1. Maclaurin and
Taylor Series

2. Maclaurin Taylor Series
KUS objectives
BAT Find the series expansion of a function where f(0) undefined using Taylors expansion formula
Starter:

3. is undefined so a series cannot be obtained
Notes 1
Maclaurin’s formula:
What would happen if ?
is undefined so a series cannot be obtained
The Mathematician Taylor got around this problem by considering a Maclaurin series expansion focused on an arbitrary point x = a
Taylor’s method requires the same steps of differentiating and evaluating, but at any point x = a, so long as the series obtained is written in powers of (x-a)

4. Imagine we have two functions, f and g, linked as shown
Note that we are using g here, but Taylor’s expansion will be in terms of f, same as Maclaurin’s
Notes 2
𝑔 𝑥+𝑎 =𝑓(𝑥)
If these are equal, the differentials will be as well
𝑔 𝑟 𝑥+𝑎 = 𝑓 𝑟 (𝑥)
𝑟=1,2,3 𝑒𝑡𝑐
Let x = 0
𝑔 𝑟 𝑎 = 𝑓 𝑟 (0)
What this means is:
If we write our function f(x) as g(x + a) instead
We have to change all the fr(0) terms to gr(a) terms (so all the differentials change as r can take any integer value)
𝑓 𝑥 =𝑓 0 + 𝑓 ′ 0 𝑥+ 𝑓′′(0) 2! 𝑥 2 + 𝑓′′′(0) 3! 𝑥 3 …
Replace ‘x’ with ‘x + a’, and all fr(0) with gr(a)
𝑔 𝑥+𝑎 =𝑔 𝑎 + 𝑔 ′ 𝑎 𝑥+ 𝑔′′(𝑎) 2! 𝑥 2 + 𝑔′′′(𝑎) 3! 𝑥 3 …

5. Notes 3
You can also change the Taylor expansion into a slightly different form…
𝑓 𝑥+𝑎 =𝑓 𝑎 + 𝑓 ′ 𝑎 𝑥+ 𝑓′′(𝑎) 2! 𝑥 2 + 𝑓′′′(𝑎) 3! 𝑥 3 …
Replace all ‘x’ terms with ‘x – a’ terms
𝑓 𝑥 =𝑓 𝑎 + 𝑓 ′ 𝑎 (𝑥−𝑎)+ 𝑓′′(𝑎) 2! (𝑥−𝑎) 2 + 𝑓′′′(𝑎) 3! (𝑥−𝑎) 3 …
Both these forms can be used under different circumstances as Taylor expansions
 You will need to decide which to use based on the question!

6. WB10 Find the Taylor expansion of ln x in powers of (x -1) up to and including the terms in (x - 1)3
𝒇 𝒙 =𝒇 𝒂 + 𝒇 ′ 𝒂 (𝒙−𝒂)+ 𝒇′′(𝒂) 𝟐! (𝒙−𝒂) 𝟐 + 𝒇′′′(𝒂) 𝟑! (𝒙−𝒂) 𝟑 …
You effectively work through in a similar way as for the Maclaurin expansion
𝑓 𝑥 = ln 𝑥
𝒇 𝟏 =𝟎
𝑓′ 𝑥 = 1 𝑥
𝒇 ′ (𝟏)=𝟏
substitute ‘1’ into these
(instead of 0 as you would have done with the Maclaurin expansion
𝑓 ′′ (𝑥)=− 1 𝑥 2
𝒇 ′′ (𝟏)=−𝟏
𝒇′′′ 𝟏 =𝟐
𝑓′′′ 𝑥 = 1 𝑥 3
ln 𝑥 =0−1× 𝑥−1 + −1 2! × 𝑥− !! × 𝑥−1 3 +…
ln 𝑥 = 𝑥−1 − 𝑥− /3 𝑥− …

7. WB11 Find the Taylor expansion of f x = 𝑥 in powers of (x -1) up to and including the terms in (x - 1)4
𝒇 𝒙 =𝒇 𝒂 + 𝒇 ′ 𝒂 (𝒙−𝒂)+ 𝒇′′(𝒂) 𝟐! (𝒙−𝒂) 𝟐 + 𝒇′′′(𝒂) 𝟑! (𝒙−𝒂) 𝟑 …
𝑓 𝑥 = 𝑥 1/2
𝒇 𝟏 =𝟏
𝑓′ 𝑥 = 1 2 𝑥 −1/2
𝒇 ′ (𝟏)= 𝟏 𝟐
𝑓 ′′ (𝑥)=− 1 4 𝑥 −3/2
𝒇 ′′ (𝟏)=− 𝟏 𝟒
substitute ‘1’ into these
(instead of 0 as you would have done with the Maclaurin expansion
𝑓 ′′′ 𝑥 = 3 8 𝑥 −5/2
𝒇′′′ 𝟏 = 𝟑 𝟖
𝑓 ′′′ 𝑥 =− 𝑥 −7/2
𝒇 ′′′ 𝟏 =− 𝟏𝟓 𝟏𝟔
f x = 𝑥 =1− 1 2 × 𝑥−1 + − ! × 𝑥− ! × 𝑥− − ! × 𝑥−1 4 +…
f x = 𝑥 = 𝑥−1 − 𝑥− 𝑥−1 3 − 𝑥− …

8. WB12 Find the Taylor expansion of e-x in powers of (x + 4) up to and including the terms in (x + 4)3
As you are asked for the expansion using powers of a bracket (x + 4), you should use this form:
𝒇 𝒙 =𝒇 𝒂 + 𝒇 ′ 𝒂 (𝒙−𝒂)+ 𝒇′′(𝒂) 𝟐! (𝒙−𝒂) 𝟐 + 𝒇′′′(𝒂) 𝟑! (𝒙−𝒂) 𝟑 …
You effectively work through in a similar way as for the Maclaurin expansion
𝑓 𝑥 = 𝑒 −𝑥
𝒇 −𝟒 = 𝒆 𝟒
substitute ‘a’ into these
(instead of 0 as you would have done with the Maclaurin expansion
𝑓′ 𝑥 = −𝑒 −𝑥
𝒇′ −𝟒 = −𝒆 𝟒
𝑓′′ 𝑥 = 𝑒 −𝑥
𝒇′′ −𝟒 = 𝒆 𝟒
𝑓′′′ 𝑥 = −𝑒 −𝑥
𝒇′′′ −𝟒 = −𝒆 𝟒
𝑒 −𝑥 = 𝑒 4 − 𝑒 4 𝑥+4 + 𝑒 𝑥+4 2 − 𝑒 (𝑥+4) 3
𝑒 −𝑥 = 𝑒 4 1− 𝑥 𝑥 (𝑥+4) 3

9. chain rule and product rule
WB 13 Express: 𝑡𝑎𝑛 𝑥+ 𝜋 4 As a series in ascending powers of x,
up to the term in x3
𝑓 𝑥+𝑎 =𝑓 𝑎 + 𝑓 ′ 𝑎 𝑥+ 𝑓′′(𝑎) 2! 𝑥 2 + 𝑓′′′(𝑎) 3! 𝑥 3 …
For f(x) you can just use tanx
The Taylor series takes into account the transformation to tan (x + π/4)
𝑓 𝑥 =𝑡𝑎𝑛𝑥
𝒇 𝝅 𝟒 =𝟏
𝑓′ 𝑥 = 𝑠𝑒𝑐 2 𝑥
chain rule
𝒇 ′ 𝝅 𝟒 =𝟐
𝑓′′ 𝑥 =
2𝑠𝑒 𝑐 2 𝑥𝑡𝑎𝑛𝑥
chain rule and product rule
𝒇 ′′ 𝝅 𝟒 =𝟒
𝑓′′′ 𝑥 =
4 𝑠𝑒𝑐 2 𝑥𝑡𝑎 𝑛 2 𝑥
+ 2 𝑠𝑒𝑐 4 𝑥
𝒇 ′′′ 𝝅 𝟒 =𝟏𝟔
𝑡𝑎𝑛 𝑥+ 𝜋 4 =(1)+(2)𝑥+ (4) 2 𝑥 𝑥 3
𝑡𝑎𝑛 𝑥+ 𝜋 4 =1+2𝑥+2 𝑥 𝑥 3

10. term in (x – π/6)2 is: 𝑠𝑖𝑛𝑥= 1 2 + 3 2 𝑥− 𝜋 6 − 1 4 𝑥− 𝜋 6 2
WB 14a
a) Show that the Taylor expansion of sin 𝑥 in ascending powers of (x – π/6) up to the
term in (x – π/6)2 is: 𝑠𝑖𝑛𝑥= 𝑥− 𝜋 6 − 𝑥− 𝜋 6 2
𝒇 𝝅 𝟔 = 𝟏 𝟐
𝑓 𝑥 =𝑠𝑖𝑛𝑥
𝒇′ 𝝅 𝟔 = 𝟑 𝟐
𝑓′ 𝑥 =𝑐𝑜𝑠𝑥
𝒇 ′′ 𝝅 𝟔 =− 𝟏 𝟐
𝑓 ′′ (𝑥)=−𝑠𝑖𝑛𝑥
𝑓 𝑥+𝑎 =𝑓 𝑎 + 𝑓 ′ 𝑎 𝑥+ 𝑓′′(𝑎) 2! 𝑥 2 + 𝑓′′′(𝑎) 3! 𝑥 3 …
𝑠𝑖𝑛𝑥= 𝑥− 𝜋 6 + − 𝑥− 𝜋 6 2
𝑠𝑖𝑛𝑥= 𝑥− 𝜋 6 − 𝑥− 𝜋 6 2

11. 𝑠𝑖𝑛𝑥= 1 2 + 3 2 2𝜋 9 − 𝜋 6 − 1 4 2𝜋 9 − 𝜋 6 2 𝑠𝑖𝑛𝑥= 1 2 + 3 2 𝜋 18
WB 14b
a) Show that the Taylor expansion of sinx in ascending powers of (x – π/6) up to the
term in (x – π/6)2 is: 𝑠𝑖𝑛𝑥= 𝑥− 𝜋 6 − 𝑥− 𝜋 6 2
b) Then use the series to find an approximation for sin40, in terms of π
𝑠𝑖𝑛𝑥= 𝑥− 𝜋 6 − 𝑥− 𝜋 6 2
The angle MUST be in radians for this process to work…
40º = 2π/9 radians (since 40º is 2/9 of 180º)
𝑠𝑖𝑛𝑥= 𝜋 9 − 𝜋 6 − 𝜋 9 − 𝜋 6 2
𝑠𝑖𝑛𝑥= 1 2 +
𝜋 18
− 𝜋
𝑠𝑖𝑛𝑥= 𝜋
− 𝜋

12. WB 15
Show that the Taylor expansion of sin⁡(𝑥 +𝜋/6) up to the term in x3 is:
𝑥− 1 4 𝑥 2 − 𝑥 3 + ….
𝒇 𝝅 𝟔 = 𝟏 𝟐
𝑓 𝑥 =𝑠𝑖𝑛𝑥
𝒇′ 𝝅 𝟔 = 𝟑 𝟐
𝑓′ 𝑥 =𝑐𝑜𝑠𝑥
𝒇 ′′ 𝝅 𝟔 =− 𝟏 𝟐
𝑓 ′′ (𝑥)=−𝑠𝑖𝑛𝑥
𝒇 ′′′ 𝝅 𝟔 =− 𝟑 𝟐
𝑓 ′′′ (𝑥)=− cos 𝑥
𝑓 𝑥+𝑎 =𝑓 𝑎 + 𝑓 ′ 𝑎 𝑥+ 𝑓′′(𝑎) 2! 𝑥 2 + 𝑓′′′(𝑎) 3! 𝑥 3 … so

13. Practice
205 D

14. Summary The following rules are provided in the Formulae Booklet:
Maclaurin series:
Standard results:
Taylor series:
2nd form:
Of course, YOU still need to memorise which one to use and when!

15. One thing to improve is –
KUS objectives BAT Find the series expansion of a function where f(0) undefined using Taylors expansion formula
self-assess
One thing learned is –
One thing to improve is –

16. END