70 Higher Engineering Mathematics
f (x) = sin x
f(0) = sin 0 = 0
f (x) = cos x
f (0) = cos 0 = 1
f (x) = −sin x
f (0) = −sin 0 = 0
f (x) = −cos x f (0) = −cos 0 = −1
f iv (x) = sin x
f iv (0) = sin 0 = 0
f v (x) = cos x
f v (0) = cos 0 = 1
f vi (x) = −sin x f vi (0) = −sin 0 = 0
f vii (x) = −cos x f vii (0) = −cos 0 = −1
Substituting the above values into Maclaurin’s series of
equation (5) gives:
sin x = 0 + x (1) +
x
2
2!
(0) +
x
3
3!
(−1) +
x
4
4!
(0)
+
x 5
5!
(1) +
x 6
6!
(0) +
x 7
7!
(−1) + · · ·· · ·
i.e. sin x = x −
x 3
3!
+
x 5
5!
−
x 7
7!
+ · · ·
Problem 4. Using Maclaurin’s series, find the
first five terms for the expansion of the function
f (x) = e 3x .
f (x) = e
3x
f (0) = e
0
= 1
f (x) = 3 e 3x
f (0) = 3 e 0 = 3
f (x) = 9 e 3x
f (0) = 9 e 0 = 9
f (x) = 27 e 3x f (0) = 27 e 0 = 27
f iv (x) = 81 e 3x f iv (0) = 81 e 0 = 81
Substituting the above values into Maclaurin’s series of
equation (5) gives:
e
3x
= 1 + x (3) +
x 2
2!
(9) +
x 3
3!
(27)
+
x 4
4!
(81) + · · ·· · ·
e
3x
= 1 + 3x +
9x 2
2!
+
27x 3
3!
+
81x 4
4!
+ · · ·
i.e. e
3x
= 1 + 3x +
9x 2
2
+
9x 3
2
+
27x 4
8
+ · · ·
Problem 5. Determine the power series for tan x
as far as the term in x 3 .
f (x) = tan x
f (0) = tan 0 = 0
f
(x) = sec
2 x
f
(0) = sec
2 0 =
1
cos 2 0
= 1
f
(x) = (2 sec x)(sec x tan x)
= 2 sec
2 x tan x
f
(0) = 2 sec
2 0 tan 0 = 0
f
(x) = (2 sec
2 x)(sec
2 x)
+ (tan x)(4 sec x sec x tan x), by the
product rule,
= 2 sec
4 x + 4 sec
2 x tan
2 x
f
(0) = 2 sec
4 0 + 4 sec
2 0 tan
2 0 = 2
Substituting these values into equation (5) gives:
f (x) = tan x = 0 + (x)(1) +
x 2
2!
(0) +
x 3
3!
(2)
i.e. tan x = x +
1
3
x
3
Problem 6. Expand ln(1 + x) to five terms.
f (x) = ln(1 + x)
f (0) = ln(1 +0) = 0
f (x) =
1
(1 + x)
f (0) =
1
1 + 0
= 1
f
(x) =
−1
(1 + x) 2 f
(0) =
−1
(1 + 0) 2 = −1
f (x) =
2
(1 + x) 3 f (0) =
2
(1 + 0) 3 = 2
f iv (x) =
−6
(1 + x) 4 f iv (0) =
−6
(1 + 0) 4 = −6
f v (x) =
24
(1 + x) 5 f v (0) =
24
(1 + 0) 5 = 24