Lab 7
The Bisection Method
and
Newton’s Method in Maple
(Classic Version for Windows)
Author: B. A. Forrest ([email protected])
Copyrighted / r1.5
Contents
1
Objectives for this Lab
i
2
Approximate Solutions of Equations
ii
3
Bisection Method
iii
4
Error in the Bisection Method
vi
5
Newton’s Method
vii
6
Lab 7 Exercises
xi
1
Objectives for this Lab
•
Understand the Bisection Method.
•
Use Maple to aid in the calculations for the Bisection Method.
•
Understand Newton’s Method.
•
Use Maple to aid in the calculations for Newton’s Method.
•
Use plots to investigate the Bisection Method and Newton’s Method in Maple.
One of the most common tasks in mathematics is to find the solutions (roots) of equations.
General solutions to linear
and quadratic equations have existed for a long time. However, it has only been since 1545 AD (about 100 years before
the “invention” of calculus) that algebraic solutions to cubic (3rd degree) equations were formalized. It is interesting to
i
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Copyrighted by B.A. Forrest ([email protected])
Newton’s Method in Maple
note that these solutions were discovered by Italian mathematicians through math contests! Although algebraic methods
for finding solutions to quartic (4th degree) equations exist, they turn out to be very complicated expressions. In fact, it is
possible to prove that there is no special formula for finding solutions to quintics (5th degree) or higher degree polynomial
equations. Instead, calculus has given us a number of methods for determining
approximate solutions
to certain equations.
Two of the methods that we will focus on in this lab are the
Bisection Method
and
Newton’s Method
.
In this lab we use Maple to help with the calculations and visualization involved in applying these methods.
2
Approximate Solutions of Equations
Suppose that we wanted to show that the equation
e
x
=

3
x
2
+ 4
has a solution in the interval [0
,
1]. We could apply the
Intermediate Value Theorem
(IVT). To do so we first introduce the
function
F
(
x
) =
e
x
+ 3
x
2

4
obtained by subtracting the function on the righthand side of the equation from the function on the lefthand side. We
then note that a point
c
is a solution of the equation if and only if
F
(
c
) = 0.
However,
F
(
x
) is a
continuous function
(can
you explain why?) and
F
(0) =
e
0
+ 3(0)
2

4 =

3
<
0
while
F
(1) =
e
+ 3

4
>
0
.
Therefore, the IVT shows that there is at least
one point
c
∈
[0
,
1] such that
F
(
c
) = 0. That is
e
c
+ 3
c
2

4 = 0
.
If we wanted a better idea of the location of
c
, we could
bisect
the original interval [0
,
1] at the midpoint
1
2
, and evaluate
F
(
1
2
).
If
F
(
1
2
)
<
0, then since
F
(1)
>
0, we would have
a solution in the new interval [
1
2
,
1].
Otherwise, if
F
(
1
2
)
>
0,
c
would be located in
the interval [0
,
1
2
].
We can then bisect the new interval, test the
midpoint and repeat the procedure.
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 Winter '09
 Prof.Smith
 Calculus, Numerical Analysis, Equations, Intermediate Value Theorem, Continuous function, Bisection Method

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