A key feature of a power series is the radius of convergence. If you have a series a₀ + a₁x + a₂x² + ... , then there is a constant R such that the series converges in the region |x| < R and diverges in the region |x| > R. That is, the region of convergence can't extend more in one direction than the other. While the function may be defined outside the radius of convergence, the series diverges there and the Taylor polynomials do not provide a reasonable approximation outside the radius of convergence. For example, the function (1 + x²)⁻¹ has Taylor series 1 − x² + x⁴ − x⁶ + ..., which has radius of convergence 1. While the function is defined for all real x, the Taylor series diverges and the Taylor polynomials do not provide an accurate approximation for |x| ≥ 1. Obviously it is important to be aware where the Taylor polynomial approximation breaks down if you are going to use it to approximate a solution to a differential equation.

In this lab, we will look at Taylor polynomials that approximate the solution of a differential equation. In this case, we won't know the exact solution, so we won't be able to compare our approximations to the actual function value in most cases. However, it usually isn't hard to see where things seem to work, and this will build some intuition about where series converge that we can justify with theorems in next Monday's lecture. We will also look at a couple of situations that illustrate dangers in using Taylor polynomial approximations that you should be aware of.

1. The first problem we will consider is (x2+1)y" + xy' + 2y = 0, y(0)=0, y'(0)=1. Launch the lab (see the link at the top of the page) and change the coefficients to match this equation. Now look at the 5, 10, 25, 50, 500 and 5000 degree Taylor polynomials. What do you think is the radius of convergence of the series solution for this problem?

2. Now change the coefficients of y and y', along with the initial values. How does changing the lower order coefficients seem to affect the radius of convergence of the power series? (You may find it easiest to check the radius of convergence if you leave the degree set at 5000).

3. Next consider the problem (.5x2+x+1)y" + (x + 1)y' + 3y = 0, y(0)=1, y'(0)=0. Where do you think the series solution for this problem converges? How does changing the lower order terms affect the radius of convergence?.

4. Based on your work so far, make a conjecture about the relationship between radius of convergence and the coefficients of the differential equation. Test your conjecture with at least two additional initial value problems.

The next two problems point out places where looking at Taylor approximations may be misleading.
5. A series solution will be valid up to the radius of convergence. The approximation provided by a finite Taylor polynomial (as opposed to the infinite Taylor series) won't work over quite as large a region. In fact, sometimes it can be difficult to impossible to discover features of the solution near the radius of convergence from examining Taylor polynomials. Consider the equation (x2+2x+1)y" + (x+1)y' + y = 0, y(0)=1, y'(0)=0. The exact solution to this initial value problem is y = cos(log(x+1)) (we will see how to find this solution next week). The radius of convergence of this equation 1, so the Taylor series represents the function in the region -1 < x < 1. Look at the behaviour of the Taylor polynomials near x = -1. Sketch how you would guess the solution to the initial value problem behaves near x = -1 based on the behaviour of the Taylor polynomials up to degree 5000. Show that the solution actually oscillates between -1 and 1 infinitely many times near x = -1. Explain why no polynomial can represent infinitely many oscillations (and in this case, it doesn't even come close).

6. Other problems may arise with the computation of the Taylor polynomials even away from the radius of convergence. Consider the equation y" + 400y = 0, y(0)=1, y'(0)=0. The true solution to this problem is y = cos(20x) of course, and the radius of convergence is infinity. Sketch the behaviour of the Taylor polynomial approximations. The difficulty you observe is caused by a build up of round-off error in the computation of the Taylor polynomial. Note that the applet uses double-precision arithmetic and is coded to minimize the number of multiplications (and hence the amount of accumulated round-off error). Sometimes the computation just doesn't want to behave (and then you try to come up with some means other than blindly computing the polynomial to find the answer). By the way, the problem here is not that 400 is too large. Anytime you try to compute a cosine using Taylor polynomials in double-precision, you will run into roundoff errors a little after 6 periods. The same behavior happens with y" + y = 0, y(0)=1, y'(0)=0, whose solution is y = cos(x), a little past x=37.5. The choice of cos(20x) was made to get the bad behavior in the graphing window.