Indy Telecom & Industrial Media

Recently, during a not particularly interesting meeting I began graphing. First, I’d draw a line between (0, 9) and (1, 0), then I’d draw a line between (0, 8 ) and (2, 0), then between (0, 7) and (0, 3)…and so on till I had drawn a line between (0, 1) and (9, 0), which resulted in an aesthetically pleasing curve.

Soon, I began to wonder if, instead of simply drawing the curve, I could express it mathematically. For example, a parabola can be seen as the set of points which are equidistant from a focus and a directrix. Was there some similar way to describe the curve I had been drawing? I decided that the first step should be to see if I could represent my curve as a function in the form of y = f(x). Unfortunately, taking that first step has proven to be more difficult than I originally expected. For example, I can express the slope of any of the lines as:

Slope = (n – MAX) / (n + 1)

and I can express the y intercept of any of the lines as

Y Intercept = MAX – n

With those two equations, I can find a general form of y_n = f(x) for any of the lines. Specifically,

y_n = x(n – MAX) / (n + 1) + MAX – n

Armed with that, I can find the point where line n crosses line n + 1, that is, the point where y_n(x) = y_n+1(x). Specifically, by using the formula for y_n above, I get:

x(n + 1 – MAX) / (n + 2) + MAX – (n + 1) = x(n – MAX) / (n + 1) + MAX – n

which, when I manipulate it as follows:

=> x(n + 1 – MAX) / (n + 2) – (n + 1) = x(n – MAX) / (n + 1) – n
=> x(n + 1 – MAX) / (n + 2) = x(n – MAX) / (n + 1) – n + (n + 1)
=> x(n + 1 – MAX) / (n + 2) = x(n – MAX) / (n + 1) + 1
=> x(n + 1 – MAX) = x(n – MAX)(n + 2) / (n + 1) + n + 2
=> x(n + 1 – MAX)(n + 1) = x(n – MAX)(n + 2) + (n + 1)(n + 2)
=> (xn + x – MAX(x))(n + 1)
=
(xn – MAX(x))(n + 2) + (n + 1)(n + 2)
=> xn² + xn – MAX(x)(n) + xn + x – MAX(x)
=
xn² - MAX(x)(n) + 2xn – 2MAX(x) + (n + 1)(n + 2)
=> xn – MAX(x)(n) + xn + x – MAX(x)
=
-MAX(x)(n) + 2xn – 2MAX(x) + (n + 1)(n + 2)
=> xn + xn + x – MAX(x) = 2xn – 2MAX(x) + (n + 1)(n + 2)
=> x – MAX(x) = – 2MAX(x) + (n + 1)(n + 2)
=> x + MAX(x) = (n + 1)(n + 2)
=> x(MAX + 1) = (n + 1)(n + 2)
=> x = (n + 1)(n + 2)/(MAX + 1)

Gives me the relatively simple statement that

y_n(x) = y_n+1(x) => x = (n + 1)(n + 2)/(MAX + 1).

Looking back at my notes (which were originally made on the back of an agenda and continued on a hotel notepad), it seems that I decided this would be a good time to go on a snipe hunt and find the distance for which any particular line was on the frontier of the curve. Using a bit more manipulation (which I may set forth in a future post), I found that, as MAX goes to infinity, the length of the longest line segment on the frontier approaches 2, while the length of the shortest line segment approaches 2^.5. This was actually something of a surprise, and it’s a result I’m glad I reached. However, it still didn’t answer how to express the curve in the form of y = f(x). My next thought for how to continue down that path is write an equation which will show which line segment I’m actually on at any point x. My guess is that it should be doable based on the formula for the crossing point of lines n and n+1 given above, though I haven’t actually done it yet. Am I right? Is it actually doable? Will it lead me to the desired y = f(x)? Only time will tell…

As a note, I normally blog about the law surrounding data privacy and information security here. Since I already have a blog devoted to those topics, my posts here will cover anything else that comes to mind.