The Unapologetic Mathematician

Mathematics for the interested outsider

The days it’s worth it

One of the things I like to do in class occasionally is a big, ornate example that pulls in a lot of stuff. It’s much more complicated than anything I’d ask the kids to do, but they can be fun. Today, since a lot of students would be preparing for the big weekend I did one such. I’ll be doing it here sooner or later, so I won’t bother with rigor or definitions, but those who have seen this sort of thing should be able to see how it goes.

Recently we covered curvature of space curves, and I went into torsion (which Stewart leaves to the exercises) and the Frenet-Serret formulæ (ditto). On Wednesday I cranked out the curvature and torsion of a helix, and showed they’re both constant. Then I went back and showed how the Frenet-Serret formulæ are secretly a separable differential equation for the Frenet frame, just like the separable equations we did in the previous calculus courses, but now done in terms of matrices!

So I gave a quick overview of matrix algebra (which, though not essential, will come in handy later), and a review of power series so I could define the exponential of a matrix as a power series. Today I started with the constant curvature matrix, integrated (multiplied by arc-length) and worked out the exponential. A judicious choice of initial frame to fix my constants of integration and I have a solution to the Frenet-Serret formulæ, from which I get a unit tangent vector. Another judicious choice of initial point to fix more constants of integration and I have the formula for a helix parametrized by arc-length, just like I’d started with on Wednesday!

Well in my first section this went over like a lead balloon. But the second.. questions galore. Checking steps of the computation, going back to remember the big picture, recognizing the duality between the curvature/torsion parameters and the radius/spacing parameters, hazarding a guess at why DNA is a helix (hint: think about how to program a machine to build a space curve with the simplest instructions possible)… and they had such fun with it. And the days like this are what keeps me going.

And there are things burning up the blathosphere which I’d love to comment on, but I’d prefer not to while applications are open. I’ll let Blake handle it, with no comment on whether I agree or disagree with him, and let you make up your own minds.

February 1, 2008 Posted by John Armstrong | Uncategorized | | 3 Comments

Some integrable functions

Okay, we know what it means for a function to be integrable (in either of the equivalent Riemann or Darboux senses), but we don’t yet know any functions to actually be integrable. I won’t give the whole story now, but a just a large enough part to work with for the moment.

The major theorem here is that a continuous function f on a closed interval \left[a,b\right] is integrable. Notice from the Heine-Cantor theorem that f is automatically uniformly continuous. That is, for any \epsilon there is some \delta so that for all x and y in \left[a,b\right] with |y-x|<\delta we have |f(y)-(x)|>\epsilon. Again, the important thing here is that we can choose our \delta independently of the point x, while continuity says \delta might depend on x.

So now we need to take our function and show that the upper and lower Darboux sums converge to the same value. Equivalently, we can show that their difference converges to zero. So given an \epsilon we want to show that there is some partition x so that the difference

\displaystyle U_x(f)-L_x(f)=\sum\limits_{i=1}^n(M_i-m_i)(x_i-x_{i-1})

is less than \epsilon, and the same is true of any refinement of this partition.

We’ll choose our partition with every slice having constant width \delta, so there are \frac{b-a}{\delta} of them. By the uniform continuity of f we can find a \delta so that for any points x and y with |y-x|<\delta we’ll have |f(y)-f(x)|<\frac{\epsilon}{b-a}. Then in particular the difference M_i-m_i will be less than \frac{\epsilon}{b-a}, while x_i-x_{i-1}=\delta and n=\frac{b-a}{\delta}. Thus the difference in the Darboux sums will be less than \epsilon, as we wanted.

What about refinements? Well, any refinement of a partition can only lower the upper Darboux sum and raise the lower one. This is because adding a point to a partition can’t raise the maximum in either of the new subintervals or lower the minimum, and in fact adding a point will usually lower the maximum and raise the minimum. So our partition has a small enough difference in the Darboux sums, and any refinement will make the difference even smaller, and thus we have the convergence we need.

Now, can we do better than continuous functions? Well, we can relax continuity at the endpoints a bit. If the function jumps to a different value at a or b than the limit seems to indicate, we can still get uniform continuity everywhere but that one point, and we’re still good. We still have problems with asymptotes where the function shoots off to infinity, like f(x)=\frac{1}{x} does at the left endpoint of \left[0,1\right].

What else? Well, we can allow a finite number of discontinuities, as long as none of them are asymptotes. If a discontinuity happens at c\in\left[a,b\right], we can choose c to be a partition point, and so on. Then a partition with these selected points is just the same as a partition on each of the continuous sections of the function in between the discontinuities, and we know that they’re all good.

Incidentally, we can use this same method of picking some of the partition points ahead of time to show another nice property of the integrals: we can break the integral in the middle of the interval and evaluate the two pieces separately, then add them together. That is, if c\in\left[a,b\right] then we have the equation

\displaystyle\int_a^bf(x)dx=\int_a^cf(x)dx+\int_c^bf(x)dx

as long as both sides are integrable.

February 1, 2008 Posted by John Armstrong | Analysis, Calculus | | No Comments