### Problem solving by changing the box and the structure of parameters

Some of my readers may recall that as a physics teacher and education researcher, I'm interested in how mathematics is used in science. The diagram that helps me think about how it works is my "four-box model" below.

The key idea is that we look at a physical system and decide that some of the measurements that we use to quantify our description of the physical world behaves in a way that reminds us of how some mathematical system works. We then treat our measured numbers as if they were mathematical quantities (modeling) and use the calculational tools that come along with the mathematical system to solve problems that we couldn't otherwise solve (processing). We then see what those solutions tell us about our physical system (interpreting) and decide whether the results are any good (evaluating). If they are, it encourages us to blend our physical concept with the mathematical one, building a rich mathematical description of the physical world. [See Redish & Kuo]

I recently had an experience that demonstrated to me that this sometimes works in unexpected ways, and in doing so, gave me some insights both into the structure of parameters in equations and into how I think about math in science.

We spent last Thanksgiving in Estes Park, Colorado with friends and relatives. Since Estes Park is at 7500 feet, since a few of us were pretty old (70's and 80's), and since we were all science geeks, one of us brought an oximeter and everyone wanted to try it. Some older folks (like me) have a tendency to respond poorly to the low levels of oxygen at high altitudes by experiencing dizziness, nausea, and headaches. The oximeter is a device that you stick on your finger and it measures how well oxygenated your blood is. If you fall below 90% you need to worry.

When I got home, I ordered one from Amazon. (It's a measuring device!

*Of course*I have to have one! It's only $15!) When my (physicist) friend Royce came to visit a couple of weeks later, I showed it to him and he asked, "How does it work?" Hmmm. Although we had asked that question in Estes Park, the answer "It measures the transmission of light through your finger and oxygenated and non-oxygenated blood absorbs the light differently" satisfied us. But, as usual, Royce likes to go deeper. We looked up some Wikipedia articles and found some interesting physics and math involved. Look it up if you like, but that's not the point of this blog.
The point of this blog is about playing with the simple algebraic equations that come up in (the first phase) of studying (a simplified version of) the absorption problem. Here's how it works.

The oximeter uses LEDs of two different colors. Oxygenated and non-oxygenated blood absorb these two colors at different rates. The red and blue curves in the figure show how oxygenated and non-oxygenated blood absorbs different frequencies of light (horizontal axis).

(Figure: By Adrian Curtin - Own work, CC BY-SA 3.0,

Suppose we have color 1 (say 700 nm) and color 2 (say 900 nm) and that oxygenated blood absorbs them at rates a

_{1}and a_{2}, while non-oxygenated blood absorbs them at rates b_{1}and b_{2}. (The values are the places where the curves cross the dotted lines.) Then suppose we have a mix of x amount of oxygenated and y amount of non-oxygenated. We want to calculate the ratio y/x. (Actually, x/(x+y), but the math there is not as interesting.)
First, let's set up the equations. Our result for colors 1 and 2 will be r

_{1}and r_{2 }where
Well, this is pretty trivial.* We have two equations in two unknowns. Solving them just requires 9

^{th}grade algebra. One way to do it is by substitution. Solve the first equation for y in terms of x, then substitute that result into the second equation and get an equation for x alone. Solve that equation for x. Then put the result for x into the first equation and get the result for y. Then take the ratio of x/y. Here's what it looks like. (You don't need to follow it through. I just want to show how messy it looks.)
This is straightforward, but tedious, especially with all the parameters to keep track of. (Amazingly, I got it right the first time! Pat self on back.) But the solution has an interesting structure to it. Here's the result:

then

This looks like both the numerator and the denominator of our result.

Usually we treat parameters in a physical problem as an "enriched number"; that is, it's not just a number. It's a number with units. But we don't typically give it more structure, as least in introductory physics.

But suppose we treat our parameters using a position vector as a metaphor for our set of parameters. Let's define a 3-dimensional parameter space, with dimensions 1, 2, 3. So (

*a*_{1},*a*_{2},*a*_{3}), (*b*_{1},*b*_{2},*b*_{3}), and (*r*_{1},*r*_{2},*r*_{3}) would be vectors in this space. We have to use a 3-space since you can't do cross products in 2-space. We'll just make*a*_{3}=*b*_{3}= r_{3}= 0.
Then we could write our two equations in two unknowns as a single vector equation with three components (the third being 0 = 0).

We want to solve for

*x*and

*y*. From our experience with cross products, we know that a vector crossed with itself is zero.** We can easily solve for

*x*and

*y*by taking cross products with the vectors

*a*and

*b*. Since the third component of each of the vectors is 0, we only have to look at the third component of the cross product since the 1 and 2 component terms will each contain a third component of some vector and therefore be equal to zero.

Similarly

and using the fact that

we find that our ratio is

Which is the same as I got before.

Somehow I find this a lot more satisfying than just churning. Why do I feel that way? The calculations both are straightforward, but the second approach feels more elegant.

I think that the answer is that the second solution doesn't just look at the problem, say, "This is a turn-the-crank algebra problem". It makes use of the symmetry in the parameter structure of the equations to imbed the simple algebra problem of numbers and symbols in a larger mathematical structure: vectors. This is a more complex mathematical structure that includes standard algebraic manipulations as a subset. Imbedding the algebra problem in vectors allows me to use the more powerful structure of vectors (and cross products).

This reminds me of the classic Gauss trick. The story goes that in 4

^{th}(?) grade the teacher wanted to keep the students quiet so he set them the task of adding all the numbers from 1 to 100 expecting it to take them a long time. Gauss came up a minute later with the answer. Instead of just adding the numbers, he paired them: 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101… and saw that there was a symmetry in the problem. Since there will be 50 such pair, the answer is 50 x 101 = 5050. The teacher saw the task as one using a particular mathematical structure: addition. Gauss reframed the task to make use of a more powerful mathematical structure, multiplication.
In both cases, we see the value of not just taking the obvious mathematical structure for granted, but of looking what additional mathematical modeling we can imbed our task in so as to have access to more powerful tools. I suspect that this kind of analysis also gives insight into the ontology of equations – what kind of things I think the elements of an equation are and what I can do with them. I hope to write more about this later.

I note that the advanced research literature is filled with examples of reframing an equation to treat it by more advanced mathematical techniques, but I can't think of too many more examples that would be obvious to a sophomore physics major. Anyone else have some examples to share?

* This is a simplified model of the system. Your finger is not just made of blood! Some processing has to be done to take out the absorption of the light by the flesh and this is done by looking at the oscillatory pattern of the signal. But since that's not our point here, we'll stick with the simpler model.

** One way of thinking about this is to consider the cross product geometrically. The cross product of two vectors has the magnitude equal to the area of the parallelogram that is created by the two vectors. If the two vectors are in the same direction, the area is zero.