Numerical Analysis for Computing
COT 4501
Spring 2005
Midterm II Solutions
April 21, 2009
1.
[35 points] Newton Error Formula:
Let
f
∈
C
2
(
I
) be given, for some interval
I
⊂ R
,
with
f
(
α
) = 0 for some
α
∈
I
. For a given
x
n
∈
I
, define
x
n
+1
=
x
n
−
f
(
x
n
)
f
′
(
x
n
)
.
Then there exists a point
ξ
n
between
α
and
x
n
such that
(
α
−
x
n
+1
) =
−
1
2
(
α
−
x
n
)
2
f
′′
(
ξ
n
)
f
′
(
x
n
)
.
(a)
[5 points]
If
ξ
n
→
α
and
x
n
→
α
as
n
→ ∞
, what does the ratio
R
n
=
α
−
x
n
+1
(
α
−
x
n
)
2
converge to?
The ratio
R
n
=
α
−
x
n
+1
(
α
−
x
n
)
2
converges to lim
n
→∞
R
n
=
−
1
2
lim
n
→∞
f
′′
(
ξ
n
)
f
′
(
x
n
)
=
−
1
2
f
′′
(
α
)
f
′
(
α
)
.
(b)
[5 points]
Evaluate
R
n
(in the limiting case as
n
→ ∞
) for
f
(
x
) = 2
e
−
βx
−
1.
For
f
(
x
) = 2
e
−
βx
−
1,
f
′
(
x
) =
−
2
βe
−
βx
and
f
′′
(
x
) = 2
β
2
e
−
βx
.
Therefore, lim
n
→∞
R
n
=
−
1
2
f
′′
(
α
)
f
′
(
α
)
=
β
2
.
(c)
[10 points]
The Newton Error Formula can be used to approximately “peek ahead” at the
solution
f
(
α
) = 0 when we are getting close to convergence. Imagine that you are executing
Newton’s method and are at time step
n
. You think you are somewhat close to the true answer.
In order to get a better estimate ˆ
α
(
n
) of the true answer
α,
you replace
ξ
n
by
x
n
in the Newton
Error Formula to get an estimate ˆ
α
(
n
). Using this approximation in the above Newton’s Error
Formula, find a solution for ˆ
α
(
n
) in terms of
x
n
+1
, x
n
, f
′′
(
x
n
) and
f
′
(
x
n
).
You are asked to use the Newton Error Formula [(
α
−
x
n
+1
) =
−
1
2
(
α
−
x
n
)
2
f
′′
(
ξ
n
)
f
′
(
x
n
)
] to find an
approximate value ˆ
α
(
n
) for the true root
α
by substituting
x
n
for the unknown
ξ
n
. Therefore
(ˆ
α
(
n
)
−
x
n
+1
) =
−
1
2
(ˆ
α
(
n
)
−
x
n
)
2
f
′′
(
x
n
)
f
′
(
x
n
)
.
1
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Define
S
n
=
−
1
2
f
′′
(
x
n
)
f
′
(
x
n
)
.
Now
(ˆ
α
(
n
)
−
x
n
+1
)
=
S
n
(ˆ
α
(
n
)
−
x
n
)
2
=
S
n
(ˆ
α
(
n
)
2
−
2ˆ
α
(
n
)
x
n
+
x
2
n
)
.
Grouping terms, we get
S
n
ˆ
α
(
n
)
2
−
(2
S
n
x
n
+ 1)ˆ
α
(
n
) + (
S
n
x
2
n
+
x
n
+1
) = 0
.
This is a quadratic equation in ˆ
α
(
n
). Solving for ˆ
α
(
n
), we get
ˆ
α
(
n
)
=
(2
S
n
x
n
+ 1)
±
radicalbig
(2
S
n
x
n
+ 1)
2
−
4
S
n
(
S
n
x
2
n
+
x
n
+1
)
2
S
n
=
(2
S
n
x
n
+ 1)
±
radicalbig
4
S
n
(
x
n
−
x
n
+1
) + 1
2
S
n
.
Defining
T
n
def
=
f
(
x
n
)
f
′
(
x
n
)
, we can rewrite the Newton update equation as
x
n
−
x
n
+1
=
T
n
from which we get that
ˆ
α
(
n
) =
(2
S
n
x
n
+ 1)
±
√
4
S
n
T
n
+ 1
2
S
n
.
It is not clear at this point whether we should take the positive or negative root. It is also not
clear whether 4
S
n
T
n
+ 1
≥
0. If the latter is not true, then the estimate ˆ
α
(
n
) becomes complex
which is unacceptable.
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 Spring '08
 Davis
 Numerical Analysis, Tn, Xn, ǫ − ǫ

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