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?

43.614000-116.202000

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