This set is due Friday, March 8, at the beginning of lecture.
- Solve problem 2 from Chapter 7. In each case, use the bisection method to approximate within a value of for which we have . Recall that in the bisection method, at each stage we have an interval and we know that and (or ). We let be the midpoint of the interval. If , we let and we are done. More likely, either , and we have , our new interval is , and we iterate the process, or , and we have , our new interval is , and we iterate the process.
- Solve problem 3 from Chapter 7. As before, approximate within .
- Find the first few convergents to , and use them to find within . Recall that the convergents of are obtained by the following process:
Define so that: is the largest integer below . Let , and let be the largest integer below . Let , and let be the largest integer below . Let , and let be the largest integer below , etc.
The first convergent to is the number . The second convergent is . The third convergent is . Etc.
The number is sandwiched between the convergents, in the sense that it is larger than the first, smaller than the second, larger than the third, smaller than the fourth, etc.
- Approximate following the following algorithm: Let be an arbitrary number that you choose, presumably not too far from . Given , we define a new approximation by the formula Check with the help a calculator that these numbers approach very quickly. Use this to find the first digits of .
- Extra credit problem: Why does the algorithm of the last problem work?