Physics 263: Problem Set #1

1. (BTM 1.2.2) The key here is to know how to expand 1/(x + Delta x) for small Delta x (since we're taking the limit as it goes to zero). We know that 1/(1+z) = 1 - z + ... for small z, so we have to put our expression in that form. Thus we want to factor out 1/x, which gives us (1/x)*[1/(1 + Delta x/x)], and we can expand the term in []'s with z = Delta x/x.
2. (BTM 1.3.2) The expansion in this problem is about x=0; the comment about expanding about the point 1 is referring to the value of the function (1+x)^p, which is equal to 1 at x=0. The idea is to apply the formula for the Taylor series; make sure you indicate what the ingredients are. For example, f'(0) is p*(1+x)^p-1 evaluated at x=0, which yields p. Don't just write down p.
3. (BTM 1.5.2) Besides identifying the zeros and the limits as x becomes very large (positive or negative), you can determine the behavior near 0 by Taylor-expanding cosh x in the denominator and then bringing it to the numerator using 1/(1+z) = 1 - z + ...
4. (BTM 1.6.2) Most of the answers are in the back of the book, but you can also check your results using MATLAB or Mathematica. Be sure to include at least one intermediate step so the grader can verify that you know what you are doing.
5. (BTM 1.6.5) The sum of probabilities must be equal to 1. So you have one equation for the ratio of the two probabilities and another for the sum. That should be enough to solve for each individually. Here's an example of calculating the average given the probabilities: finding the average number obtained after throwing an ordinary six-sided die many times. If you threw it N times, with N a large number, then you would expect to get N/6 1's, where 1/6 is the probability of a 1, which we'll define as P(1). And so on with the other numbers. So the average number is: <N> = (1/N)[P(1)*N*1 + P(2)*N*2 + P(3)*N*3 + P(4)*N*4 + P(5)*N*5 + P(6)*N*6] and we see that the N's drop out. This leaves the average as the sum of the probabilities for a given result times the value of that result.