A details explanation about Taylor's and Maclaurin's series with variety of examples are included in this slide. The aim is to give the viewer the basic knowledge about the topic.

1. C.K.PITHAWALA COLLEGE OF
ENGINEERING & TECHNOLOGY, SURAT
Branch:- computer 1st Year (Div. D)
ALA Subject:- Calculus
ALA Topic Name:- Power series, Taylor’s & Maclaurin’s series
Group No:- D9
Student Roll No Enrolment No Name
403 160090107051 Sharma Shubham
421 160090107028 Naik Rohan
455 160090107027 Modi Yash
456 160090107054 Solanki Divyesh
Submitted To
Gautam Hathiwala

2. Power Series
Taylor’s and Maclaurin’s Series

3. Introduction to Taylor’s series & Maclaurin’s series
› A Taylor series is a representation of a function as an infinite sum of
terms that are calculated from the values of the function’s derivatives at
a single point.
› The concept of Taylor series was discovered by the Scottish
mathematician James Gregory and formally introduced by the English
mathematician Brook Taylor in 1715.
› A Maclaurin series is a Taylor series expansion of a function about zero.
› It is named after Scottish mathematician Colin Maclaurin, who made
extensive use of this special case of Taylor series.

4. Statement of Taylor’s series
If 𝑓 𝑥 + ℎ is a given function of h which can be expanded into a
convergent series of positive ascending integral power of h then,
𝑓 𝑥 + ℎ = 𝑓(𝑥) +ℎ𝑓′
𝑥 + ℎ2
𝑓′′
𝑥
1
2!
+
ℎ3 𝑓′′′ 𝑥
1
3!
+. . . . . . . . .
ℎ 𝑛
𝑛!
𝑓 𝑛 𝑥 +. . . . . . .

5. Proof of Taylor’s series
› Let 𝑓(𝑥 + ℎ) be a function of h which can be expanded into a convergent series of
positive ascending integral powers of h then
𝑓 𝑥 + ℎ = 𝑎 𝑜 + 𝑎1ℎ + 𝑎2ℎ2
+ 𝑎3ℎ3
+. . . . . . . . . .
Differentiating w.r.t. h successively,
(1)
𝑓′
𝑥 + ℎ = 𝑎1 + 𝑎2. 2ℎ + 𝑎3. 3ℎ2
+. . . . . . . . . .
𝑓′′ 𝑥 + ℎ = 𝑎2. 2 + 𝑎3. 6ℎ+. . . . . . . . . .
and so on.
(2)
(3)

6. Putting h=0 in Eq. (1) (2) & (3),
𝑎0 = 𝑓 𝑥
𝑎1 = 𝑓′ 𝑥
𝑎2 = 𝑓′′ 𝑥 and so on
Substituting 𝑎0, 𝑎1, 𝑎2 in Eq.(1) we get,
𝑓 𝑥 + ℎ = 𝑓(𝑥) +ℎ𝑓′
𝑥 + ℎ2
𝑓′′
𝑥
1
2!
+
ℎ3 𝑓′′′ 𝑥
1
3!
+. . . . . . . . .
ℎ 𝑛
𝑛!
𝑓 𝑛 𝑥 +. . . . . . .
This is known as Taylor’s series.

7. Putting 𝑥 = 𝑎 and ℎ = 𝑥 − 𝑎 in the series, we get Taylor’s series in the powers
of 𝑥 − 𝑎 as,
𝑓 𝑥 = 𝑓(𝑎) +(𝑥 − 𝑎)𝑓′
𝑎 + (𝑥 − 𝑎)2
𝑓′′
𝑎
1
2!
+
(𝑥 − 𝑎)3 𝑓′′′ 𝑎
1
3!
+. . . . . . . . .
(𝑥−𝑎) 𝑛
𝑛!
𝑓 𝑛 𝑎 +. . . . . . .
NOTE : To express a function in ascending power of 𝑥, express h in terms of 𝑥.

8. Statement of Maclaurin’s series
If 𝑓 𝑥 is a given function of 𝑥 which can be expanded into a convergent
series of positive ascending integral power of 𝑥 then,
𝑓 𝑥 = 𝑓(𝑥) +ℎ𝑓′
0 + ℎ2
𝑓′′
0
1
2!
+
ℎ3 𝑓′′′ 0
1
3!
+. . . . . . . . .
ℎ 𝑛
𝑛!
𝑓 𝑛 0 +. . . . . . .

9. Proof of Maclaurin series
› Let 𝑓(𝑥) be a function of 𝑥 which can be expanded into positive ascending integral
powers of 𝑥 then
𝑓 𝑥 = 𝑎 𝑜 + 𝑎1 𝑥 + 𝑎2 𝑥2
+ 𝑎3 𝑥3
+. . . . . . . . . .
Differentiating w.r.t. 𝑥 successively,
(1
)
𝑓′
𝑥 = 𝑎1 + 𝑎2. 2𝑥 + 𝑎3. 3𝑥2
+. . . . . . . . . .
𝑓′′ 𝑥 = 𝑎2. 2 + 𝑎3. 6𝑥+. . . . . . . . . .
and so
on.
(2)
(3)

10. Putting 𝑥 =0 in Eq. (1) (2) & (3),
𝑎0 = 𝑓 0
𝑎1 = 𝑓′
0
𝑎2 = 𝑓′′
0 and so on
Substituting 𝑎0, 𝑎1, 𝑎2 in Eq.(1) we get,
𝑓 𝑥 = 𝑓(0) +𝑥𝑓′
0 + 𝑥2
𝑓′′
0
1
2!
+ 𝑥3
𝑓′′′
0
1
3!
+. . . . . . . . .
𝑥 𝑛
𝑛!
𝑓 𝑛
0 +. . . . . . .
This is known as Maclaurin’s series.

11. › The Taylor’s series and Maclaurin’s series gives the expansion of a function 𝑓(𝑥) as a
power series under the assumption of possibility of expansion of 𝑓 𝑥 .
› Such an investigation will not give any information regarding the range of values 𝑥 for
which the expansion is valid.
› In order to find the range of values of 𝑥, it is necessary to examine the behaviour of 𝑅 𝑛,
where 𝑅 𝑛 is the Remainder after n terms.
We have,
𝑓 𝑥 = 𝑓(𝑎) + 𝑥 − 𝑎 𝑓′ 𝑎 + 𝑥 − 𝑎 2 𝑓′′ 𝑎
1
2!
+
𝑥 − 𝑎 3 𝑓′′′ 𝑎
1
3!
+. . . . . . . . .
𝑥 − 𝑎 𝑛−1
𝑛 − 1 !
𝑓 𝑛−1 𝑎 + 𝑅 𝑛
Where 𝑅 𝑛 is the remainder after n terms defined as,
(1)

12. 𝑅 𝑛 =
𝑥−𝑎 𝑛
𝑛!
𝑓 𝑛 𝜀 : 𝑎 < 𝜀 < 𝑥.
when this expansion (1) converges over a certain range of value of 𝑥 that is
𝑅 𝑛 → 0 𝑎𝑛𝑑 𝑛 → ∾ then the expansion is called Taylor series of 𝑓(𝑥) expanded
about a with the range values of 𝑥. (also known as 𝑟𝑎dius of convergence) for
which the expansion is valid.

13. Examples of Taylor’s series
Example 1.
Prove that: 𝑓 𝑚𝑥 = 𝑓 𝑥 + 𝑚 − 1 𝑥𝑓′
𝑥 +
𝑚−1 2
2!
𝑥2
𝑓′′(𝑥) + ⋯
Solution
𝑓 𝑚𝑥 = 𝑓 𝑚𝑥 − 𝑥 + 𝑥 = 𝑓 𝑥 + 𝑚 − 1 𝑥
By Taylor’s series,
𝑓 𝑥 + ℎ = 𝑓(𝑥) +ℎ𝑓′
𝑥 + ℎ2
𝑓′′
𝑥
1
2!
+ ℎ3
𝑓′′′
𝑥
1
3!
+. . . .
Putting ℎ = 𝑚 − 1 𝑥,
𝑓 𝑥 + 𝑚 − 1 𝑥 = 𝑓 𝑚𝑥 = 𝑓 𝑥 + 𝑚 − 1 𝑥𝑓′ 𝑥 +
𝑚−1 2
2!
𝑥2 𝑓′′(𝑥) + ⋯

14. Example 2:
Prove that:
𝑓
𝑥2
1 + 𝑥
= 𝑓 𝑥 −
𝑥
1 + 𝑥
𝑓′
𝑥 +
𝑥2
2! 1 + 𝑥 2 𝑓′′(𝑥) −
𝑥3
3! 1 + 𝑥 3 𝑓′′′(𝑥) … …
Solution:
𝑥2
1 + 𝑥
= 𝑥 −
𝑥
1 + 𝑥
By Taylor’s series,
𝑓 𝑥 + ℎ = 𝑓(𝑥) +ℎ𝑓′ 𝑥 + ℎ2 𝑓′′ 𝑥
1
2!
+ ℎ3 𝑓′′′ 𝑥
1
3!
+. . . .
Putting ℎ = −
𝑥
1+𝑥

15. 𝑓 𝑥 −
𝑥
1 + 𝑥
= 𝑓
𝑥2
1 + 𝑥
𝑓
𝑥2
1 + 𝑥
= 𝑓 𝑥 −
𝑥
1 + 𝑥
𝑓′ 𝑥 +
𝑥2
2! 1 + 𝑥 2
𝑓′′ 𝑥
−
𝑥3
3! 1 + 𝑥 3
𝑓′′′(𝑥) … …
Hence proved.
Example 3:
Express 𝑓 𝑥 = 2𝑥3 + 3𝑥2 − 8𝑥 + 7 in terms of 𝑓(𝑥 − 2)
Solution:
𝑓(𝑥) = 2𝑥3
+ 3𝑥2
− 8𝑥 + 7

16. By Taylor’s series,
𝑓 𝑥 = 𝑓(𝑎) +(𝑥 − 𝑎)𝑓′ 𝑎 + (𝑥 − 𝑎)2 𝑓′′ 𝑎
1
2!
+
(𝑥 − 𝑎)3 𝑓′′′ 𝑎
1
3!
+. . . . . . . . .
(𝑥−𝑎) 𝑛
𝑛!
𝑓 𝑛 𝑎 +. . . . . . .
Putting 𝑎 = 2,
𝑓 𝑥 = 𝑓(2) +(𝑥 − 2)𝑓′
𝑎 + (𝑥 − 2)2
𝑓′′
2
1
2!
+ (𝑥 − 2)3
𝑓′′′
2
1
3!
+. . . . . . (1)
𝑓(𝑥) = 2𝑥3 + 3𝑥2 − 8𝑥 + 7, 𝑓 2 = 16 + 12 − 16 + 7 = 19
𝑓′ 𝑥 = 6𝑥2 + 6𝑥 − 8, 𝑓′ 2 = 24 + 12 − 8 = 28
𝑓′′ 𝑥 = 12𝑥 + 6, 𝑓′′ 2 = 24 + 6 = 30
𝑓′′′
𝑥 = 12, 𝑓′′′
2 = 12 and so on.

17. Substituting in Eq.(1),
𝑓 𝑥 = 19 + 𝑥 − 2 28 + (𝑥 − 2)2
30
2!
+ (𝑥 − 2)3
12
3!
+. . . . . .
𝑓 𝑥 = 19 + 𝑥 − 2 28 + 15(𝑥 − 2)2
+ 2(𝑥 − 2)3
+. . . . . .
Example 4.
Express 5 + 4 𝑥 − 1 2
− 3 𝑥 − 1 3
+ 𝑥 − 1 4
in ascending powers of x.
Solution:
Let, 𝑓 𝑥 − 1 = 5 + 4 𝑥 − 1 2
− 3 𝑥 − 1 3
+ 𝑥 − 1 4
𝑓 𝑥 = 5 + 4𝑥2
− 3𝑥3
+ 𝑥4
By Taylor’s series,
𝑓 𝑥 + ℎ = 𝑓(𝑥) +ℎ𝑓′ 𝑥 + ℎ2 𝑓′′ 𝑥
1
2!
+ ℎ3 𝑓′′′ 𝑥
1
3!
+. . . .

18. putting ℎ = −1,
𝑓 𝑥 − 1 = 𝑓(𝑥) +(−1)𝑓′ 𝑥 + −1 2 𝑓′′ 𝑥
1
2!
+ −1 3 𝑓′′′ 𝑥
1
3!
+. . . .
= 5 + 4𝑥2 − 3𝑥3 + 𝑥4 + −1 8𝑥 − 9𝑥2 + 4𝑥3 + −1 2(8 − 18𝑥 +

19. Solution:
Let, 𝑓 𝑥 +
𝜋
4
= tan 𝑥 +
𝜋
4
𝑓 𝑥 = tan 𝑥
By Taylor’s series,
𝑓 𝑥 + ℎ = 𝑓(𝑥) +ℎ𝑓′
𝑥 + ℎ2
𝑓′′
𝑥
1
2!
+ ℎ3
𝑓′′′
𝑥
1
3!
+. . . .
Putting 𝑥 =
𝜋
4
, ℎ = 𝑥,
𝑓
𝜋
4
+ 𝑥 = 𝑓(𝜋/4) +𝑥𝑓′ 𝜋/4 + 𝑥2 𝑓′′ 𝜋/4
1
2!
+ 𝑥3 𝑓′′′ 𝜋/4
1
3!
+. . . . (1)
𝑓 𝑥 = tan 𝑥 , 𝑓 𝜋/4 = tan 𝜋/4 = 1
𝑓′
𝑥 = sec2
𝑥, 𝑓′
𝜋/4 = sec2
𝜋/4 = 2

20. 𝑓′′
𝑥 = 2 sec 𝑥 . sec 𝑥 tan 𝑥 , 𝑓′′ 𝜋
4
= 2 tan
𝜋
4
+ 2 tan3 𝜋
4
= 4
= 2 1 + tan2 𝑥 tan 𝑥 ,
= 2 tan 𝑥 + 2 tan3
𝑥
𝑓′′′ 𝑥 = 2 sec2 𝑥 + 6 tan2 𝑥 sec2 𝑥 , 𝑓′′′ 𝜋
4
= 2 + 8 tan2 𝜋
4
+ 6 tan4 𝜋
4
= 16
= 2 1 + tan2
𝑥 + 6 tan2
𝑥 1 + tan2
𝑥
= 2 + 8 tan2 𝑥 + 6 tan4 𝑥
𝑓4
𝑥 = 16 tan 𝑥 . sec2
𝑥 + 24 tan3
𝑥 . sec2
𝑥 ,
𝑓4 𝜋
4
= 16 tan
𝜋
4
. sec2 𝜋
4
+ 24 tan3 𝜋
4
. sec2 𝜋
4
=80 and so on.

21. Substituting in Eq.(1),
𝑓
𝜋
4
+ 𝑥 = 1 + 𝑥 2 + 𝑥2
4
2!
+ 𝑥3
16
3!
+ 𝑥4
80
4!
. . . .
tan
𝜋
4
+ 𝑥 = 1 + 2𝑥 + 2𝑥2 +
8
3
𝑥3 +
10
3
𝑥4 …
Now tan 43 𝑜
= tan(45 𝑜
− 2 𝑜
)
= tan
𝜋
4
−
2𝜋
180
= tan
𝜋
4
− 0.0349
=1 + 2(0.0349) + 2(0.0349)2
+
8
3
(0.0349)3
+
10
3
(0.0349)4
…
= 0.9326 𝑎𝑝𝑝𝑟𝑜𝑥.

22. Maclaurin series expansion of some standard functions
𝑒 𝑥
= 1 + 𝑥 +
𝑥2
2!
+
𝑥3
3!
+ ⋯
sin 𝑥 = 𝑥 −
𝑥3
3!
+
𝑥5
5!
− ⋯
cos 𝑥 = 1 −
𝑥2
2!
+
𝑥4
4!
− ⋯
tan 𝑥 = 𝑥 +
𝑥3
3
+
2𝑥5
15
+ ⋯
log 1 + 𝑥 = 𝑥 −
𝑥2
2!
+
𝑥3
3!
…
(1 + 𝑥) 𝑚= 1 + 𝑚𝑥 +
𝑚 𝑚 − 1
2!
𝑥2 +
𝑚 𝑚 − 1 (𝑚 − 2)
3!
𝑥3 + ⋯

23. Examples of Maclaurin’s series
Example 1.
𝑒𝑥𝑝𝑎𝑛𝑑 5 𝑥 𝑢𝑝𝑡𝑜 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑡ℎ𝑟𝑒𝑒 𝑛𝑜𝑛 𝑧𝑒𝑟𝑜 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑟𝑖𝑒𝑠.
Solution :
Let 𝑓 𝑥 = 5 𝑥
By Maclaurin’s series,
𝑓 𝑥 = 𝑓 0 + 𝑥𝑓′
𝑥 +
𝑥2
2!
𝑓′′
𝑥 … … (1)
𝑓 𝑥 = 5 𝑥
, 𝑓 0 = 50
= 1
𝑓′ 𝑥 = 5 𝑥 log 5, 𝑓′ 0 = 50 log 5 = log 5
𝑓′′
𝑥 = 5 𝑥
log 5 2
𝑓′′
0 = 50
log 5 2
= log 5 2
Substituting in Eq.(1),
𝑓 𝑥 = 5 𝑥
= 1 + 𝑥 log 5
𝑥2
2!
log 5 2
… …

24. OR
Using Exponential series expansion,
𝑓 𝑥 = 5 𝑥 = 𝑒log 5 𝑥
= 𝑒 𝑥 log 5
= 1 + 𝑥 log 5 +
𝑥 log 5 2
2!
+ ⋯
Example 2.
𝑖𝑓 𝑥 = 𝑦 −
𝑦2
2
+
𝑦3
3
−
𝑦4
4
+ ⋯ prove that,
𝑦 = 𝑥 +
𝑥2
2
+
𝑥3
3
+
𝑥4
4
+ ⋯ and conversely.
Solution :
𝑥 = log(1 + 𝑦)
1 + 𝑦 = 𝑒 𝑥
𝑦 = 𝑒 𝑥 − 1

25. By using exponential series expansion we get,
𝑦 = 𝑥 +
𝑥2
2
+
𝑥3
3
+
𝑥4
4
+ ⋯
Conversely,
𝑦 = 𝑒 𝑥 − 1
𝑒 𝑥
= 1 + 𝑦
𝑥 = log(1 + 𝑦)
= 𝑦 −
𝑦2
2
+
𝑦3
3
−
𝑦4
4
+ ⋯
Example 3.
Prove that tan−1
𝑥 = 𝑥 −
𝑥3
3
+
𝑥5
5
−
𝑥7
7
+ ⋯

26. Solution :
Let 𝑦 = tan−1 𝑥
𝑑𝑦
𝑑𝑥
=
1
1 + 𝑥2
= 1 + 𝑥2 −1
= 1 − 𝑥2
+ 𝑥4
− 𝑥6
+ ⋯
Integrating the Eq.(1),
𝑦 = 𝑐 + 𝑥 −
𝑥3
3
+
𝑥5
5
−
𝑥7
7
+ ⋯
tan−1 𝑥 = 𝑐 + 𝑥 −
𝑥3
3
+
𝑥5
5
−
𝑥7
7
+ ⋯
Putting 𝑥 = 0,
tan−1 0 = 𝑐,
𝑐 = 0
Hence, tan−1 𝑥 = 𝑥 −
𝑥3
3
+
𝑥5
5
−
𝑥7
7
+ ⋯

27. Example 4.
Expand sec−1 1
1−2𝑥2
Solution :
Let, 𝑦 = sec−1 1
1−2𝑥2
Putting 𝑥 = 𝑠𝑖𝑛θ,
𝑦 = sec−1
1
1 − 2𝑠𝑖𝑛2θ
= sec−1
1
cos 2θ
= sec−1 sec 2θ
= 2θ
2 sin−1 𝑥
Using expansion series of sin−1
𝑥,
𝑦 = 2 𝑥 +
𝑥3
6
+
3𝑥5
40
+ ⋯

28. Example 5.
Prove that
𝑒 𝑒 𝑥
= 𝑒 1 + 𝑥 + 𝑥2
+
5𝑥3
6
+ ⋯
Solution :
𝑒 𝑒 𝑥
= 𝑒
1+𝑥+
𝑥2
2!
+
𝑥3
3!
+⋯
= 𝑒𝑒 𝑥+
𝑥2
2!
+
𝑥3
3!
+⋯
= 𝑒 1 + 𝑥 +
𝑥2
2!
+
𝑥3
3!
+ ⋯ +
1
2!
𝑥 +
𝑥2
2!
+ ⋯
2
+
1
3!
𝑥 + ⋯ 3
…

29. = 𝑒 1 + 𝑥 + 𝑥2
1
2
+
1
2
+ 𝑥3
1
6
+
1
2
+
1
6
+ ⋯
= 𝑒 1 + 𝑥 + 𝑥2
+
5𝑥3
6
+ ⋯
Example 6.
Expand (1 + sin 𝑥)
Solution :
(1 + sin 𝑥) = sin
𝑥
2
+ cos
𝑥
2
=
𝑥
2
−
1
3!
𝑥
2
3
+ ⋯ + 1 −
1
2!
𝑥
2
2
+
1
4!
𝑥
2
4
− ⋯
= 1 +
𝑥
2
−
𝑥2
8
−
𝑥3
48
+
𝑥4
384
− ⋯

30. End of Presentation
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