M629 Lecture 4

"Lecture" for Week 4

Homework procedures

Homework papers should be e-mailed (or, when necessary, faxed) to Valentina Vega (vvega@math.tamu.edu), with a copy to me. Normally I will not look at them until after she has graded them; in fact, I may not see them at all if no controversy arises. Major papers (such as the book report) should be e-mailed directly to me.

Valentina is grading papers on a scale of 100 and posting the grades on Vista as soon as she can. She also attempts to provide some feedback by e-mail; but typing comments is very time-consuming, so they will be sparser than they would be if we could return a paper copy of your work to you. You are always free to ask questions about the grading.

Mathematical background for this week

I have no particular expertise on China and India, but if I were lecturing on this material I would include more background on some of the mathematical topics that arise. Here's some help; feel free to ask for more (and with luck, you may actually get it).

Modular arithmetic (congruences). An expository paper of mine on the Chinese remainder theorem is attached to the assignments page.
Square (and cube) root algorithm. (See message on Vista.) See this addendum.
Newton's interpolation formula. This is mentioned on p. 151. Chapter 25 of the famous "Handbook of Mathematical Functions ..." of the National Bureau of Standards (whose history is recounted in an article in the most recent American Mathematical Monthly, by the way) states two interpolation formulas of Newton, of which the relevant one seems to be the "divided difference" formula.

Parabola and triangle

I promised to say something about Archimedes's parabola area formula. This is from the Fall '03 lectures, mutatis mutandis.

Let A, B, C be three points on a parabola, as on p. 9 of Allen, "Archimedes", such that the tangent line at B is parallel to the secant between A and C. This is equivalent to Katz's condition (pp. 70-71) that B is "that point of the curve whose perpendicular distance to AC is the greatest."

With apologies to Archimedes, we shall take advantage of what we've learned from Descartes and Leibniz. By displacing and scaling the coordinates (x' = ex + f, y' = gy + h) we can assume that the parabola is in the standard form, y = x². (The areas will change by the determinant of the scaling transformation, but only the ratios of areas matter in the problem.) Let A = (a,a²), etc. The slope of the secant is

(c² - a²)/(c - a) = a + c,

and the slope of the tangent is 2b, so b = (a + c)/2 -- i.e., B is midway between A and C in abscissa! (This step may not be strictly necessary, but the new condition is certainly easier for my post-Cartesian mind to embrace.)

We can now reformulate the problem as integral calculus. Let y = f(x) be the equation of the secant line, y = g(x) be the equation of the broken line forming the other two sides of the triangle, and y = h(x) be the equation of the parabola. Let I be the operation of definite integration between a and c. Then the area of the triangle is If - Ig, and the area of the parabolic segment is If - Ih. What makes this problem so interesting to me is that it now reduces to a famous fact of numerical analysis:

If the graph of a function k passes through A, B, and C (equally spaced in abscissa), then If is the crudest approximation to Ik by the trapezoidal rule, Ig is the next crudest trapezoidal approximation, and Ih is the crudest approximation to Ik by Simpson's rule. (In particular, Simpson's rule is exact if k is quadratic -- i.e., k = h.) Romberg's formula states

Ih = (4/3)Ig - (1/3)If.

Thus you can evaluate Simpson's rule by doing the trapezoidal rule with more and more points until the numbers look good, then applying Romberg's formula to get an answer much better than either of your last two trapezoidal results! Intuitively, the formula says that the coarser trapezoid result is in error in the same direction as the finer one, but even more so; so you improve your estimates by combining them in a weighted average where the coarser one is penalized by a negative coefficient. (If you have Stewart's Calculus (the textbook we use at TAMU), you will find a similar discussion in the section on numerical integration, but using the midpoint approximation instead of the finer trapezoidal one.) Romberg's formula can be proved by simple algebra if you evaluate the three integrals, or it can be derived more abstractly by seeking a linear combination of the two trapezoidal approximations that makes the error term of order (step size)² cancel out. In any case, trivial algebra shows that Romberg's formula is equivalent to Archimedes's.