Homework #4
Solutions for problems from
Section 2.5, Section 2.6
can be found in the solution
to Homework #3.
Section 3.1
1.
( )
Three iterations are required.
We have
( ) ( )
The
first iteration
gives us:
( )
Therefore the new interval is
[ ] [ ]
The
second iteration
gives us:
( )
Therefore the new interval is
[ ] [ ]
The
third iteratio
n gives us:
(
)
Therefore the new interval is
[ ] [
]
2a.
( )
Desired accuracy is
We have
( ) ( )
The
first iteration
gives us:
( )
Therefore the new interval is
[ ] [ ]
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Length of the interval
therefore we need to continue.
The
second iteration
gives us:
(
)
Therefore the new interval is
[ ] [
]
Length of the interval
therefore we need to continue.
The
third iteratio
n gives us:
(
)
Therefore the new interval is
[ ] [
]
Length of the interval
therefore we need to continue.
The
fourth iteratio
n gives us:
(
)
Therefore the new interval is
[ ] [
]
Length of the interval
desired accuracy is achieved. It took
4
iterations to get
accuracy of
. This result coincides with formula
( )
( )
( )
3a, f.
Here is the MATLAB code (2 functions):
First m.file
%Function funvalue returns value of function f at point x
function [ f ] = funvalue(x)
% function for problem 3a
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 Spring '08
 Rottman
 Tail recursion, Control flow, Continuous function, New interval, funvalue

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