Series Expansion and Multivariable Calculus

Module 2 - Series expansion and multivariable calculus


2.1 Taylor’s and Maclaurin’s series for one variable
Introduction:
𝑥−𝑎 (𝑥−𝑎)2 (𝑥−𝑎)3
Taylor’s series: 𝑓(𝑥) = 𝑓(𝑎) + 𝑓′(𝑎) + 𝑓′′(𝑎) + 𝑓 ′′′ (𝑎) + ⋯
1! 2! 3!
𝑥 𝑥2 𝑥3
Maclaurin’s series: 𝑓(𝑥) = 𝑓(0) + 1! 𝑓′(0) + 𝑓′′(0) + 3! 𝑓 ′′′ (0) + ⋯
2!

Maclaurin’s series is a Taylor’s series expansion of a function at the origin.
Problems:
1. Obtain the Maclaurin’s series expansion for the following functions:
𝒙𝟐 𝒙𝟑 𝒙𝟒
√𝟏 + 𝐬𝐢𝐧 𝟐𝒙 = 𝟏 + 𝒙 − 𝟐
− 𝟔
+ 𝟐𝟒 + ⋯ (May 22)

sin x  cos x 
2
f ( x )  1  sin2x  1  2sin x cos x  sin2 x  cos2 x  2sin x cos x 


𝑦 = sin 𝑥 + cos 𝑥 𝑓(0) = 1
𝑦1 = cos 𝑥 − sin 𝑥 𝑓 ′ (0) = 1
𝑦2 = − sin 𝑥 − cos 𝑥 = −𝑦 𝑓 ′′ (0) = −1
𝑦3 = −𝑦1 𝑓′′′(0) = −1
𝑦4 = −𝑦2 𝑓 ′𝑣 (0) = 1
𝑦5 = −𝑦3 𝑓 𝑣 (0) = 1

By Maclaurin’s series,

𝑥 𝑥2
𝑓(𝑥) = 𝑓(0) + 1! 𝑓 ′ (0) + 𝑓"(0) + ⋯
2!


x x2 x3 x4
1  sin2x  1  (1)  (1)  (1)  (1) 
1! 2! 3! 4!

x2 x3 x 4
 1  sin2x  1  x    
2! 3! 4!




Dr. Narasimhan G, RNSIT 1

, 2. 𝐬𝐞𝐜 𝒙 upto 𝒙𝟒 term
𝑦 = sec 𝑥 𝑓(0) = 1
𝑦1 = sec 𝑥 tan 𝑥 = 𝑦 tan 𝑥 𝑓′(0) = 0
𝑦2 = 𝑦 sec 2 𝑥 + 𝑦1 tan 𝑥 = 𝑦 3 + 𝑦 tan2 𝑥 𝑓′′(0) = 1
= 𝑦 3 + 𝑦(sec 2 𝑥 − 1) = 𝑦 3 + 𝑦 3 − 𝑦
= 2𝑦 3 − 𝑦
𝑦3 = 6𝑦 2 𝑦1 − 𝑦1 𝑓 ′′′ (0) = 0
𝑦4 = 12𝑦𝑦12 + 6𝑦 2 𝑦2 − 𝑦2 𝑓 𝑖𝑣 (0) = 5

By Maclaurin’s series,
𝑥 𝑥2 x3 𝑥4
𝑓(𝑥) = 𝑓(0) + 1! 𝑓 ′ (0) + 𝑓′′(0)+ 3! 𝑓′′′(0) + 𝑓 𝑖𝑣 (0) + …
2! 4!
𝑥2 𝑥4 𝑥2 5𝑥 4
sec 𝑥 = 1 + 0 + (1)+0 + (5) + ⋯ = 1 + +
2! 4! 2 24



3. 𝐥𝐨𝐠(𝟏 + 𝒆𝒙 ) up to 3rd degree term.
𝑦 = log(1 + 𝑒 𝑥 ) 𝑓(0) = log 2
𝑒𝑥 1
𝑦1 = 1+𝑒 𝑥 𝑓′(0) = 2
(1+𝑒 𝑥 )𝑒 𝑥 −𝑒 𝑥 𝑒 𝑥 1
𝑦2 = (1+𝑒 𝑥 )2
= 𝑦1 − 𝑦1 2 𝑓 ′′ (0) = 4

𝑦3 = 𝑦2 − 2𝑦1 𝑦2 𝑓′′′(0) = 0
1
𝑦4 = 𝑦3 − 2(𝑦1 𝑦3 + 𝑦22 ) 𝑓 𝑖𝑣 (0) = − 8

By Maclaurin’s series,
𝑥 𝑥2 x3 𝑥4
𝑓(𝑥) = 𝑓(0) + 1! 𝑓 ′ (0) + 𝑓′′(0)+ 3! 𝑓′′′(0) + 𝑓 𝑖𝑣 (0) + …
2! 4!
𝑥 1 𝑥2 1 x3 𝑥4 1
log(1 + 𝑒 𝑥 ) = log 2 + 1! (2) + ( ) + 3! (0) + (− 8)
2! 4 4!
𝑥 𝑥2 𝑥4
= log 2 + 2 + − 192
8




Dr. Narasimhan G, RNSIT 2

, 𝒙𝟐 𝒙𝟑 𝒙𝟒
4. Prove that 𝐥𝐨𝐠(𝟏 + 𝒙) = 𝒙 − + − +⋯
𝟐 𝟑 𝟒

𝑦 = log(1 + 𝑥) 𝑓(0) = 0
1
𝑦1 = 1+𝑥 𝑓′(0) = 1
1
𝑦2 = − (1+𝑥)2 = −𝑦12 𝑓′′(0) = −1

𝑦3 = −2𝑦1 𝑦2 𝑓′′′(0) = 2
𝑦4 = −2𝑦1 𝑦3 − 2𝑦22 𝑓 𝑖𝑣 (0) = −6

By Maclaurin’s series,
𝑥 𝑥2 x3 𝑥4
𝑓(𝑥) = 𝑓(0) + 𝑓 ′ (0) + 𝑓′′(0)+ 𝑓′′′(0) + 𝑓 𝑖𝑣 (0) + …
1! 2! 3! 4!
𝑥 𝑥2 x3 𝑥4
log(1 + 𝑥) = 0 + 1! − + 3! (2) + (−6) + …
2! 4!
𝑥2 𝑥3 𝑥4
=𝑥− + − +⋯
2 3 4



𝟐𝒙𝟑 𝟏𝟔𝒙𝟓
5. 𝐭𝐚𝐧 𝒙 = 𝒙 + + +⋯
𝟑! 𝟓!

𝑦 = tan 𝑥 𝑦(0) = 0
𝑦1 = sec 2 𝑥 = 1 + tan2 𝑥 = 1 + 𝑦 2 𝑦1 (0) = 1
𝑦2 = 2𝑦𝑦1 = 2𝑦 + 2𝑦 3 𝑦2 (0) = 0
𝑦3 = 2𝑦12 + 2𝑦𝑦2 𝑦3 (0) = 2
𝑦4 = 4𝑦1 𝑦2 + 2𝑦1 𝑦2 + 2𝑦𝑦3 = 6𝑦1 𝑦2 + 2𝑦𝑦3 𝑦4 (0) = 0
𝑦5 = 6𝑦22 + 6𝑦1 𝑦3 + 2𝑦1 𝑦3 + 2𝑦𝑦4 𝑦5 (0) = 16
By Maclaurin’s series,
x x2 x3
f(x) = f(0) + 1! f ′ (0) + 2! f ′′ (0) + 3! f ′′′ (0) + ⋯
x3 x5
tan x = 0 + x + 0 + + 0 + 5! (16) + ⋯
3
x3 x5
=x+ + 5! (16) + ⋯
3




Dr. Narasimhan G, RNSIT 3

, 𝒙𝟐 𝟐𝒙𝟒 𝟏𝟔𝒙𝟔
6. 𝐥𝐨𝐠 𝐬𝐞𝐜 𝒙 = + + +⋯
𝟐! 𝟒! 𝟔!

𝑦 = log sec 𝑥 𝑦(0) = 0
1
𝑦1 = sec 𝑥 sec 𝑥 tan 𝑥 = tan 𝑥 𝑦1 (0) = 0

𝑦2 = sec 2 𝑥 = 1 + tan2 𝑥 = 1 + 𝑦12 𝑦2 (0) = 1
𝑦3 = 2𝑦1 𝑦2 𝑦3 (0) = 0
𝑦4 = 2𝑦22 + 2𝑦1 𝑦3 𝑦4 (0) = 2
𝑦5 = 4𝑦2 𝑦3 + 2𝑦1 𝑦4 + 2𝑦2 𝑦3 = 6𝑦2 𝑦3 + 2𝑦1 𝑦4 𝑦5 (0) = 0
𝑦6 = 6𝑦2 𝑦4 + 6𝑦32 + 2𝑦1 𝑦5 + 2𝑦2 𝑦4 𝑦6 (0) = 16
By Maclaurin’s series,
x x2 x3 x4 x5
f(x) = f(0) + 1! f ′ (0) + 2! f ′′ (0) + 3! f ′′′ (0) + 4! f ′v (0) + 5! f v (0) …
x2 x4 x6
log sec x = 0 + 0 + 0 + 2! + 0 + 4! (2) + 0 + 6! (16) + ⋯
x2 2x4 x6
= + + 6! (16) + ⋯
2! 4!



Home work:
𝐞𝐱
7. up to 3rd degree term.
𝟏+𝒆𝒙
𝑒𝑥 (1+𝑒 𝑥 )𝑒 𝑥 −𝑒 2𝑥 𝑒𝑥
Hint: 𝑦 = 1+𝑒 𝑥 , 𝑦1 = (1+𝑒 𝑥 )2
= (1+𝑒 𝑥 )2 = 𝑦12
1 𝑥 𝑥3
Ans: 2 + 4 − 48 + ⋯
𝒙𝟐 𝟑𝒙𝟒 𝟖𝒙𝟓
8. 𝒆𝐬𝐢𝐧 𝒙 = 𝟏 + 𝒙 + − − +⋯
𝟐! 𝟒! 𝟓!

𝐻𝑖𝑛𝑡: 𝑦 = 1, 𝑦1 = 1, 𝑦2 = 1, 𝑦3 = 0, 𝑦4 = −3, 𝑦5 = −8
−𝟏 𝒙 𝒙𝟐 𝒙𝟑 𝟕𝒙𝟒
9. 𝒆𝐭𝐚𝐧 =𝟏+𝒙+ − − +⋯
𝟐 𝟑! 𝟒!

𝐻𝑖𝑛𝑡: 𝑦 = 1, 𝑦1 = 1, 𝑦2 = 1, 𝑦3 = −1, 𝑦4 = −7, 𝑦5 = 5
𝒙 𝒙𝟐 𝒙𝟒
10. 𝐥𝐨𝐠(𝟏 + 𝒆𝒙 ) = 𝐥𝐨𝐠 𝟐 + 𝟐 + − 𝟏𝟗𝟐 + ⋯
𝟖
1 1 1
𝐻𝑖𝑛𝑡: 𝑦 = log 2 , 𝑦1 = , 𝑦2 = , 𝑦3 = 0, 𝑦4 = −
2 4 8




Dr. Narasimhan G, RNSIT 4