RI Maclaurin Series C7C Lect Notes

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ______________________________________________ Chapter 7C: Maclaurin Series Page 1 of 24 Chapter 7C: Maclaurin Series SYLLABUS INCLUDES  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 (i) repeated differentiation, e.g. sec x (ii) repeated implicit differentiation, e.g. 32 2 2yyy x x  (iii) using standard series, e.g. e( c o s 2)x x , 1ln 1 x x     Range of values of x for which a standard series converges  Concept of Maclaurin’s series as an approximation of a function  Small angle approximations: sin xx , 2 cos 1 2 xx , tan xx PRE-REQUISITES  Differentiation Techniques  Use of the Binomial Theorem for positive integer n  Use of the notations !n and n r    Partial Fractions CONTENT 1 Maclaurin Series 1.1 Derivation of Maclauri n Series by Differentiation 1.2 Standard Series Expansion of ex , sin x , cosx , ln(1 ) x 2 Binomial Expansion 2.1 Binomial Expansion for Positive Integral Index (Self-Reading) 2.2 Standard Series Expansion of (1 ) nx for any rational n 3 Applications of Series Expansion of a Function 3.1 Estimation 3.2 Linear Approximations 3.3 Trigonometric Series – Small Angle Approximations Appendix

Raffles Institution H2 Mathematics 2025 Ye ar 5 _________________________________________________________________________________________________________ _____________________________________________ Chapter 7C: Maclaurin Series Page 2 of 24 0a  f0 1a  f0 2a  f0 2!  3a   3 f0 3! … … na  f0 ! n n 1 Maclaurin Series For any polynomial of the form   23 01 2 3f n nx aa x a xa x a x     , there is a relationship between the coefficients 01, ,..., naa a and derivatives at 0x , i.e.       ()f 0 ,f 0 ,...,f 0 n . For example, for the cubic function  23 f 282 4 x xxx   , where 01 2 3 11 10, , , 28 2 4aa a a   , we observe that   0f0 0 a       2 1 2 (3) (3) 3 11f , f 024 8 2 11 1f , f 0 2 244 4 8 11 1f , f 0 6 644 2 4 xxxa xx a x a                 So  0 f0 ,a  1 f0 ,a  2 1 f0 ,2a   (3) 3 1 f0 .3!a  We shall now find a polynomial to approximate standard functions (for x in an interval about 0). 1.1 Derivation of Maclaurin Series by Differentiation Given a function f , for x in an interval around 0, if xf can be expressed as   23 01 2 3f, n nxa a x a xa x a x      then   0f0 a   21 12 3 1f2 3 f 0 n nx aa x a x n a x a             2 23 2f2 3 2 1 f 0 2 n nx aa x n n a x a                  3 33 3 3f3 2 1 2 f 0 3 2 n nx an n n a x a         Hence we deduce that, after differentiating function f n times, we will get   f0 ! n na n . Thus, if xf can be expanded as a power series (where all terms are non-negative powers of x), for a given range of x including zero, then (3) ( ) 23f (0) f (0) f (0)f( ) f( 0 ) f ( 0 ) 2! 3! ! n nx= + x + x x x n     Note: the expression on the RHS of xf is known as the Maclaurin Series , where )0(f )(n is the nth derivative of f w.r.t x , evaluated at 0x . f( 0 ) f( 0 ) f( 0 ) (3)f( 0 ) … ()f( 0 )n 0xy  0 d d x y x  2 2 0 d d x y x  3 3 0 d d x y x  … 0 d d n n x y x  f( )yx

Raffles Institution H2 Mathematics 2025 Ye ar 5 _________________________________________________________________________________________________________ _____________________________________________ Chapter 7C: Maclaurin Series Page 3 of 24 Example 1 Use the Maclaurin Theorem (3) ( ) 23f (0) f (0) f (0)f ( ) f (0) f (0) 2! 3! ! n nx= + x + x x x n     to find the first four non-zero terms of (a) ex , (b)  1 21 x , and (c) cos 2x . Solution: (a) Let f( ) e xx            (3) (3) fe f 0 1 fe f 0 1 fe f 0 1 fe f 0 1 x x x x x x x x            By Maclaurin Theorem, 23 23 e1 2! 3! 111 26 x xx+x + +x + x x       (b) Let  1 2f( ) 1x x f0 1  f x    1 2 111f ' 022 x     f x    3 2 111f 044 x        (3)f x    5 (3)2 331f 088 x    By Maclaurin Theorem,  231 2 23 11 311 24 2 8 3 ! 11 11 28 1 6 xxxx xx x                   (c) Let f( ) c o s2x x  f0 1    f2 s i n 2f 0xx    0    f4 c o s 2 f 0xx     4       33 f8 s i n 2f 0xx  0       44 f1 6 c o s 2f 0xx  16       55 f3 2 s i n 2 f 0xx   0       66 f6 4 c o s 2 f 0xx   64 By Maclaurin Theorem,    23 45 6 24 6 46 4 01 6 0cos 2 1 0 2! 3! 4! 5! 6! 2412 34 5 xx x x x x x xx x             

Raffles Institution H2 Mathematics 2025 Ye ar 5 _________________________________________________________________________________________________________ _____________________________________________ Chapter 7C: Maclaurin Series Page 4 of 24 Example 2 Given that sin e xyx , show that xx yy x cos ed d2  . By further differentiation of this result, obtain the first four terms in the Maclaurin series for y . Solution: sin e 1 xyx  2 sin e xyx  Differentiate w.r.t x,  d2c o s ed d2e c o s 2d x x yyx x yyx x    Differentiate (2) w.r.t x,  2 2 2 dd22 e s i n 3dd xyy yxxx      Differentiate (3) w.r.t x,  22 3 22 3 dd dd d42 2 e c o s 4dd dd d xyy yy y yxxx xx x       Substitute 0x , (1): 1y  (2):  dd21 1 1 1dd yy x x   (3):   22 22 dd 121 21 1 0 dd 2 yy xx     (4): 33 33 11 d d 3422 1 122 2 dd yy xx      By Maclaurin Theorem,    31 22 23 23 11 2! 3! 111 ... 44 yx x x xx x        

Raffles Institution H2 Mathematics 2025 Ye ar 5 _________________________________________________________________________________________________________ _____________________________________________ Chapter 7C: Maclaurin Series Page 5 of 24 Some problems are more efficiently solved when implicit differentiation is applied. Example 3 Given the curve 3 0yy x , obtain the Maclaurin series for y , in ascending powers of x , up to and including the term in 3x . Solution:  3 01yy x  Differentiate (1) w.r.t x, 2 dd31 0dd yyy x x    2 d31 1 2 d yy x  Differentiate (2) w.r.t x,   22 2 2 dd31 6 0 3dd yyyy xx     Differentiate (3) w.r.t x,   332 2 2 32 2 dd d d dd31 6 1 2 6 0 4dd d d dd yy y y yyyy y xx x x xx       Substitute 0x , (1): 22( 1) 0 0 since 1 0yy y y    (2):  dd30 1 1 1 dd yy x x  (3):    22 22 22 dd30 1 60 1 0 0dd yy x x   (4):     33 23 33 dd30 1 60 1 20 61 0 6dd yy x x     By Maclaurin Theorem, 336 ... ...3!yx x xx   Example 4 Given that y is a function of x such that 2 2 2 dd(1 9 ) 9 dd yyx mx yxx  and the Maclaurin series for y is 2313 9x nx x  +…, where m and n are some constants, find the values of m and n. Solution: 2 2 2 dd(1 9 ) 9 (1)dd yyxm x yxx   Differentiate w.r.t x,   32 2 2 32 2 dd d d d19 1 8 9 ( 2 )dd d d d yy y y yxx m m xxx x x x     Given Maclaurin series 2313 9 . . . ,yx n x x  

Raffles Institution H2 Mathematics 2025 Ye ar 5 _________________________________________________________________________________________________________ _______________________________________