Wednesday, December 23, 2009

Mediated Algebra Project: Success!

Three instructors used the Mediated Algebra Project (MAP) materials during the Fall 2009 semester.  Kathie Yoder taught a section that met early afternoon twice weekly , and Kathy Yoshiwara and I taught the two sections that met mid-morning four-days-a-week.

We now have evidence that MAP students learn more than cohorts in other sections of the Pierce College intermediate algebra course:  On the departmental Math Exit Test (MET), our three sections all scored at least 2.5 standard errors above the department mean.

But we had significant setbacks during the semester.

We had numerous technical difficulties.  Many of the WeBWorK problems I had authored had coding errors and/or needed refinement in wording or formatting.  And most of the WeBWorK exercises taken from the national WeBWorK library were poor fits for our project and had to be rewritten or removed from our problem sets during the semester.

Our sets of video tutorials--intended to help with drill and skill exercises--had many gaps in content.  And yet we were not given sufficient space on our school's server to store the videos created  by our faculty for the MAP.  Instead, our IT department arranged that only a subset of those videos would be accessible at any one time.

All three instructors found that the project's classroom activities and clicker questions required more time than was available in a class meeting.  Some of the activities, or the clicker questions, or both would go unused in each lesson.

We heard complaints about our WeBWorK assignments, the insufficiency of available videos, and the amount of work we asked the students to do both in and outside of class.

But the students who persisted in MAP averaged much higher on a department-graded common exam than students from the other sections of intermediate algebra.

Tuesday, December 15, 2009

The Correlation Coeffiicent as cosine theta

Mathematicians define the dot product between vectors  \vec{v}= (v_{1}, v_{2}, \, \ldots \, , v_{n}) and  \vec{w}= (w_{1}, w_{2}, \, \ldots \, , w_{n}) as

\vec{v} \cdot \vec{w} = v_{1} w_{1} + v_{2} w_{2} + \, \cdots \, + v_{n} w_{n}

On the other hand, the alternate geometric definition for the dot product popular with physicists is

\vec{v} \cdot \vec{w} = \left|\left|{\vec{v}\right|\right| \,\left|\left|{\vec{w}\right|\right| \,\cos \, \theta

\cos \, \theta = \frac{\vec{v} \cdot \vec{w}}{\left|\left|{\vec{v}\right|\right| \,\left|\left|{\vec{w}\right|\right|

And statisticians define Pearson's correlation coefficient r so that

r = \frac {\sum (x_{i} - \bar{x})(y_{i} - \bar{y}) }  {\sqrt{\sum (x_{i} - \bar{x})^2}  \sqrt{ \sum (y_{i} - \bar{y})^2}}

Thus if we set  \vec{v} = (x_1 - \bar{x}, x_2 - \bar{x},\, \ldots \, , x_n - \bar{x}) and  \vec{w} = (y_1 - \bar{y}, y_2 - \bar{y},\, \ldots \, , y_n - \bar{y}) , then r = \cos \,\theta.

The idea is to think not of n ordered pairs (x1, y1), (x2, y2), ..., (xn, yn), but rather to think of two vectors in n-dimensional space. When the vectors are pointing in the same direction, the angle between them is zero and the correlation coefficient is cos 0 = 1. When the vectors point in opposite directions, the correlation coefficient is the cosine of a straight angle, r = -1. And when the vectors are orthogonal, the correlation coefficient is the cosine of a right angle, r = 0.

The only tricky part is that the two n-dimensional vectors are not the vectors \vec{x} and  \vec{y}, the vectors containing all the x_{i} and y_{i} respectively.  Instead, the necessary two n-dimensional vectors are the \vec{v} and \vec{w} defined above.

And nicely, the least-squares regression line for the (x_i , y_i ) data is y = mx + b, where  m= r \frac{\left|\left|\vec{w}\right|\right|}{\left|\left|\vec{v}\right|\right| } and b = \bar{y} - m \bar{x}.  (Notice that the variance \sigma_{x}^{2} = \frac{\vec{v} \cdot \vec{v}}{n}, so m can also be written as  m= r \frac{\sigma_y}{\sigma_x}.

One typically derives the least-squares regression line by finding m and b that minimize  \sum  (m x_i +b - y_i )^2.  But one can alternatively use the n-dimensional vector point of view, where the coefficients m and b correspond to the solution of the vector equation m\vec{x} + b\vec{1} = \hat{y}.  The vector \vec{1}= (1, \, 1, \, \ldots \, , \, 1) is the vector of all 1's and the vector \hat{y}  is the orthogonal projection of the vector  \vec{y} onto the space spanned by \vec{x} and \vec{1}.

Monday, December 7, 2009

Generating Pythagorean Triples

The 5 millennia old clay tablet designated Plimpton 322 contains a trig table. The second and third columns represent a leg and hypotenuse of a right triangle with positive integer sides. The rows are arranged in approximately equal steps of angle.

The existence of such a table suggests that the Babylonians were adept at producing Pythagorean triples (integers a, b, and c satisfying a2 + b2 = c2), a trick which is also useful to many algebra, geometry, and trig teachers attempting to create exercises with nice values.

Every positive Pythagorean triple can be generated by choosing positive integers u and v with u > v and setting a = 2uv, b = u2 - v2, and c = u2 + v2 (or by scaling such a triple by a positive integer). We'll derive that fact below. (Pythagorean triples with no common factor are called primitive Pythagorean triples, and all the primitive Pythagorean triples are generated when u and v are relatively prime with exactly one of them being odd.)

It's straightforward to verify that the a, b, and c so defined do form a Pythagorean triple. And conversely, if a, b, and c form a Pythagorean triple, then (a/c, b/c) is a point on the unit circle x^2+y^2=1, so the positive Pythagorean triples can be mapped onto the rational points of the unit circle that lie in the first quadrant.

The line y = 1 + mx will intersect the unit circle at (0,1) and also at a point in the first quadrant when the slope m is between -1 and 0. In fact, we can find the x-coordinate of the second intersection point by solving the equation x2 + (1 + mx)2 = 1--we find that x=\frac{-2m}{1+m^2}, so y=1-\frac{2m^2}{1+m^2}.

Thus the second intersection point is a rational point if m is rational. Of course the slope between (0,1) and any rational point on the unit circle is rational, so we have a 1-1 correspondence between positive rational points on the unit circle and rational slopes between -1 and 0.

We now assume that m is a rational number between -1 and 0, so we can write m = -v/u, where u and v are positive integers with u > v. Then the second intersection point we found above has the form

\left(\frac{-2(\frac{-v}{u})}{1+(\frac{-v}{u})^2}, \,\, 1- \frac{2(\frac{-v}{u})^2}{1+(\frac{-v}{u})^2}\right) = \left(\frac{2uv}{u^2+v^2}, \frac{u^2-v^2}{u^2+v^2}\right).

Thus every rational point on the unit circle can be written in this form. In particular, every primitive Pythagorean triple a, b, and c can be expressed as above in terms of u and v.

Thursday, December 3, 2009

Edublog Awards

My nominations for the 2009 Edublog Awards ( are:

Best resource sharing blog:
Best educational tech support blog:

Many thanks to Maria Andersen for providing a wonderful resource!