This document provides an overview of Maclaurin series and examples of their use. It begins by stating the competencies around Maclaurin series, including deriving and stating the series for various functions. Two examples are given of expanding exponential and logarithmic functions as Maclaurin series. The document concludes by listing standard Maclaurin series expansions for common functions and providing trial questions for practice.

1. Maclaurin Series
for DES II 2019/2020
By
MUJUNGU HERBERT
National Teachers’ College
Kabale
MUJUNGU HERBERT (National Teachers College Kabale) May 15, 2020 1 / 8

2. Maclaurin Series
Competences
1 State and derive Maclaurin series
2 Obtain Maclaurin’s series for various functions
3 Derive and state the exponential and logarithmic functions.
MUJUNGU HERBERT (National Teachers College Kabale) May 15, 2020 2 / 8

3. Maclaurin’s series
Question By comparing the Taylor’s series given by;
f(x) =
f(a)
0!
+
f (a)
1!
(x − a) +
f (a)
2!
(x − a)2
+
f (a)
3!
(x − a)3
+
f(4)
4!
(x − a)4
+ ...
and Maclaurin’s series given by;
f(x) =
f(0)
0!
+
f (0)
1!
x +
f (0)
2!
x2
+
f (a)
3!
x3
+
f(4)
4!
x4
+ ... (1)
State the conditions applied in the Taylor’s theorem to obtain the
Maclaurin’s theorem.
MUJUNGU HERBERT (National Teachers College Kabale) May 15, 2020 3 / 8

4. Example 1
Use Maclaurin’s theorem to expand ex in ascending powers of x as far as
the x5 term.
MUJUNGU HERBERT (National Teachers College Kabale) May 15, 2020 4 / 8

5. Example 1
Use Maclaurin’s theorem to expand ex in ascending powers of x as far as
the x5 term.
Solution
MUJUNGU HERBERT (National Teachers College Kabale) May 15, 2020 4 / 8

6. Example 1
Use Maclaurin’s theorem to expand ex in ascending powers of x as far as
the x5 term.
Solution
Let f(x) = ex
⇒ f(0) = e0 = 1
f (x) = ex
⇒ f (0) = 1
f (x) = ex
⇒ f (0) = 1
f (x) = ex
⇒ f (0) == 1
f (x) = ex
⇒ f (0) == 1
f (x) = ex
⇒ f (0) == 1
ex
=
1
0!
+
1
1!
x +
1
2!
x2
+
1
3!
h3
+
1
4!
x4
+
1
5!
x5
+ ...
MUJUNGU HERBERT (National Teachers College Kabale) May 15, 2020 4 / 8

7. Example 2 Use Maclaurin’s theorem to expand ln(1 + x) in ascending
powers of x as far as x5.
MUJUNGU HERBERT (National Teachers College Kabale) May 15, 2020 5 / 8

8. Example 2 Use Maclaurin’s theorem to expand ln(1 + x) in ascending
powers of x as far as x5. Solution
MUJUNGU HERBERT (National Teachers College Kabale) May 15, 2020 5 / 8

9. Example 2 Use Maclaurin’s theorem to expand ln(1 + x) in ascending
powers of x as far as x5. Solution
Let f(x) = ln(1 + x) ⇒ f(0) = ln(1 + x) = 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 (x) = −
6
(1 + x)4
⇒ f (0) = − 6
(1+0)4 = −6
f (x) =
24
(1 + x)5
⇒ f (0) = 24
(1+0)5 = 24
ln(1 + x) =
0
0!
+
1
1!
x +
−1
2!
x2
+
2
3!
x3
+
−6
4!
x4
+
24
5!
x5
+ ...
MUJUNGU HERBERT (National Teachers College Kabale) May 15, 2020 5 / 8

10. Example 3
Express 17o11 in radians correcting to one significant figure. Use the
approximation; sin x ≈ x − x3
3! + x5
5! , to express 17o11 to four significant
figures.
MUJUNGU HERBERT (National Teachers College Kabale) May 15, 2020 6 / 8

11. Example 3
Express 17o11 in radians correcting to one significant figure. Use the
approximation; sin x ≈ x − x3
3! + x5
5! , to express 17o11 to four significant
figures.
Solution
MUJUNGU HERBERT (National Teachers College Kabale) May 15, 2020 6 / 8

12. Example 3
Express 17o11 in radians correcting to one significant figure. Use the
approximation; sin x ≈ x − x3
3! + x5
5! , to express 17o11 to four significant
figures.
Solution
Recall that ;
1 = 1 arcmin =
1
60
degrees
and 1 degree =
π
180
radians
17o
11 = 17o
+ 11 ∗
1
60
o
≈
17.1833o
17.1833o
≈ 0.3 radians
sin 0.3 ≈ 0.3 − 0.33
3!
+ 0.35
5!
sin 0.3 ≈ 0.2955
By comparison, from the calculator,
I have 0.2954
MUJUNGU HERBERT (National Teachers College Kabale) May 15, 2020 6 / 8

13. Trial Questions
1 Given that the functions f (x) and co + c1x + c2x2 + c3x3 + c4x4 + ...
have the same value when x = 0, and equal successive derivaties
when x = 0, deduce the first four terms of the Maclaurin expansion of
f (x) in ascending powers of x.
2 By using Maclaurin’s expansions for ex , ln(1 + x), cos x and sin x,
write down the first four terms of the expnasions of the following in
ascending powers of x:
(a). e2x , (b). ln(1 − x), (c). cos x2, (d) sin x
2
3 By substracting the expansion of ln(1 − x) from that of ln(1 + x)
deduce that ln 1+x
1−x = x + x3
3 + x5
5 + x7
7 + ...
4 Find approximations of the following;
(a). e0.4, (b). ln(1.2), (c). cos 0.3, (d) sin 0.2
5 Use Maclaurin’s theorem to show that if, x5 and higher powers of x
are neglected. ln x + (1 + x2) = x − 1
6x3
MUJUNGU HERBERT (National Teachers College Kabale) May 15, 2020 7 / 8

14. Standard Expanssion of
Maclaurin’s series
• 𝑒 𝑥 = 1 + 𝑥 +
𝑥2
2!
+
𝑥3
3!
+ ⋯
• 𝑒−𝑥
= 1 − 𝑥 +
𝑥2
2
−
𝑥3
3
+ ⋯
• ln 1 + 𝑥 = 𝑥 −
𝑥2
2
+
𝑥3
3
−
𝑥4
4
+ ⋯
• ln 1 − 𝑥 = −𝑥 −
𝑥2
2
−
𝑥3
3
−
𝑥4
4
− ⋯
• 𝑙𝑛
(1+𝑥)
(1−𝑥)
= 2𝑥 +
2
3
𝑥3
+
2
5
𝑥5
+ ⋯
• 𝑠𝑖𝑛𝑥 = 𝑥 −
𝑥3
3!
+
𝑥5
5!
−
𝑥7
7!
+ ⋯
• 𝑐𝑜𝑠𝑥 = 1 −
𝑥2
2!
+
𝑥4
4!
−
𝑥6
6!
+ ⋯
• 𝑡𝑎𝑛𝑥 = 𝑥 +
𝑥3
3
+
2𝑥5
15
+
17𝑥7
315
+
62𝑥9
2835
+ ⋯
! !

15. • 𝑠𝑒𝑐𝑥 = 1 +
𝑥2
2
+
5𝑥4
24
+
61𝑥6
720
+ ⋯
• sin−1
𝑥 = 𝑥 +
𝑥3
6
+
3𝑥5
40
+
5𝑥7
112
+ ⋯
• cos−1
𝑥 =
𝜋
2
+ 𝑥 +
𝑥3
6
+
3𝑥5
40
+
5𝑥7
112
+ ⋯
• tan−1 𝑥 = 𝑥 −
𝑥3
3
+
𝑥5
5
−
𝑥7
7
+ ⋯
• 𝑠𝑖𝑛ℎ𝑥 = 𝑥 +
𝑥3
3!
+
𝑥5
5!
+
𝑥7
7!
+ ⋯
• 𝑐𝑜𝑠ℎ𝑥 = 1 +
𝑥2
2!
+
𝑥4
4!
+
𝑥6
6!
+ ⋯

16. with God, Learning
Continues
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MUJUNGU HERBERT (National Teachers College Kabale) May 15, 2020 8 / 8