Monthly Archives: December 2011

The Problem with the Test

Following is a snip from a 12.19.2011 post by James K. to the AP Calculus EDG .  It was a response to a teacher’s frustration with her district increasing the number of students in her AP Calculus course and then evaluating her performance in part on the scores her students receive AP exam.  Sorry for the length of the block quote, but it’s worth absorbing.

I have some separate thoughts about tests and teacher evaluation, which I’m hoping to carve out time for sometime soon, but I’m eager to develop the conversation around the following comments:

[citing another post] … I do get students with weaker and weaker math backgrounds.  How weak? I have student who can’t evaluate sin(pi/3), and a student [who] can’t understand it’s the same to multiply a number by 1/6 and to divide it by 6. How did they pass precalculus? …. It feels nice to be able to say “I’m always willing to teach those willing to learn”, but how do you teach a class with huge gap in students’ capability? Teach fast, and you get lots of complaints. Teach slower, and the more capable students suffer. They (the more capable) are the ones that pay the price when we say “I’m always willing to …” [end internal citation]

What’s at the heart of this discussion is differentiating instruction: that is, developing and implementing activities (which can include lecture, as just one type of activity) which provide rich, effective learning for a wide variety of students.  This requires acknowledging not only that students come to the class with diverse background knowledge and conceptual frames, but also that they are likely to leave the class with different knowledges and conceptual frames– in short, serving [intellectually] diverse students … means valuing significant growth at least as much as valuing specific pre-determined objectives.

This is, I believe, the greatest and most tragic failure of mathematics instruction over the past 4000 years: the teaching of mathematics almost invariably has been designed and implemented around an assumption that the cognitive development of mathematical ability is linear in nature, with the overwhelming result that math courses are substantially about distinguishing successful students (those who are able to reproduce a specific set of mathematical procedures and algorithms in novel contexts) from unsuccessful students (who are unable to do so) … as opposed to being about the development of better mathematical thinking in all of the students in the course.

That’s gotten a bit better, just in the past 100 years or so, but the culture is still overwhelmingly focused on determining who can “do the math” (and, implicitly, who cannot).  I recognize that, particularly within the context of an AP course (where the desired outcomes of the course are richly and rigorously defined by an external examination), it’s hard to embrace the notion that it’s okay for students to come away from a lesson, a unit, and the entire course as a whole with different degrees of knowledge and understanding, and even with different ways of *approaching* their own understanding.  That said, I think it’s the work that we ought to be doing:  Teaching is about fostering growth.  It’s certainly easier to do with a roughly uniform group– but I don’t let my students get away with ignoring the hard stuff, and I don’t let myself get away with it, either.  That means we ought to be teaching at both (really, all) levels, simultaneously: give the weaker students what THEY need, while also giving the more capable students what THEY’RE ready for, all at once.  Not easy, as I say, and especially hard to do through lecture (which is why that’s just one of the types of activity we should be using…)

I certainly haven’t achieved all I want to do with respect to differentiating instruction in my classes.  I’ve also been part of numerous discussions like the one James advances above.  His argument for reaching all students where thy are with what they need is, in my opinion, the great calling of all teachers.

What struck me as unique about James’ post was his focus on interpreting the outcome of an AP Course.  I’ve never clearly thought about it in this way, but the standardization of a single international test for all students with only five possible outcomes (scores ranging 1 to 5) does drive one to believe that all students (if they would just try) will exit the course with the same curricular and comprehension results.  I’ve worked for years (with varying success) to meet my students in their areas of need and have tried to promote their unique perspectives.  I hope all of my students (no matter what their mathematical background) say my classes push them to think in new ways and encourage them to apply what they have learned in ways unique to their own perspectives.  What I hadn’t fully considered was the pervasive way external tests can leverage teachers away from that ideal.

So, I recommit myself to

the notion that it’s okay for students to come away from a lesson, a unit, and the entire course as a whole with different degrees of knowledge and understanding, and even with different ways of *approaching* their own understanding,

especially when teaching courses with external final assessments.

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Why teach statistics

The following is from the AP Statistics EDGClick here to access the original 10.30.11 post.

George Cobb … posted … on how statistics differs from mathematics and how that impacts the teaching of statistics by mathematics teachers. It was a great note but I felt that the “bad guy” was perhaps not entirely mathematics but the way in which mathematics has often been taught. While everyone the world over gets the same answer to a long division problem, if you look at other cultures and other times, you will find that the algorithms used have varied widely. Each has it’s justification, but those are rarely taught to children. This is why I refused to learn it in fourth grade;-) While some lament the substitution of an electronic device for the hallowed algorithm of their youth, both that algorithm and the plastic calculator are cultural artifacts that come and go while mathematical truth remains. In graduate level mathematics, definitions are chosen for a reason, though again that is rarely shared with the learner. What are the pros and cons and historical roots of the definitions of “real number” given by Cantor and Dedekind? So George’s comments made me think of a quote from another of my mentors that used to hang on my office door. It’s taken me this long to find it on an old Windows backup and move it to my Linux computer.

Robert B. Davis was an MIT-trained mathematician who gave up research to work with inner city children. The paragraphs below I assembled from a variety of his writings over the years — most of them never widely available.

“Whereas science and creative mathematics are essentially and necessarily tentative, uncertain and open-ended, the traditions of elementary school teaching in many instances are authoritative, definite, absolute, and certain.

Such a view is incompatible with science, with mathematics — or, for that matter, with nearly any serious body of thought. In fact many questions have no answer, some questions have many (equally good) answers, and some questions have approximate answers but no perfect answers. The “tolerance of ambiguity” that is required of anyone who would see the world realistically is a severe demand for some teachers.

It should be emphasized that the difficulty here is with some teachers, not with children. Children know that they live with incompleteness and uncertainty; scientists know that no other state is available to living human beings. Unfortunately, teachers have all too often been taught that every question has exactly one right answer, and that the child is entitled to know what it is — or, perhaps, should even be required to know what it is.”

One reason to learn and teach statistics is so we can become better mathematics teachers.

Truth.

Nested surds

A few months ago, I noted the following problem on a separate ‘blog for a textbook a colleague and I created to fill a need at our school.

Assuming the pattern continues forever, what is the value of \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}} ?

As you know, I encourage my students to be creative and solve problems in ways that make the most sense to them.  Following are three different approaches my students used when presented with this problem.  All three are essentially identical invocations of the problem’s self-similarity, but I liked the differences in how they approached the problem.

STOP!!!
SOLUTION ALERT!!
DO NOT READ ANY FURTHER IF YOU WANT TO SOLVE THESE PROBLEMS ON YOUR OWN.

Approach 1:
Notice that other than the outermost square root and addition, the entire problem is repeated inside the problem.  Therefore, if X=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}, then

X=\sqrt{1+X}
X^2-X-1=0
X=\frac{1\pm\sqrt{5}}{2}

The value of X is obviously positive, so X=\frac{1+\sqrt{5}}{2}=\phi, the golden ratio!  How’s that for an unexpected surprise?!

Approach 2:
Two students thought in terms of sequences.  Each assumed a first term of a_1=\sqrt{1} and iterated using a_n=\sqrt{a_{n-1}}.

Recognizing the decimal equivalent of \phi, the first student concluded that the iterated surd appeared to be the golden ratio, but said in a class discussion the following day that while he had compelling evidence for an answer, he knew he hadn’t proven his solution.

Approach 3:
My second student who thought of a sequence approach reached the a_n=\sqrt{a_{n-1}} statement and realized that if she was able to iterate forever and this problem actually had an answer, then there would be a point where L=a_{n-1}=a_n where L is the limit of the iteration.  At this point, she wrote L=\sqrt{1+L} and proceeded as in Approach 1 to get L=\phi.

Some readers might argue that Approaches 1 and 3 really aren’t distinct, but I maintain that the students who used Approach 1 (the clear majority) were thinking purely algebraically and took advantage of the problem’s self-similarity, while the girl who used Approach 3 was thinking numerically and found her answer using limits.

Minor Extension:
For fun, I also assigned 1+\displaystyle\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}.  At first, this problem looks very different from the one discussed above, but its closed form delighted some (and disturbed others!)  with its connection.  I hope you enjoy!

Three Little Geometry Problems

Here are three variations of geometry problems I got from @jamestanton on Twitter.

  1. The numerical measure of a rectangle’s area and perimeter are equal (P=A) (obviously the units are different).  If the rectangle’s sides have integer lengths, what are the dimensions of the rectangle?
  2. The numerical measure of a box’s surface area and volume are equal (V=SA).  If the box’s sides have integer lengths, what are its dimensions?
  3. The numerical measure of a right triangle’s area and perimeter are equal (P=A).  If the triangle’s sides have integer lengths, what are its dimensions?

To give a complete solution to a math problem, remember that you must

  1. show that your proffered solution(s) is (are) correct, and
  2. shows that no other solutions exist.

Find convincing arguments that you have found all of the solutions for each.  While my solutions are shown below, I eagerly welcome suggestions for any other approaches.

STOP!!!
SOLUTION ALERT!!
DO NOT READ ANY FURTHER IF YOU WANT TO SOLVE THESE PROBLEMS ON YOUR OWN.

Problem 1:  P=A for a rectangle
Let the a=length and b=width.  Without loss of generality, assume a\le b.  Then, a\cdot b=2(a+b) \Longrightarrow \frac{1}{2}=\frac{1}{a}+\frac{1}{b} which implies 2<a\le 4.

If a=3, b=6.  If a=4, b=4.  Thus, the only rectangles for which P=A are a 3×6 and a square with side 4.

Problem 2:  P=SA for a box
Let the a=length, b=width, and c=height.  Without loss of generality, assume a\le b\le c.  Then, a\cdot b\cdot c=2(a\cdot b+a\cdot c+b\cdot c) \Longrightarrow \frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} which implies 2<a\le 6.

It wasn’t worth it to find all these by hand, so I wrote a quick spreadsheet to find all c values for given a and b values under the condition a\le b\le c.


Therefore, there are 8 such boxes with integer dimensions: 3x7x42, 3x8x24, 3x9x18, 3x10x15, 4x5x20, 4x6x12, 5x5x10, and 6x6x6.

Problem 3:  P=A for a right triangle
Let the legs be a and b and the hypotenuse be c.  Perhaps there is a way to employ the technique I used on the first two problems, but my first successful solution invoked a variation on Euclid’s formula:  For some integer values of k, m, and n with m\le n, a=k\cdot (m^2-n^2), b=2k\cdot m\cdot n, and c=k\cdot (m^2+n^2) will form all Pythagorean triples (although not uniquely).

Because \frac{1}{2} a\cdot b=a+b+c, Euclid’s formula gives

\begin{tabular}{ r c l }  \(2\) & \(=\) & \(\frac{\displaystyle a\cdot b}{\displaystyle a+b+c}\) \\  & \(=\) & \(\frac{\displaystyle [k\cdot (m^2-n^2)]\cdot [2k\cdot m\cdot n]}{\displaystyle [k\cdot (m^2-n^2)]+[2k\cdot m\cdot n]+[k\cdot (m^2+n^2)]}\) \\  & \(=\) & \(\frac{\displaystyle 2k^2mn(m^2-n^2)}{\displaystyle 2km(m+n)}\) \\  & \(=\) & \(kn(m-n)\)  \end{tabular}

What started out feeling like a difficult search for unknown side lengths has been dramatically simplified.  2=kn(m-n) can only happen if

  1. k=2, n=1, m-n=1 \rightarrow m=2 which Euclid’s formula converts to a=6, b=8, c=10 .
  2. k=1, n=2, m-n=1 \rightarrow m=3 which Euclid’s formula converts to a=5, b=12, c=13 .
  3. k=1, n=1, m-n=2 \rightarrow m=3, but this leads to a repeat of the first solution.

Therefore, there are only two right triangles with the property P=A:  the 6-8-10 and the 5-12-13 right triangles.

Again, any other solution approaches are encouraged and will be posted.

I can guess ANY polynomial with only 2 points

I thought this was a brilliant solution to a surprising problem. At first glance, it seems impossible…

Given the coordinates of only two points of your choosing from an unknown polynomial, could you guess the exact equation of that polynomial?

The problem as originally stated is here and a few different explanations are here.  For the computations, I think the use of technology is a no-brainer.

Thought Variations and Tests as Learning Tools

I love seeing the different ways students think about solving problems.  Many of my classes involve students analyzing the pros and cons of different approaches.

As an example, a recent question on my first trigonometry test in my precalculus class asked students to find all exact solutions to 3sin^2x-cos^2x=2 in the interval 0\leq x\leq 2\pi.  Admittedly, this is not a complicated problem, but after grading several standard approaches to a solution, one student’s answer (Method 3 below) provided a neat thinking alternative.

As an assessment tool, I don’t view any test as a final product.  While optional, all of my students are encouraged to complete corrections on any test question which didn’t receive full credit.  For me, corrections always require two parts:

  1. Specifically identify the error(s) you made on the problem.
  2. Provide a correct solution to the problem.

My students usually take their tests on their own, but after they are returned, they are encouraged to reference any sources they want (classmates, notes, me, the Web, anyone or anything …) to address the two requirements of test corrections.  The point is for my students to learn from their misunderstandings using any source (or sources) that work for them.  Because students are supposed to do self-assessments, I intentionally don’t provide lots of detail on my initial evaluation of their work.

To show their different approaches, I’ve included the solutions of three students.  Complete solutions  are shown so that you can see the initial feedback I offer.  If there’s interest, I’m happy to provide examples of student test corrections in a future post.

Method 1:  Substitution–By far the most common approach taken.  This student solved sin^2x+cos^2x=1 for sin^2x and substituted.  Others substituted for cos^2x.  [You can click on each image for a full-size view]

This solution started well, but she had an algebra error and an angle identification problem.

Method 2:  Elimination–The same Pythagorean identity could be added or subtracted from the given equation.  After talking yesterday with the student who created this particular solution, I was told that he initially completed the left column and attempted the work in the right column as a check at the end of the period.  After committing the same algebra error as the student in method one, he realized at the end of the test that something was amiss when the cosine approach provided an answer different from the two he initially found using the sine approach.

After conversations with classmates yesterday, he caught his algebra error and found the missing answer.  He also corrected the units issue.

[I’m not sure whether I should even care about the units here and am seriously considering removing the 0\leq x\leq 2\pi restriction from future questions.  With enough use in class, they’ll eventually catch on to radian measure.]

Method 3:  Creation–This approach was used by only one student in the class and uses the same Pythagorean identity.  The difference here is that he initially moved the cos^2x term to the other side and then added an additional 3cos^2x to both sides to create a 3 on the left using the identity.  Nothing like this had been discussed in class, and I was quite interested to learn that the student wasn’t even sure his approach was valid.  What I particularly liked was that this student created an expression in his solution rather than eliminating expressions given in the initial equation as every other student in the class had done.  It reflected a mantra I often repeat in class:  If you don’t like the form of a problem (or want a different form), change it!

Also notice how he used an absolute value in the penultimate line rather than the more common \pm.

Again, nothing especially deep about any of these, but I learn so much from watching how students solve problems.  Hopefully they gain at least as much from each other when comparing each others solutions during corrections.