ACJC 2025 Maclaurin Series Summary

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Maclaurin Series / Summary / Page 1 of 2 SUMMARY: Maclaurin Series Maclaurin Series (in MF27) 2 ()f ( ) f (0) f (0) f (0) ... f (0) ...2! ! n nxxxx n = + + + + + Note: ()f ( )n x denotes the nth derivative of f ( )x . Implicit differentiation is usually required when doing repeated differentiation of an equation involving the derivatives. Standard Expansions (in MF27) 2( 1)(1 ) 1 ... 2! ( 1)...( 1) , 1 ! n r nnx nx x n n n r xxr −+ = + + + − − ++ +  23 e 1 ... ... 2! 3! ! n x x x xx n= + + + + + + 3 5 7 sin ... 3! 5! 7! x x xxx= − + − + 2 4 6 cos 1 ... 2! 4! 6! x x xx= − + − + 2 3 4 ln(1 ) ... 2 3 4 x x xxx+ = − + − + (–1  x  1) • The standard expansions can be applied when x is replaced by kx or kx , where k is a constant. • For expressions like ,xk+ it needs to be simplified before applying the standard expansion. Examples 23 3 3 3e e e e 1 ... 2! 3! ! n xx x x xx n + = = + + + + + + 2 3 4 ln(2 ) ln 2 1 ln 2 ln 122 1 1 1ln 2 ...2 2 2 3 2 4 2 xxx x x x x    + = + = + +            = + − + − +           • Take note of the range of validity for each of the above expansion. Convergence & Approximation In general, there are two ways to improve the approximation of a function using Maclaurin series • Comparing two Maclaurin series of a function consisting of different number of terms, the Maclaurin series with the greater number of terms gives a better approximation to the function. • For a given Maclaurin series of a function, the approximation to the function gets better when the x value is closer to 0. Binomial Series (in MF27) 2( 1)(1 ) 1 ... 2! ( 1)...( 1) ,! where 1 and is any rational number. n r nnx nx x n n n r xr xn −+ = + + + − − +++  • This is an infinite series and it is valid for 1x  (i.e. 11 x−   ). • For the expansion of () nax+ , it is advisable to change it to the form (1 ) ny+ as follows: (i) If x is small, the expansion is in ascending powers of x: ( ) 1 1 n n nn xxa x a a aa    + = + = +       . The expansion is valid for 1x a  3 3! xyx=− yx= sinyx= 3 5 7 3! 5! 7! x x xyx= − + − 35 3! 5! xxyx= − + 4 3 2 1 -1 -2 -3 -4 -2 2 4

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Maclaurin Series / Summary / Page 2 of 2 Special Binomial Series to remember: 1. ( ) 1 1 x − + = 2 3 41 ... ( 1) ... rrx x x x x− + − + + + − + 1x   11 x−   2. ( ) 1 1 x − − = 2341 ... ... rx x x x x+ + + + + + + 1x−  11 x−   We may apply these special series when finding the binomial expansions of ( )1 n kx where 1n=− . (ii) If x is large, then a x is small, and the expansion is in descending powers of x: ( ) 11 n nn naax a x x xx    + = + = +       The expansion is valid for 1a x  . • When using binomial series for approximation: (i) if the value of x is given, then just substitute it into the expansion. (ii) if not, then explore and look for a value within the range of validity. • If the denominator can be factorised, then change to Partial Fractions first. Example ( ) ( ) 22 1 12 1 1 1 2 23 23 7 5 1 1 4 21(2 )(1 ) (1 ) 2 (1 ) 4(1 ) 2 1 (1 ) 4(1 )2 1 1 ...2 2 2 2 1 ... x xxx x x x x x x xx x x x x x x − −− − − − − − = + −−+− + + = − + + − + = − + + − +     = + + + +         + − + − + 2( 2)( 3)4 1 2 ... ... 2!xx −−− − + + = Small Angle Approximations When x is in radians and x is small (i.e. 3x & higher powers of x can be neglected), In general, if x is small and kx is small, where k is a constant, we have Knowledge of geometrical results related to circles and triangles (e.g. arc length, area of sector, sine rule, cosine rule etc) are required. Sine Rule: sin sin sin a b c A B C== Cosine Rule: 2 2 2 2 cosa b c bc A= + − Example Given that  is a sufficiently small angle, show that 2 πsin 6 1 3 3 .cos 2 2 2 4    +  + + ( ) ( ) ( )( ) 2 2 122 22 2 2 2 π ππ 13sin sin cos cos sin cos sin6 66 22 cos 2 cos 2 cos 2 1 1 312 2 2 112 2 1 3 1 122 2 4 1 3 1 1 2 2 ...2 2 4 1 3 3 2 2 4                 −  + + + == −+ − = + − −   + − + + +   + + sin xx 21cos 1 2xx− tan xx sinkx kx 21cos 1 ( ) 2kx kx=− tankx kx