how can i prove this?
Answers
Answer:
he approximation is given as
e
x
=
1
+
x
+
x
2
2
.
The function is given as:
f
(
x
)
=
e
x
It can be expanded as:
f
(
x
)
=
f
(
a
)
+
f
′
(
a
)
x
+
f
′′
(
a
)
2
!
x
2
+
.
.
.
f
n
(
a
)
n
!
x
n
Calculating for
a
=
0
:
f
(
x
)
=
f
(
0
)
+
f
′
(
0
)
x
+
f
′′
(
0
)
2
!
x
2
+
.
.
.
f
n
(
0
)
n
!
x
n
f
(
x
)
=
e
0
+
e
0
⋅
x
+
1
2
e
0
⋅
x
2
+
R
2
(
x
)
f
(
x
)
=
1
+
x
+
x
2
2
+
R
2
(
x
)
Here
R
2
(
x
)
is the remainder.
Recall the Taylor Inequality:
|
R
n
(
x
)
|
≤
M
|
x
−
a
|
n
+
1
(
n
+
1
)
!
Here
M
is the maximum value at
x
=
1
4
,
n
is the order.
Substitute 2 for
n
, 0 for
a
as series is centered at 0 and
e
0.1
for
M
in the above inequality:
|
R
3
(
x
)
|
≤
e
1
/
4
|
0.1
−
0
|
2
+
1
(
2
+
1
)
!
≤
1.284
×
0.1
3
3
!
≤
2.14
×
10
−
4
Thus, the maximum error is
2.14
×
10
−
4
when
x
<
1
4
.
Explanation: