UCB MATH 128A2, SUMMER 2009: FINAL EXAM SOLUTIONS
JUSTIN BLANCHARD
1.
(a) Prove that the bisection method on [1
,
2] will converge to a root of
x
3

x

1.
(b) Find a bound on the absolute error in approximating this root after
n
iterations.
(c) How many iterations are necessary to ensure the relative error is at most 2

5
?
(Your answer does not have to be exact.)
Solution:
(a) This function is continuous. Its values at the endpoints are

1
<
0 and 5
>
0. Therefore, the
interval brackets a root and the bisection method will converge.
(b) On the
n
th iteration, the root is known to lie in an interval of width
2

1
2
n

1
. Its distance from this
interval’s midpoint can be at most
1
/
2
n
. (The answer 1
/
2
n
+1
is also acceptable, depending
on how the phrase “after
n
iterations” is interpreted.)
(c) Since the root lies in [1
,
2], the relative error is at most equal to the absolute error. By (b), we
must use
n
= 5
iterations.
2.
(a) Find an algebraic expression for the unique root (in
R
) of
f
(
x
) =
x
2

2
/x
.
(b) Newton’s method searches for this root using the iteration
p
n
+1
=
g
(
p
n
).
Find
g
(
x
). Simplify your answer.
(c) Is the convergence linear? Is it at least quadratic?
(d) Find the exact order of convergence.
Solution:
(a)
x
2

2
/x
= 0
⇐⇒
x
3

2 = 0
⇐⇒
x
= 2
1
/
3
.
(b)
g
(
x
) =
x

f
(
x
)
f
0
(
x
)
=
x

x
2

2
/x
2
x
+2
/x
2
=
x

x
x
3

2
2
x
3
+2
=
x
(
x
3
+ 4)
2(
x
3
+ 1)
(c) The convergence is at least quadratic, for either of the following reasons: 2
1
/
3
is a simple root
of
f
(
x
), as
f
0
(2
1
/
3
) = 2
·
2
1
/
3
+ 2
/
2
2
/
3
6
= 0.
Alternatively,
g
0
(2
1
/
3
) = 0 after the calculation
g
0
(
x
) =
(4
x
3
+4)2(
x
3
+1)

x
(
x
3
+4)(6
x
2
)
2
2
(
x
3
+1)
2
=
8(
x
3
+1)
2

6
x
3
(
x
3
+4)
2
2
(
x
3
+1)
2
.
(d) (This was intended to be a harder problem.)
By definition, the order of convergence is
α
iff lim
n
→∞
p
n
+1

2
1
/
3
p
n

2
1
/
3
=
λ
6
= 0.
If
p
n
=
x
, this
fraction’s numerator is
g
(
x
)

2
1
/
3
=
x
4

2
4
/
3
x
3
+4
x

2
2(
x
3
+1)
.
Factor out (
x

2
1
/
3
) as many times
as possible:
(
x

2
1
/
3
)
3
(
x
+2
1
/
3
)
2(
x
3
+1)
. (Alternatively, Taylorexpand
x
4

2
4
/
3
x
3
+ 4
x

2 with center
x
= 2
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 Spring '10
 tao/analysis
 Math, Numerical Analysis, Method, local truncation error, UCB MATH 128A2

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