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,