Math 558 - Q&A

2/9/99 I was thinking of the exercise you talked about in class the other day, which deals with the equivalent of xn where: x(n+1) = xn*(1-xn). You said that we should not assume that xn is asymptotic to a/n^b, and we should prove the result without using the hint.

What I meant to say was that you should first follow the hint, i.e. assume that xn ~ a/n^b and determine the coefficients a and b. You will obtain xn ~ 1/n. This reasoning is heuristic because it assumes that xn has the desired form. If you do this, you will get partial credit. To get full credit, you have to go further and show that xn ~ 1/n without making an assumption as before. This part is more difficult. A hint I mentioned in class was to relate xn to yn, where yn given by the map y(n+1) = yn / ( 1 + yn).

But it seems to me that, since the definition of asymptotic equivalence is: lim eq*xn =1 as n tends to infinity (eq represents the equivalent expression),

(no, xn ~ eq means that lim xn/eq = 1, not lim eq*xn =1)

showing that xn is between two sequences equivalent to 1/n or showing directly that the definition is satisfied, is the same. Am I right?

Yes, that is correct. To show that xn ~ 1/n, you could show lim n * xn = 1, or find two sequences yn, zn, for which you know that yn, zn ~ 1/n, and for which you can show zn < xn < yn. The yn I mentioned above will work, but the zn is a bit trickier.

1/25/99 In the case of second order equations d/dt(d/dt(x(t))) = f(x) in which f does not depend on t, why is it that the equilibrium points X are those where f(X) = 0?

An equilibrium point is a solution in which x(t) is constant. In both types of equations, x'=f(x) and x''=f(x), the condition that determines the equilbrium points is f(X)=0. For the 2nd order equation x''=f(x), you need to supply initial data x(0) and x'(0). Now suppose that x(0)=X. If x'(0)=0, then x(t)=X for all t. However, you might have x'(0) ne 0, in which case x(t) moves away from X.

1/25/99 In the case of problem 1.8, the fact that the plane in which the ring lies is spinning means that the particle never has 0 kinetic energy in three dimensions, even at its equilibria positions.

The particle may be at rest at a point on the ring. In that sense, the particle may have zero kinetic energy with respect to its motion on the ring.

Can I just find the zeroes of d/dt(d(theta)/dt) and assume those are the equilibria?

Yes.

1/23/99 In general, since much of the answer is given in the back of the book for each problem, are we always supposed to derive the given answer?

Yes, if the answer is given in the back of the book, you should derive it, or justify it.

1/23/99 In 1.6, I can't figure out how to get the equilibrium points X as a function of a. I found a horrible-looking general formula for the roots of a cubic polynomial in x (in this case, F(a,x) = x^3 + a^3 - 3ax). The book says "F(a,x) = 0 has asymptote x + a = -1" -- but the equation for the roots of the cubic is so bad that I don't see how I would ever figure out X=f(a), and that leaves me unable to figure out where the bifurcations occur. Do you have any advice?

There are various ways to approach this. Using the formula for the roots of a cubic is a natural idea, but it doesn't help much in practice as you found. In this problem, you don't need to find X=f(a) explicitly, just the shape so that you can draw the bifurcation diagram. You know that a cubic function g(x)=F(a,x) may have 0, 1, or 2 critical points depending on the parameter a. My hint is to find these critical points and plot the graph of g(x) in each of the three generic cases.

Another possibility is to use the Matlab "contour" command to plot F(a,x)=0. You would still have to justify the answer on the homework, but at least you'd know what it looks like.