Discrete Math - Day 5

This was an interesting class. I assigned the first portfolio problem. These portfolio problems are designed to have students fully explain their reasoning and justify their results. I allow students to make revisions on their work until the get the results correct. These portfolio problems become mentor text that shows students what they need to do when responding to questions. Students do not receive credit for a portfolio problem until it is 100% correct.

I also use a grading standard of Essential Correct (E), Partially Correct (P), and Incomplete (I). An E says that students "get it" and know what they are doing and can communicate their results and reasoning. There may be minor issues but that these are things that a student could, in essence, self-correct. A P indicates a student knows what they are doing but gets stuck, does not explain their reasoning, or would need some prodding or help through questioning to move on. An I indicates the student cannot proceed without a lot of assistance or provides an answer without any explanation of their thinking or justification of their results.

All graded work is scored using E, P, and I. On a test, a student getting all P's would receive a grade of C. A student getting all E's would receive an A. A student receiving a mix of E's and P's would fall somewhere in between.

After going through the portfolio problem requirements and grading, we started on the day's work. This involved using values in a table to explore functional relationship. We started with linear equations. I had students create linear tables and look at the difference in the y-value. Students quickly saw that the difference was equal to the coefficient. After confirming this worked for other linear equations I posed the question as to why this would happen. Some students related it to slope but didn't come up with any definitive response. I left this as an open question and then asked if this result would always work.

A student wondered if it would work with quadratic equations and so we explored the situation for two specific equations. They saw that the first difference wasn't a constant but that if you took the second difference the result was. They also noticed that it looked like you should multiply the coefficient times the exponent. We explored this for a couple of more quadratics.

The question again was would this hold and we proceeded to explore this for cubic equations. Students found that the third difference was constant and the third difference looked like (coefficient) x (exponent) x 2. Students were stumped about what there was a 2 being multiplied.

I recapped what we learned so far, that linear equations had a first difference that was constant, quadratic equations had a second difference that was constant, and that cubic equations had a third difference that was constant. We also knew that the difference took on the form of (coefficient) x (exponent) x (something) where the something was a value of 2 for cubic equations.

I asked the class what this would mean for a quartic equation and they said that the fourth difference should be a constant. We used y = x⁴ to verify this. Students confirmed that the fourth difference was constant and saw that its value was 24. I then referenced our other result of 24 = (coefficient) x (exponent) x (something) = 1 x 4 x (something) = 1 x 4 x 6.

Several students noticed that 6 was the (exponent) x (something) component of the cubic equation and suggested writing out the value as 24 = 1 x 4 x 3 x 2. At this point someone suggested to write a one at the end of the sequence and there were several aha's that the coefficient was being multiplied by the factorial of the exponent, i.e. 24 = 1 x 4!.

I then explained that if a polynomial equation had as its largest term a_nxⁿ then the n^th difference would be
a_n x n!.

We then looked at the first few pentagonal numbers to see what we could tell about the equation. The values we had were 1, 5, 12, and 22. Students saw that the second difference was a constant and its value was 3. There was some confusion as to whether that meant it was a quadratic or cubic. I said that the second difference was 2 and asked what this meant. Most came to the realization that this meant it had to be a quadratic equation. I then asked what the coefficient was. Again, the fact that the coefficient was a fraction threw them off a bit. I wrote out that a x 2! = 3 and asked what a had to equal. They said it had to be 1.5 or 3/2. I then said that this meant the equation for pentagonal numbers started with the term 1.5x².

I created a table and talked about the contribution that 1.5x² made to the y-value. We subtracted this contribution out and saw that the results decreased by .5 each time. This meant we had a linear component of the form -.5x. Repeating the process we saw that the result was always 0. The provided an equation for pentagonal numbers of: P_n = 1.5x² - .5x. This can be written as P_n = 3x(x-1)/2 the more traditional form of the equation.

I then had students summarize what they had learned about taking differences and how these results could be used to find equations.

Visit the summary page for a student's perspective on the class and to view the lesson slides.