"There is no one way"

Thursday, December 19, 2013

Asilomar Report, Part 2

Read about my morning at the Asilomar meeting of the California Math Council here.

My afternoon was taken up with function diagrams. First, I attended Martin Flashman's presentation on this topic, then I made my own presentation, and finally I had dinner with Martin. (If you know nothing about function diagrams, read no further. Or find out about them on my Web site, and/or on these previous posts before reading on.)

My talk went exceedingly well, though as usual I was not able to get to all the things I had hoped to cover. Along the way, I ended up learning a couple of new things. I will share those here with any function diagram aficionados out there.

1. During my talk, I showed how the linked function diagrams representation of the composition of functions helps understand the chain rule. One of the participants asked about the location of the focus of the composite function, in the case of the composition of two linear functions. I was unable to give an answer, but I did think about it after getting home. The question is hard to formulate clearly, because the location of the focus depends on the distance between the x and y number lines. But the composition of functions is represented by linked function diagrams, like this:


So is the question about the location of the focus assuming a standard distance between the number lines, including in the case of the composite function? Or is it about the location of the focus in the linked diagrams representation?

At home, I investigated the case of two functions f(x) and g(x) with magnifications greater than 1, and I assumed a standard distance d between the x and y number lines. After doing some algebraic manipulation, I found out that if f's focus is at a distance a to the left of its x number line, and g's focus is at a distance b of its x number line, then the focus for the composite function is at a distance ab / (d+a+b) of its x number line. Or, another way to put it is that if the distance we are looking for is c, we have

1/c = 1/a + 1/b + d/ab

Again, this is if the distance between the x and y number lines is d in all three diagrams.

I may well have made a mistake, but I did check this answer in one particular case, and it worked. If you have a different answer, let me know!

2. In the talk, I also discussed the point/line duality in the case of linear function diagrams, leading to the idea that in the world of function diagrams, the solution of a system of linear functions is the line joining the two focuses (applet) — the dual of the situation in the Cartesian representation, where the solution is the point at the intersection of two lines. Later, I showed how the input-output lines of the function f(x)=1/x remains tangent to an ellipse. (Check out this animation.) What I learned from Martin over dinner, was that in the real projective plane, conics exist as sets of points, in the familiar representation, or as envelopes of a set of lines, the dual situation. In this case, the ellipse is simply the dual of the corresponding hyperbola in the Cartesian representation! That totally made my day.

--Henri

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