This page is intended as a place to exchange ideas about teaching the Elementary Mathematical Models course. I am just beginning it, so there is very little here now.

Data Tables and Position Numbers

Difference Equations and Recursion

Quadratic Formula and Vertices of Graphs

Recognizing Mixed Models

From Doug Ensley

In the very beginning of the class this semester (Spring 98), I introduced the idea of a data table in the following format:

I continued to use this format repeatedly throughout the semester. This has several advantages. First, it allows the introduction of n into difference equations in a natural way, first in a concrete form using numerical values, then later symbolically. For example, if the following pattern is observed:

it is natural to formulate this description: each term is found by adding its position number to the preceding term. That later can be expressed as a difference equation with variable subscripts. Indeed, the position number provides a handy realization of the subscript idea: the subscript attached to a term is nothing other than the position number for that term. A second advantage of using a data table in this form is that it gives a nice distinction between difference and functional equations: the difference equation uses preceding terms (above in the table) to calculate a term, whereas the functional equation uses only the position number (to the side in the table). A third benefit is that it ties in nicely with a graph of the sequence, and with the idea of creating a table of ordered pairs for graphing. The position number terminology is useful in formulating verbal descriptions of patterns reflected by difference equations, especially in the unit on quadratic growth.

The worksheet called Number Patterns 1 uses tables and position numbers to introduce students to number patterns.

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The worksheet called Number Patterns 2 uses these ideas to introduce students to subscripts and difference equations.

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 Quadratic Formula and Vertices of Graphs

This semester, I developed a new way to approach the derivation of the quadratic formula. When I reached that point in the course, the students had already been finding features of the graphs of quadratic functions, such as the axis of symmetry, the vertex, and the x intercepts, using the coefficients (a, b, and c for the function ax² + bx + c). They had experience using the quadratic formula. My objective was to give them some understanding of where the formula comes from. I will outline the presentation below, in much the same way it was done in class, working with a specific example.

Suppose we are graphing the equation y = 3x² - 8x + 2. Using a formula discussed earlier in the course (-b/2a ), we find the axis of symmetry to be x = 8/6 = 4/3. To find the vertex, we substitute 4/3 for x and compute y = -10/3. These features are included in a graph, along with the y intercept at 2, and the symmetric point reflected across the axis of symmetry. The next step is to find the x intercepts. We know these will be equally distant from the axis of symmetry, but we don't know how far away from the axis they will be. So, introduce the variable w (for width) which will represent the distance from the axis to the x intercept on either side. In particular, one of the x intercepts will have to be at 4/3 + w. If that is substituted for x in the equation, the result will have to be y = 0. This gives rise to the equations

Now this equation can be easily solved for w using squareroots, but observe that the two numerical coefficients are familiar values: 3 is the coefficient of x² in the original equation, and -10/3 is the y coordinate we found for the vertex. It is easy to see that the steps above will always lead to these results: if you have found the vertex (x₀ ,y₀ ) while graphing the equation y = ax² + bx + c, then the x intercepts (if they exist) can always be found to the right and left of the axis of symmetry, a distance w = (-y₀ /a)^1/2 . In particular, for the example problem, once the vertex (4/3, -10/3) has been found, it is easy to compute the x intercepts as (4/3) + (10/9)^1/2 and (4/3) - (10/9)^1/2, rather than recomputing things with the quadratic formula. (I am using the exponent 1/2 solely due to the formatting limitations on this web page. In class these results are expressed using radicals.)

These results can be expressed as a formula using just a, b, and c, if one chooses. The coordinates of the vertex will be (-b/2a) and a(-b/2a)²+b(-b/2a) + c, and these can be used in place of 4/3 and (-10/9) in the computations above. The result is pretty complicated, but can be algebraically simplified. And that is one way to derive the quadratic formula.

A colleague pointed out that this development depends on knowing the location for the axis of symmetry, which is generally derived through completing the square. In our class, the axis of symmetry formula had been discovered empirically in a laboratory session exploring graphs on computers. This is hardly the same as a logical derivation. However, the approach used above does show that the x intercepts are symmetrically placed relative to the axis of symmetry that we found from the empirically deduced result. Thus, the derivation demonstrates that the axis of symmetry formula is correct, at the same time that the quadratic formula is derived. Of course, to really demonstrate the symmetry of the graph in a valid way, it is ultimately necessary to complete the square, expressing the equation in the form y = a(x-x₀ )² + y₀ . But I have found that many students can not make much sense out of that kind of development. The approach sketched above seems a little less abstract than completing the square, and for most of my students, it seemed to make sense.

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For arithmetic growth, quadratic growth, and geometric growth, the EMM book includes simple tests by which each type of model can be recognized: constant differences for arithmetic growth; constant second differences for quadratic growth; and constant growth factors for geometric growth. There is a similar test for mixed models: if the first differences of a sequence exhibit geometric growth, then the sequence is defined by a mixed model. Put another way, if the sequence of differences are observed to have constant growth factors, then the original sequence follows a mixed model difference equation.

This fact can be used in a test very similar to the test for constant second differences used for quadratic growth. However, rather than computing second differences by subtracting successive terms in the sequence of first differences, compute growth factors by dividing successive terms. If the results are all constant, then the original sequence is an example of a mixed model.

To verify the validity of the test, suppose that for any n, the ratio

(a_n+2 - a_n+1)/(a_n+1 - a_n) = r

is constant. Algebraic rearrangement shows that

a_n+2 - ra_n+1 = a_n+1 - ra_n

Since this holds for any n, it must be that a_n+1 - ra_n is constant, say d. But that shows that

a_n+1 = ra_n + d

and verifies that the sequence a_n satisfies a mixed model difference equation. In a similar way, it is easy to verify the converse.

For a real data set, it is unlikely to observe this exactly, so the idea would be to look for first differences that approximately grow by a constant growth factor. If that is observed, it is an indication that a mixed model will probably be a reasonably good fit. The assumed constant growth factor defines the parameter r. The other parameter, d, can be determined through trial and error to fit the data as well as possible. A mathwright module called demodel is useful for this purpose.

One example of interest (no pun intended) here is the computation of successive balances when repeated payments are made to a savings account, or against a loan. It is an easy matter to compute the first several terms in such a sequence from an understanding of interest, with no insight about the operation of a mixed model. For example, suppose you make payments of $100 per month to a savings account paying .5% interest each month. The original balance is 100. After one month, you add $.50 interest and another $100 deposit to obtain a balance of $200.50. After another month add $1.0025 interest and a $100 deposit for a balance of $301.5025. Continuing in this fashion, it is easy to generate several additional terms. The sequence that results will exactly follow a mixed model, and so will exactly satisfy the test described above.

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Doug Ensley (home page), of Shippensburg University, Shippensburg, PA created some Supplementary Problems for use in a course in algebra. Doug made the following comments.

I should tell you that the course I used this for is a non-credit course to remediate poor algebra skills in preparation for a traditional college algebra course. We are moving toward splitting this course so that it also serves as a "competency" course for students who are not eventually required to take the college algebra course. I used your book in anticipation of this change, but I felt I had to supplement with a fair bit of "drill" problems for the students going on to college algebra. Typically these students just need more practice, but I believe using your book gave them a better sense of what the algebra is for and so hopefully they will retain it longer.

I just wanted to give that disclaimer to explain the flavor of the supplementary problems on my web page. My only innovation for you to pass along to others teaching the course is to have students bring graphs and data from newspapers and magazines as part of their homework and then turn those into the exercises for the next class period. This idea made the creation of new problems pretty painless.

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