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Math 3110 Homework 3 Solutions
Exercise 4.1.1
(i)
Claim.
If
a
n
→
L
and
b
n
→
M
, then
a
n
+
b
n
→
L
+
M
.
Proof.
Let
± >
0 and let
d
n
=
a
n

L
and
e
n
=
b
n

M
denote the error terms for the given
sequences.
Since
a
n
→
L
and
b
n
→
M
, then by the errorform principle,
d
n
→
0 and
e
n
→
0. So there
exist
p
and
q
such that for all
n
≥
p
,

d
n

<
±
2
, and for all
n
≥
q
,

e
n

<
±
2
. Let
N
= max
{
p, q
}
.
Then for all
n
≥
N
,

d
n
+
e
n
 ≤ 
d
n

+

e
n

<
±
2
+
±
2
=
±
Hence,
d
n
+
e
n
→
0. But
d
n
+
e
n
=
a
n
+
b
n

(
L
+
M
), so by the errorform principle, we
have
a
n
+
b
n
→
L
+
M
.
(ii)
Claim.
If
a
n
→
L
, then
ca
n
→
cL
.
Proof.
Let
± >
0 and denote the error term of the given sequence by
e
n
=
a
n

L
. Notice
that since

c

+1
>
0, then
±

c

+1
>
0. Since
a
n
→
L
, then by the errorform principle,
e
n
→
0.
So there is some
P
such that for all
n
≥
P
,

e
n

<
±

c

+1
. Then for all
n
≥
P
,

ce
n

=

c

e
n

<

c

±

c

+ 1
< ±.
Hence,
ce
n
→
0. But
ce
n
=
ca
n

cL
, so by the errorform principle, we have
ca
n
→
cL
.
1
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View Full DocumentExercise 4.3.3
Using Newton’s method to ﬁnd
M
, the unique positive zero of
f
(
x
) =
x
2
+
x

1:
(i)
We give the recursion formula for
a
n
+1
in terms of
a
n
.
Since
a
n
+1
=
a
n

f
(
a
n
)
f
0
(
a
n
)
, then
a
n
+1
=
a
n

a
n
2
+
a
n

1
2
a
n
+ 1
=
2
a
n
2
+
a
n

a
n
2

a
n
+ 1
2
a
n
+ 1
=
a
n
2
+ 1
2
a
n
+ 1
.
(ii)
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 '08
 RAMAKRISHNA
 Math

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