendicular to k passing through the midpoint of (0, 0) and (a, b), and n will be the line perpendicular to k passing through (a, b). Then the rotation can be written as r_l o r_k, and the translation can be written as r_m o r_l. Notice that the inverse translation can be written r_l o r_m -- but it can also be written as r_m o r_n (as either l and m, or m and n, are the correct distance apart). So we write it:
T = trans o rot o trans^-1
= r_m o r_l o r_l o r_k o r_m o r_n
= r_m o I o r_k o r_m o r_n
= r_m o r_k o r_m o r_n
= r_k o r_m o r_m o r_n (as reflections in perpendicular mirrors commute)
= r_k o I o r_n
= r_k o r_n
which is the composite of reflections in perpendicular mirrors intersecting at (a, b). And so T is in fact the rotation centered at (a, b), which is what we were expecting.
Once again, though, this is not the sort of symbolic manipulation I'd want my students to see. But then, what should we expect students to do when faced with a PARCC question where they have to rotate around a point other than the origin?
Most likely, this is something that can wait until we discuss the Slope Formula -- especially since it's this rotation that leads to the slopes of perpendicular lines. For now, one can consider such rotations only informally -- after all, the PARCC questions usually include graphs, so students might be able to perform the rotations just by counting units on the graph, rather than using an algebraic formula or manipulating mirrors and symbols.
But with rotations, we only perform a precious few of them. The only rotations that appear on the PARCC and other Common Core tests are those of magnitude 90, 180, or 270. Yet we've seen a few of these rotations centered at points other than the origin on the PARCC.
We'll begin with rotations that are centered at the origin, though. Just as we used the Two Reflections Theorem for Translations yesterday, today we'll use the Two Reflections Theorem for Rotations. So to perform the rotation of 180 degrees about the origin, we compose two reflections in mirrors that intersect at the origin, at an angle of half of 180, or 90 degrees. The obvious choices for mirrors are the x- and y-axes. We've already proved that the reflection image of (x, y) in the x-axis is (x, -y) and the reflection image of (x, y) in the y-axis is (-x, y). It doesn't matter in which order we compose these as reflections in perpendicular mirrors always commute. So we prove that the rotation image of (x, y) centered at the origin and of magnitude 180 degrees is (-x, -y).
Now our other common rotation magnitude is 90 degrees -- and this time, it will make a difference whether it's clockwise or counterclockwise. The angle between the mirrors will now have to be half of 90, or 45 degrees. There's one mirror to consider that will help us with a 45-degree angle -- the line whose equation is y = x.
We've hinted at several proofs involving reflection over the line y = x. Let's look at the quadrilateral whose vertices are (0, 0), (a, 0), (a, a), and (0, a). We can show that this figure is a kite.
So now we can apply the properties of a kite -- the Kite Symmetry Theorem. The diagonal of our kite running from (0, 0) to (a, a) bisects the angle between the x- and y-axes -- and since we know that the angle between the axes is 90 degrees, the diagonal must form a 45-degree angle with each axis. And reflecting across this symmetry diagonal must map the axes to each other and x = a to y = a.
Recall that at this point, we don't know the equations of lines, so we aren't yet certain that the graph of y = x is even a line (which we'd better figure out before trying to use it as a mirror). But we see that the value of a in the above proof is arbitrary -- it's true for every single real number a (although in case a is negative, we should probably say that the kite has sides of length |a|, not a). Therefore every single point of the graph of y = x lies on the bisector of the angle between the axes -- that is, the graph of y = x is exactly that line. And reflecting in that line maps x = a to y = a and vice versa -- that is, it switches x and y. Therefore the image of (a, b) must be (b, a).
Now that we know how to reflect in the line y = x, let's use it to perform a 90-degree rotation. It's probably easiest just to start with the reflection in y = x first, so (x, y) maps to (y, x). As for the second mirror, it depends on whether we want to go clockwise or counterclockwise. To go clockwise, the second mirror must be 45 degrees clockwise of the first mirror, y = x. That is the x-axis, and to reflect in it, we change the sign of the second coordinate. So (y, x) reflected in the second mirror is the point (y, -x), so mapping (x, y) to (y, -x) rotates points 90 degrees clockwise. To go counterclockwise, the second mirror must be 45 degrees counterclockwise of the first mirror, y = x. That is the y-axis, and to reflect in it, we change the sign of the first coordinate. So (y, x) reflected in the second mirror is the point (-y, x), so mapping (x, y) to (-y, x) rotates points 90 degrees counterclockwise.
Notice that some of the PARCC questions mention 270-degree rotations -- for example, there was a released question that mentions a 270-degree clockwise rotation. Of course, a 270-degree clockwise rotation is equivalent to a 90-degree counterclockwise rotation, so it maps (x, y) to (-y, x). If students forget this, they can still take half of 270 degrees to get 135 degrees clockwise, and they can see that 135 degrees clockwise from the line y = x is still the y-axis, just as it would have been if they'd gone 45 degrees counterclockwise instead.
Last year, I created a quick worksheet to help students perform any of the reflections and rotations mentioned in this post. (This was late in the year when we were covering PARCC questions, but now I'm giving this lesson much earlier.) It takes the coordinate plane labeled with the positive x-, negative x-, positive y-, and negative y-axes. Students can then perform the rotations on the axes to see what happens. For example, let's try our 270-degree clockwise rotation. After we rotate the paper 270 degrees clockwise, we see that where the x ought to be, we see -y instead, and where the y ought to be, it's +x. Thus the image must be (-y, x).
Okay, so we've taken care of all the rotations centered at the origin, But on the PARCC, there are questions with rotations centered at other points. These questions that I've seen direct the students to take a triangle ABC and rotate it around one of its vertices -- let's say C. Well that makes things a little easier, since then the rotation image of C is C itself. So then there are only two points that we need to find, A' and B'.
It's possible, in principle, to find formulas to determine the image of (x, y) under reflections in mirrors parallel to the axes and rotations centered at points other than the origin. We've seen, for example, that the point (x, y) reflected in the line x = a is (2a - x, y). An interesting question is, where exactly does the 2 in 2a - x come from?
To find out, we notice that if we were reflecting in the y-axis (which is parallel to x = a), then the point (x, y) is mapped to (-x, y). Now that extra 2a term looks just like a horizontal translation of exactly 2a units.
So somehow, our reflection in the line x = a appears to be the composite of a reflection in the y-axis and a horizontal translation. (This is not a glide reflection between the mirror is perpendicular to the direction of translation -- we found out last year that such a composite yields a simple reflection, not a glide reflection.)
Using symbols, let's call the composite transformation T. It is the composite of a y-axis reflection, which we'll call r_y, and a horizontal translation of 2a units, H_2a:
T = H_2a o r_y
But the horizontal translation is itself the composite of two reflections. The two mirrors here must be vertical mirrors spaced exactly half of 2a, or a units apart. We might as well let the two mirrors be the y-axis itself and the line x = a.
T = H_2a o r_y
= r_(x = a) o r_y o r_y
= r_(x = a) o I
= r_(x = a)
which is exactly what we wanted -- a reflection in the line x = a.
Likewise, we see that the reflection in the line y = b maps (x, y) to (x, 2b - y). The composite of both reflections is a 180-degree rotation about the point (a, b), which maps (x, y) to (2a - x, 2b - y) -- and that's also the composite of a 180-degree rotation about the origin and yet another translation.
Now 90-degree rotations about points other than the origin are even trickier, because now we'd have to reflect about mirrors with equations like y = x + b -- and we don't even know that's a line yet. The algebra involved in this reflection gets very messy.
Of course, if we try to visualize the rotation, another composite transformation jumps at us. To perform a 90-degree rotation (either clockwise or counterclockwise) about the point (a, b), it appears that we can first perform the translation that maps (a, b) to (0, 0), then perform the rotation centered at the origin, and finally translate (0, 0) back to (a, b).
This seems to work, but is there any reason why it should? Let's use symbols again -- in order to remember what the symbols stand for, we let "rot" stand for the rotation and "trans" stand for the translation mapping (0, 0) to (a, b). Then trans^-1 can stand for the inverse translation -- the one mapping (a, b) to (0, 0). This gives us:
T = trans o rot o trans^-1
This composite has a name in classes like linear algebra and above -- conjugation. That is, we are conjugating the rotation by the translation.
We now want to rewrite both the translation and origin-rotation with two mirrors each. And as usual, we want to choose the mirrors strategically so that some of the reflections cancel out. For the translation, we'll let k be the line joining the points (0, 0) and (a, b). Then l will be the line perpendicular to k passing through the origin, m will be the line perpendicular to k passing through the midpoint of (0, 0) and (a, b), and n will be the line perpendicular to k passing through (a, b). Then the rotation can be written as r_l o r_k, and the translation can be written as r_m o r_l. Notice that the inverse translation can be written r_l o r_m -- but it can also be written as r_m o r_n (as either l and m, or m and n, are the correct distance apart). So we write it:
T = trans o rot o trans^-1
= r_m o r_l o r_l o r_k o r_m o r_n
= r_m o I o r_k o r_m o r_n
= r_m o r_k o r_m o r_n
= r_k o r_m o r_m o r_n (as reflections in perpendicular mirrors commute)
= r_k o I o r_n
= r_k o r_n
which is the composite of reflections in perpendicular mirrors intersecting at (a, b). And so T is in fact the rotation centered at (a, b), which is what we were expecting.
Once again, though, this is not the sort of symbolic manipulation I'd want my students to see. But then, what should we expect students to do when faced with a PARCC question where they have to rotate around a point other than the origin?
Most likely, this is something that can wait until we discuss the Slope Formula -- especially since it's this rotation that leads to the slopes of perpendicular lines. For now, one can consider such rotations only informally -- after all, the PARCC questions usually include graphs, so students might be able to perform the rotations just by counting units on the graph, rather than using an algebraic formula or manipulating mirrors and symbols.
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