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?