TMJC Term 4 Revision Chapter 9 Maclaurin Series Solutions

Term 4 Revision Topical Quick Check: Maclaurin Series TMJC 2024 Page 1 of 13 JC1 H2 Mathematics (9758) Term 4 Revision Topical Quick Check Chapter 9 Maclaurin Series Revision Guide Page 2 Section 1: Using Maclaurin’s expansion (MF27): Steps for use of Maclaurin’s expansion 1) Through multiple implicit differentiation, obtain equations involving 23 23 d d d, , ,...d d d y y y x x x 2) Substitute 0x= to obtain y. Subsequently, substitute 0x= and y to obtain d d y x . Continue with the substitutions to obtain 23 23 dd , ,...dd yy xx . 3) Substitute 2 2 dd, , ,...dd yyy xx into [Note: ( )f0 refers to y when 0x= , ( )f0 refers to d d y x when 0x= , …] Revision Guide Page 4 Example 1: 2012 YJC/1/8 Objectives: 1. Finding Maclaurin Series using Differentiation (Implicit Differentiation). 2. Use of previous result and standard series. It is given that ( ) 1ln 1 tan 2yx −=+ . Show that (i) ( ) 2 d1 4 2e d yyx x −+= , [2] (ii) ( ) 22 2 2 d d d1 4 8 0 d d d y y yxx x x x  + + + =  . [2] Hence find the Maclaurin series for y, up to and including the term in 2x . [3] Deduce the Maclaurin series for 11 tan 2ln 1 xy x − +=  − , up to and including the term in 2x . [3] 2 ' '' ( )f ( ) f (0) f (0) f (0) f (0)2! ! n nxxxx n= + + + + + 2 ' '' ( )f ( ) f (0) f (0) f (0) f (0)2! ! n nxxxx n= + + + + +

Term 4 Revision Topical Quick Check: Maclaurin Series TMJC 2024 Page 2 of 13 1 Solution (i) ( ) 1ln 1 tan 2yx −=+ 1e 1 tan 2y x−=+ Diff. wrt. x: ( ) 2 2 d1e (2)d 1 4 d1 4 2e (shown)d y y y xx yx x − = + += (ii) ( ) 2 d1 4 2e d yyx x −+= Diff. wrt. x: ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 22 22 2 22 22 2 22 2 2 d d d1 4 8 2e d d d d d d1 4 8 1 4d d d d d d1 4 8 1 4 0d d d d d d1 4 8 0 (shown)d d d yy y yxx x x x y y yx x x x x x y y yx x x x x x y y yxx x x x −+ + =− + + =− +  + + + + =   + + + =  When x = 0, 2 2 dd0, 2, 4dd yyy xx= = =− 22 2 ...y x x = − + (*) ( ) ( ) ( ) 1 1 22 2 1 tan 2ln 1 ln 1 tan 2 ln 1 12 2 2 33 2 x x xx x x x x xx − −  +  − = + − −  − − − −  =− Learning Point: Simplify the equation so that it is easier to differentiate Carry out implicit differentiation on the given result Use Maclaurin’s expansion formula in MF27: 2 ' '' ( )f ( ) f (0) f (0) f (0) f (0)2! ! n nxxxx n= + + + + + For ( ) 1ln 1 tan 2 x−+ : from (*) For ( )ln 1 x− : Use standard series for ( )ln 1 x+ in MF27, and replace x by –x 2 3 1 ( 1)ln(1 ) 23 rrx x xxx r +−+ = − + − + + ‘Deduce’: Relate to Maclaurin’s series for ( ) 1ln 1 tan 2 x−+ found earlier

Term 4 Revision Topical Quick Check: Maclaurin Series TMJC 2024 Page 3 of 13 Let’s Try Now! 1 EJC Promo 9758/2022/Q2 It is given that 2e cos .xyx= (a) Show that 2 2 dd 4 5 .dd yy yxx =− [3] (b) Find the Maclaurin series for y up to the term in 2x . [2] (c) Hence, show that the Maclaurin series for e cosx x is 211, 4xx++ up to the term in 2.x [2] (a) 2e cosxyx= Differentiate wrt x: ( ) ( ) 2 2 2 d e sin 2d d e sin 2 (1) e d cosx x xy xx y xx x y = − + =− + −−−−−− Differentiate wrt x: ( ) ( ) 2 2 2 2 2 2 2 2 sin e d2 ) dd 22dd dd 22dd dd 4 5 (show d n e cos dd xxy y y xx y x y xx y x x y x y y yx  − − =− + =− − + =  −   (From (1): ( ) 2 de sin 2 d x yxy x=− ) (b) When 0,x= ( )20 1e cos0yy= = = , ( ) 0 2d e sin 0 2 1d y x =− + = , ( ) ( ) 2 2 3d 4 2 5 1d y x = − = Thus 2 2 ...2! 31 2 ... 1 2 32y x x xx = + + + = + + + Differentiate using Product Rule Rewrite your equation in terms of y using the substitution 2e cosxyx= given in the question. Concept: Use Maclaurin Theorem to find the Maclaurin Series required ➔ ( ) ( ) ( ) ( ) ( ) 2 () f f 0 f 0 f 0 f 0 2! ! n nxxxx n  = + + + + + (in MF 27)

Term 4 Revision Topical Quick Check: Maclaurin Series TMJC 2024 Page 4 of 13 (c) ( ) ( ) 2 2 1 2 2 2 1 2 2 22 2 e cos e cos 1 ... 11 1 221 ...2 2! 311 4 ... 48 11 ... 32 2 3322 4 22 xx x xx x x x x x x x x xx = =+   −        = + + +       = + + − + + + + =+ + + + Rewrite the expression such that the original expression 2e cosx x from (a) appears See 1 2231 2 ... 2xx++ as 1 2231 2 ... 2xx ++  and use the Binomial Expansion in MF27 ...! )1)...(2)(1(...!2 )1(1)1( 2 ++−−−++−++=+ rn xr rnnnnxnnnxx

Term 4 Revision Topical Quick Check: Maclaurin Series TMJC 2024 Page 5 of 13 Revision Guide Page 2 Section 2: Using Standard Series Series expansion Validity range all x all x all x Revision Guide Page 6 Example 3: 2011 NYJC Prelim/P1/3 Objectives: 1. Use of standard series (MF27) to find Maclaurin Series. 2. Use of Binomial Theorem when denominator is a polynomial (usually linear or quadratic). By using standard series expansions, find the Maclaurin series for ( ) ( )f ln e cos 2= xxx up to and including the term in 2x . [3] Given that the first two non -zero terms in the Maclaurin series for ( )f x are equal to the first two non-zero terms in the series expansion of 2x a bx− , find a and b, where a and b are constants. [4] ( )  ++−−++−++=+ rn xr rnnnxnnnxx ! )1()1( !2 )1(11 2 1x  ++++++= !!3!21e 32 r xxxx r x ( )  ++ −+−+−= + !12 )1( !5!3sin 1253 r xxxxx rr ( )  +−+−+−= !2 )1( !4!21cos 242 r xxxx rr  +−+−+−=+ + r xxxxx rr 132 )1( 32)1ln( 11 − x

Term 4 Revision Topical Quick Check: Maclaurin Series TMJC 2024 Page 6 of 13 3 Solution ( ) ( ) ( ) ( ) ( ) 2 2 2 f ln e cos 2 ln e ln ln . l 2 cos 2 2 n 1 2 ... +... 2 1 .. x x x x x xx x xx x = =+  =+  =+ −+ −+ −= ( ) 1 1 1 1 2 2 2 2 1 2 1 2 1 [Expand till so as to compare the t erms to find and ] x x a bxa bx bxa x a bxxaa bx x x a baa − − − − =−− =−  =−  +  22 2 2 2 2 2xb x x x xa bx a a  + = −− Comparing like terms: 2 1a =  2a = 2 2 2b a =−  2 4ba=− =− For cos 2x : Use standard series for cos x in MF27, and replace x by 2x ( ) 2 4 2 ( 1)cos 1 2! 4! 2 ! rrx x xx r −= − + − + + For ( ) 2ln 1 2 x− : Use standard series for ( )ln 1 x+ in MF27, and replace x by 22x− ( ) 2 3 1 ( 1)ln 1 23 rrx x xxx r +−+ = − + − + + Make first term of the expansion 1 and use the standard series in MF27 ( ) 2( 1) ( 1) ( 1)11 2! ! n rn n n n n rx nx x x r − − − ++ = + + + + + Use ln ln lnab a b=+

Term 4 Revision Topical Quick Check: Maclaurin Series TMJC 2024 Page 7 of 13 Let’s Try Now! 2 MI PU2 P2 Promo 9758/2022/Q6(modified) It is given that ( )ln cosyx= . (i) Find the Maclaurin series for ( )ln cos x up to and including the term in 4x . [2] (ii) Hence, by substituting π 3x= , show that 24ππln 2 18 972+ . [2] Q2 MI PU2 P2 Promo 9758/2022/Q6(modified) (i) ( ) 2 2 4 2 2 44 4 2 4 4 24 2 ln ln ln 1 ... ...2! 2 24 8 ... 2 cos 1 ... ! 2! 4! 2! 4! 2 12 2 4! ! 4! xx xx x xxx x x x x x x −+ −+ − + −= =+     + = − + =− + − − +  + =− +   (ii) Substitute 3x = into the Maclaurin series in (i): ( ) ( ) ( ) 24 24 24 1 24 24 33ln cos ...3 2 12 1ln 2 18 972 ln 2 18 972 ln 2 18 972 ln 2 18 972       −           =− − +  − − − − − − − + Use the Standard Series of cos x in MF27 Use the Standard Series of ( )ln 1 x+ and replace x by 24 2! 4! xx−+ Subst 3x = into both LHS and RHS of the Maclaurin Series.

Term 4 Revision Topical Quick Check: Maclaurin Series TMJC 2024 Page 8 of 13 Revision Guide Page 3 Section 4: The Binomial Theorem where n is a POSITIVE INTEGER and When is a NEGATIVE INTEGER OR A FRACTION , the binomial expansion of is an infinite series which is valid only for . The following formula is given in the formulae list (MF27): where Note: The coefficient of must be written as Revision Guide Page