ACJC 2025 Maclaurin Series Lecture Notes

1 7 MACLAURIN SERIES SYLLABUS • Standard series expansion of (1 ) nx+ for any rational n, ex , sin x , cos x and ln(1 ) x+ • Derivation of the first few terms of the Maclaurin series by - repeated differentiation, e.g. sec x - repeated implicit differentiation, e.g. 3 2 2 2y y y x x+ + = − - using standard series, e.g. e cos 2x x , 1ln 1 x x +  − • Range of values of x for which a standard series converges • Concept of “approximation” • Small angle approximations: sin xx , 21 2cos 1xx− , tan xx

ACJC 2025/26 H2 Mathematics (9758) 2 CONTENTS 0 Introduction: Power Series ............................................................... 3 1 Maclaurin Series 1.1 Derivation of Maclaurin Series ............................................... 3 1.2 Some Standard Series Expansions .......................................... 7 2 Binomial Series 2.1 Recap: Partial Fractions ........................................................ 15 2.2 Binomial Expansion for Positive Integer Indices ................. 16 2.3 Binomial Series for Rational Indices that are not Positive Integers .................................................................................. 16 2.4 Small Angle Approximation ................................................. 22 Annex A: Partial Fractions .................................................................... 27 Annex B: Practice Questions ................................................................. 28

7 Maclaurin Series 3 LECTURE 1 Lesson Outline • Introduction to power series • Maclaurin series: derivation, standard series expansions & some basic examples 0 INTRODUCTION: POWER SERIES Generally, functions can be complicated, thereby making calculations and manipulations with them tough. Thankfully, many functions can be represented as an “infinite polynomial”, called a power series. This enables many functions to be well approximated by an appropriate polynomial. It is useful since polynomials are easier to evaluate, differentiate and integrate. Power series are applied in many fields, including Physics and Economics, as a way of simplifying computations. Formally, a power series is an infinite series of the form 23 0 1 2 3 n na a x a x a x a x+ + + + + + where the coefficients 0 1 2 3, , , , , , na a a a a are constants, e.g. 2 4 61 5 6 3 7x x x x+ + + + + The following geometric series are examples of power series, 234 2 3 4 1 1 , 11 1 1 ... , 11 x x x x xx x x x x xx = + + + + + − = − + − + − + 1 MACLAURIN SERIES 1.1 Derivation of Maclaurin Series Let f ( )x be a function of x. Suppose f ( )x can be expanded as a power series of x, 2 3 4 0 1 2 3 4f ( )x a a x a x a x a x= + + + + +

ACJC 2025/26 H2 Mathematics (9758) 4 Problem To find 0 1 2 3, , , , , , na a a a a so that the power series expansion is completely determined. Solution 2 3 4 0 1 2 3 4 0f ( ) f(0)x a a x a x a x a x a= + + + + +  = 23 1 2 3 4 1 1f ( ) 2 3 4 f (0) f (0)x a a x a x a x a a  = + + + +  =  = 2 2 3 4 2 2 f (0)f ( ) 2 (2)(3) (3)(4) f (0) 2 2!x a a x a x a a  = + + +  =  = 3 4 3 3 f (0)f ( ) (2)(3) (2)(3)(4) f (0) (2)(3) 3!x a a x a a  = + +  =  = (4) (4) (4) 4 4 4 f (0)f ( ) (2)(3)(4) f (0) (2)(3)(4) 4!x a a a= +  =  = Hence, the Maclaurin series of f ( )x is () 23f (0) f (0) f (0) f (0)f ( ) f (0) 1! 2! 3! ! r rx x x x x r   = + + + + + + Note The general formula for the Maclaurin series of f ( )x is given in MF27 as 2 ()f ( ) f (0) f (0) f (0) f (0)2! ! n nxxxx n  = + + + + + Example 1 Find the Maclaurin series expansion of the following functions up to and including the term in 4.x (a) ln(1 ) x+ (b) cos x Solution (a) Let f( ) ln(1 )xx= +  f(0) ln1 0== 1f ( ) f (0) 1(1 )x x  =  =+ 2 1f ( ) f (0) 1(1 )x x − =  =−+ 3 2f ( ) f (0) 2(1 )x x  =  =+ (4) (4) 4 2(3)f ( ) f (0) 2(3)(1 )x x −=  =−+

7 Maclaurin Series 5 Using Maclaurin series formula above: 23 2 3 4 2 3 4 f (0) f (0) f (0)f ( ) f (0) 1! 2! 3! 1 1 2 2(3)0 1! 2! 3! 4! 1 1 1 2 3 4 x x x x x x x x x x x x   = + + + + −−= + + + + + = − + − + (b) Let f ( ) cosxx= f(0) cos0 1== f ( ) sin f (0) sin 0 0xx =−  =− = f ( ) cos f (0) cos0 1xx =−  =− =− f ( ) sin f (0) sin 0 0xx =  = = (4) (4)f ( ) cos f (0) cos0 1xx=  = = Using Maclaurin series formula above: 23 2 3 4 24 f (0) f (0) f (0)f ( ) f (0) 1! 2! 3! 0 1 0 11 1! 2! 3! 4! 111 2! 4! x x x x x x x x xx   = + + + + −= + + + + + = − + + ■ The Maclaurin series of a function, when it converges, gives an equivalent form of the function. In practice, the first few terms of a Maclaurin series provide a sufficiently accurate approximation of the function . How does the number of terms in the Maclaurin series affect the level of accuracy of approximation? Does the approximation work for all values of x? Learning Experience Activity Access the GeoGebra worksheet at http://ggbtu.be/mez563VJB

ACJC 2025/26 H2 Mathematics (9758) 6 Example 3 5 7 sin 3! 5! 7! x x xxx= − + − + . Using GC, compare the graph of sinyx= with the graphs of the following equations. • yx= • 3 3! xyx=− • 35 3! 5! xxyx= − + • 3 5 7 3! 5! 7! x x xyx= − + − Observations 1. The graphs of the Maclaurin series (for sin x ) approximate the graph of sinyx= . This approximation gets better as the number of terms in the Maclaurin series increases. 2. The graphs are close to that of sinyx= when x is near 0. Thus, Maclaurin series can be used to approximate a function in the vicinity of 0x= . Learning points In general, there are two ways to improve the approximation of a function using Maclaurin series: • As the number of terms in a Maclaurin series increases, the level of accuracy of the approximation increases. • For a given Maclaurin series of a function, the approximation to the function gets better when the x value is closer to 0. y x O 1 –1 1

7 Maclaurin Series 7 1.2 Some Standard Series Expansions Note that the following expansions are standard expansions in MF27 under “Maclaurin expansion”: ( ) ( ) ( ) ( )21 1 ... 111 2! ! n rn n n n n rx nx x x r − − − ++ = + + + + + for 1x  23 e 1 for all values of 2! 3! ! r x x x xxx r= + + + + + + ( ) ( ) 3 5 7 2 1 sin 1 for all values of 3! 5! 7! 2 1 ! r rx x x xx x x r + = − + − + + − + + ( ) ( ) 2 4 6 2 cos 1 1 for all values of 2! 4! 6! 2 ! rrx x x xxx r= − + − + + − + ( ) ( ) 2 3 4 1 ln 1 1 for 1 1 2 3 4 r rx x x xx x x r + + = − + − + + − + −   Using the appropriate standard expansions and substitutions for x, we can write down the following expansions. ( ) ( ) ( ) ( ) ( ) 23 23 e1 2! 3! ! 11 for all values of 2! 3! ! r x r r x x xx r xxxxx r − − − −= + − + + + + + −= − + − + + + ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 42 2 2 2 122 4 6 8 2 2 ln 1 1 2 3 4 2 3 4 r r r x x x x xx r x x x xx r +− − − − − = − − + − + + − + =− − − − − − − 22for 1 1 0 1 1 1x x x− −     −   ( ) 2 3 4 2 3 4 ln 2 ln 2 1 ln 2 ln 1 22 2 2 2ln 2 2 2 3 4 ln 2 2 8 24 64 xxx x x x x x x x x     + = + = + +                        = + − + − + = + − + − + for 1 1 2 22 x x−   −   Why not expand from “ ( )++ln 1 (1 ) x "?

ACJC 2025/26 H2 Mathematics (9758) 8 Example 2 Using the standard results given in the List of Formulae (MF27), show that for small values of x, ( ) ( ) 23771 2 e ln 1 2 1 3 22 xx x x x x−+ + +  + − + and state the range of validity. Solution ( ) ( )1 2 e ln 1 2xxx −+ + + ( ) ( ) ( ) 2323 221 2 1 2 2! 3! 2 3 xxxxx x x = + − + − + + − + −    23 2 3 2 3 81 ( 2 ) 2 2 2 2 6 3 xxx x x x x x x   = + − + + − + − + + − + +       237713 22x x x + − + (shown) Range of validity: 111 2 1 22xx−   −   ■ Example 3 It is given that ( )sin ln(1 )yx=+ . (i) Show that