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Sunday, September 17, 2006

Moral: When you write an exam, check your problems.

Mathematics: Differential Equations

I almost had some egg on my face. In differential equations class, we had just covered the basic existance and uniqueness theorems for linear first order differential equations. That is, if one is given y' = f(t, y) where f is continuous at (t0, y0) and continuous on some open rectangle containing (to, yo) then the differential equation y'=f(t,y), y(t0) = y0 has at least one solution. Furthermore, if the partial derivative of f(t,y) with respect to y is similarly continuous at (t0, y0), then the solution is unique.

Example: y' = t*exp(y), y(t0) = y0 has a unique solution for any initial condition,
y' = t*y^(2/3) has a soluton of all initial conditions, but fails to have a unique solution for any initial condition y(t0) = 0, as the partial derivative with respect to y is t*(2/3)*y^(-1/3) which fails to be continuous when y = 0.

So, on the test I gave y' = y^(2/3)*ln(t)*tan(y) and asked if there was a guarantee for a unique solution at y(2) = 0.

Answer: not that obvious! taking the partial derivative with respect to y we get:

ln(t) (y^(2/3) * sec^2(y) + (2/3)^y(-1/3)*tan(y)). Pluggin in t = 2 causes no difficulty, but what about y = 0? The first term gives 0 (no problems) but what about the second?

Actually, if you go through the proof (of the uniqueness theorem), it is the fact that we have some boundedness condition for the partial with respect to y (or a Lipshitz condition, at least).

Note what happens if we apply L'Hopital's rule to the second term:
lim as y goes to zero of (2/3)*(tan(y)/y^(1/3)) = (2/3)(sec^2(y)/ (1/3)y^(-2/3)) = (2/3)(3)sec^2(y)y^(2/3) = 2*1*0 = 0 !!!

In other words, technically speaking, the partical derivative may not have been continuous, but the singularity is removable and hence the uniqueness condition holds.

I didn't anticipate this at all and was lucky that I caught this prior to passing out the exams. Moral: test your problems ahead of time.

Friday, September 01, 2006

New Calculators: Brain still required I teach lots of calculus as I am a small-time college mathematics professor.

By "small time" I mean that my research never put me in contention for an endowed professorship anywhere!

One of the biggest changes we've had to calculus teaching over the past 15 years or so is the introduction of calculators that do symbolic calculus; for exampe, if you wanted to find:
integral(e^2*x dx) you could just enter it and obtain (1/2)e^(2x). You'd have to add the "c", of course.

So, after a recent class, one student brought me the following:

integral((e^x )*sin(2*x) dx) and obtained:
(1/pi^2 + 8100)*e^x*(-90*pi*cos(2*x) + 8100*sin(2*x)) + C

What did the student do wrong?