Well, perhaps I muddied the water for some of you. Sorry about that. In terms of skills, when you solve |u|<a where u is any expression and a is another expression [I thought it was for constants, but it turns out to always work], you can solve by finding the intersection (and) of the solutions to u<a and to -u<a, which we write u<a and -u<a. When you solve |u|>a, you find the union (or) of the solutions to u>a and to -u>a, which we write u>a or -u>a.

My explanation in the supplemental handout was to motivate why this works. After all, the course is not just about skills, but it is also about justification. The absolute value is a piecewise-defined function. That is, there are different rules depending on the value of the expression being worked with. The skill-based method that works for absolute value does not work for other piecewise-defined functions. But thinking about each of the "pieces" separately and joining them properly will always work.

Since the handout was a first edition, I'm curious where you found the biggest issues.

## Wednesday, August 27, 2008

## Monday, August 25, 2008

### Solving Equations Graphically

So, in class today, some of you may have been wondering about the computer program that I was using. This utility is called Grapher and it is installed in any recent Mac OS computer. You'll find it in the Utilities folder within Applications.

I was wondering then whether there is a similar resource available for Windows computers as well. Doing a Google search on "Graphing Calculator" I found the following possibility: GraphCalc. I don't have immediate access to check this out, so I'd certainly welcome some comments here as to how well it works.

Now that you have something to work with (and a calculator will work as well, just a little slower), here is something interesting to notice. To solve ax=sin(x), we need to plot y=ax and y=sin(x). You also need to choose a value of a. In Grapher, you would add a New Equation (Cmd-Opt-N) like a=0.25. You now need to find where these graphs intersect.

In class, we learned that we can create new equations that have the same solutions by performing the same operation to both sides (other than division, where we worry about division by zero). So we could get a new equation like a=sin(x)/x. Now we plot y=a and y=sin(x)/x. If you add these as two new graphs instead of getting rid of the old plots, you can compare the two equations graphically. Here is the plot:

I used different colors to distinguish which intersections I was looking for.

For the original equation (ax=sin(x), shown in red), we see there are three intersections. For the new equation (a=sin(x)/x, shown in green), we only have two intersections. But those two intersections agree exactly with the original (see the highlighted intersection at the same value of x marked by circles), and the third corresponds to x=0 which disappeared because we divided both sides by x.

I was wondering then whether there is a similar resource available for Windows computers as well. Doing a Google search on "Graphing Calculator" I found the following possibility: GraphCalc. I don't have immediate access to check this out, so I'd certainly welcome some comments here as to how well it works.

Now that you have something to work with (and a calculator will work as well, just a little slower), here is something interesting to notice. To solve ax=sin(x), we need to plot y=ax and y=sin(x). You also need to choose a value of a. In Grapher, you would add a New Equation (Cmd-Opt-N) like a=0.25. You now need to find where these graphs intersect.

In class, we learned that we can create new equations that have the same solutions by performing the same operation to both sides (other than division, where we worry about division by zero). So we could get a new equation like a=sin(x)/x. Now we plot y=a and y=sin(x)/x. If you add these as two new graphs instead of getting rid of the old plots, you can compare the two equations graphically. Here is the plot:

I used different colors to distinguish which intersections I was looking for.

For the original equation (ax=sin(x), shown in red), we see there are three intersections. For the new equation (a=sin(x)/x, shown in green), we only have two intersections. But those two intersections agree exactly with the original (see the highlighted intersection at the same value of x marked by circles), and the third corresponds to x=0 which disappeared because we divided both sides by x.

### Calculus I -- Welcome Fall 2008

Welcome to JMU and your first semester of calculus. I'm excited this semester to be teaching calculus again. I hope that you're excited as well.

So, while I was preparing for class, I came across MIT's open courseware program. This is a pretty amazing collection of knowledge that is freely accessible. Feel free to browse it. In particular, I found a text written by Gilbert Strang that will be an outstanding parallel reference for our course this year. I'm especially impressed with how he makes the text conversational in style rather than the more typical dry style of math textbooks.

We'll be pushing through the first chapter of our official textbook very quickly as it should be a review of mathematics that you have already taken.

See you in class!

So, while I was preparing for class, I came across MIT's open courseware program. This is a pretty amazing collection of knowledge that is freely accessible. Feel free to browse it. In particular, I found a text written by Gilbert Strang that will be an outstanding parallel reference for our course this year. I'm especially impressed with how he makes the text conversational in style rather than the more typical dry style of math textbooks.

We'll be pushing through the first chapter of our official textbook very quickly as it should be a review of mathematics that you have already taken.

See you in class!

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