MATHEMATICS(Calculus)Pre-U TextStrategicSTPM2022 – 2024• Purified STPM Syllabus• Scheme of Assessment Starting from Semester 1 STPM Examination 2026SEMESTER2> Info > Video> InfographicDIGITAL RESOURCES• Tan Guan Hin• Lee Yoon Woh• Lim Wen Li• Soon Pai EeNew! Based onFull Solutions for All Questions QR CODEFull Solutions for All Questions QR CODE
viChapterLimits and Continuity 7 17.1 Limits 7.2 Continuity ChapterDifferentiation 8 8.1 Derivatives and rules of differentiation 8.2 Curve sketching 8.3 Rates of change and optimisation problems ChapterIntegration 9 9.1 Indefinite and definite integrals 9.2 Techniques of integration 9.3 Areas between curves and volumes of solids of revolution ChapterDifferential Equations 10 10.2 First-order linear differential equations ChapterMaclaurin Series 11 11.1 Representations of functions as Maclaurin series 11.2 Applications of Maclaurin series ChapterNumerical Methods 12 16912.1 Numerical solution of equations 12.2 Numerical integration STPM Model Paper (954/2) Answers CONTENTSFull solutions for all questionshttps://qr.pelangibooks.com/?u=FullsolutionsSem2
147ChapterSubtopic Outline11.1 Representations of functions as Maclaurin series11.2 Applications of Maclaurin seriesMACLAURIN SERIES11Bilingual Keywordsconvergence - penumpuanMaclaurin series - siri MaclaurinMaclaurin's theorem - teorem Maclaurinpower series - siri kuasaMaclaurin series was named after Scottish mathematician, Colin Maclaurin in the 18th century. The series enables us to express complicated functions as an infinite sum of simpler and manageable polynomial terms. Maclaurin series facilitate the determination of the limits of uncomputable complicated functions and the evaluation of definite integrals of complex functions.Physicists use Maclaurin series to predict and approximate the height of the projectile's motion under the gravitational influence. Engineers make use of Maclaurin series to estimate the deflection of beams and deformation of structures in order to design structures that can withstand loads.Actuaries employ Maclaurin series to determine the elasticity of demand for a product with the variation of prices. Where analytical solutions are difficult to obtain, Maclaurin series are used in financial models to approximate compound in functions, risk functions or utility functions. Computer scientists use Maclaurin series to compute approximate values of transcendental functions such as ex, In x, sin x and cos x to enable faster speed computational on CPU.147
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series148Chapter 11 11.1 Representations of functions as Maclaurin seriesLearning outcomes:(a) Use repeated differentiation to obtain the first few terms of the Maclaurin series of a function;(b) Identify the standard Maclaurin series for (1 + x)n, where n ∈ , ex, ln(1 + x), sin x and cos x;(c) Use the standard Maclaurin series and the multiplication of Maclaurin series to find the series expansion of a function;(d) Use differentiation or integration of a power series to find the first few terms of the series expansion of a function.Representations of functions as power seriesA power series is a series of the form∞Σn = 0cnxn = c0 + c1x + c2x2 + … + cnxn + …where x is a variable and cn are constants called the coefficients of the series.If lim sn , where sn = c0 + c1x + c2x2 + . . .+ cnxn, exists, then the series is said to be convergent. The set of of x for which the series is convergent is called the interval of convergence.If all the coefficients of the power series are equal to 1, the power series becomes∞Σn = 0xn = 1 + x + x2 + … + xn + …This is an infinite geometric series which has first term 1 and common ratio x. If has sum 11 – x provided –1 , x , 1. The interval of convergence is (–1, 1).Thus, 11 – x= 1 + x + x2 + … + xn + …, provided –1 , x , 1. This example shows that a function may be expanded as a power series in a single variable.We can represent a function by its power series expansion as long as the single variable, x, is within the interval of convergence.In general, a power series of the form∞nΣ = 0cnxn = c0 + c1x + c2x2 + … + cnxn + …converges at x = 0.Maclaurin seriesMaclaurin's theorem states that if a function f(x) is infinitely differentiable at x = 0, then it can be represented as a power series: f(x) = f(0) + xf(0) + x22! f(0) + … + xnn! f(n)(0) + …n→∞INFOMaclaurin Series
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series149Let f be a function that can be expressed as a power series in x,i.e. f(x) = c0 + c1x + c2x2 + … + cnxn + …where x is a variableand cn are the coefficients to be determined.Then f(x) can be differentiated term by term and f(n)(0) exists for every positive integer n.We have f(x) = c0 + c1x + c2x2 + … + cnxn + …Substituting x with 0,f(0) = c0\\ c0 = f(0)Differentiating f(x) and substituting x with 0,f9(x) = c1 + 2c2x + 3c3x2 + 4c4x3 + … + ncnxn – 1 + f9(0) = c1\\ c1 = f9(0)Differentiating f9(x) and substituting x with 0,f0(x) = 2c2 + 6c3x + 12c4x2 + … + n(n – 1)cnxn – 2f0(0) = 2c2\\ c2 = f0(0)(2)(1) = f0(0)2!Differentiating f0(x) and substituting x with 0, f -(0) = 6c3\\ c3 = f-(0)6 = f-(0)(3)(2)(1) = f-(0)3!By performing further differentiation and substituting x = 0,we have cn = f (n)(0)(n)(n – 1)(n – 2)(n – 3) …1 = f (n)(0)n! AHence, f(x) = f(0) + f9(0)x + f0(0)2!x2 + … + f (n)(0)n!xn + …The infinite series is known as the Maclaurin series of the function f. If f(x) is undefined at x = 0, i.e. f(0) = ∞ or –∞, then f(xNotice that 1x , ln x and cot x cannot be represented by Maclaurin series because they are undefined at x = 0.
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series150Chapter 11 Example 1Find the Maclaurin series for ex.Solution:Let f(x) = exThen, f9(x) = ex,f0(x) = ex, …In general, f(n)(x) = ex, n > 1Thus f(0) = 1, f9(0) = 1, f0(0) = 1, f-(0) = 1, …The Maclaurin series for ex isex = f(0) + f9(0) x + f0(0)2!x2 + f -(0)3!x3 + …= 1 + x + x22! + x33! + ··· It can be shown, by a convergence test called the ratio test, that series converges for all real values of The interval of convergence of the series is (–∞, ∞).Example 2Find the Maclaurin series for sin x.Solution:Let f(x) = sin x Then, f9(x) = cos x, f0(x) = –sin x, f-(x) = –cos x, …Thus f(0) = 0, f9(0) = 1, f0(0) = 0, f-(0) = –1, …The Maclaurin series for sin x issin x = f(0) + f9(0) x + f0(0)2!x2 + f -(0)3!x3 + … = x – x33! + x55! – x77! + …The interval of convergence is (–∞, ∞).INFO
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series151Example 3Find the Maclaurin series for ln (1 + x).Solution:Let f(x) = ln (1 + x)Then, f9(x) = 11 + x,f0(x) = – 1(1 + x)2 ,f-(x) = 2(1 + x)3 , AThus f(0) = 0, f9(0) = 1, f0(0) = –1, f -(0) = 2, …The Maclaurin series for ln (1 + x) isln (1 + x) = f(0) + f9(0) x + f0(0)2!x2 + f-(0)3!x3 + …= x – x22 + x33 – …The interval of convergence is (–1, 1].Example 4Find the Maclaurin series for (1 + , .Solution:Let f(x) = (1 + x)kThen f(x) = k(1 + x)k – 1f(x) = k(k – 1)(1 + x)k – 2f-(x) = k(k – 1)(k – 2)(1 + x)k – 3AThus, f(0) = 1, f(0) = k, f(0) = k(k – 1), f-(0) = k)(k – 1)(k – 2), …The Maclaurin series for (1 + x)k is(1 + x)k = 1 + kx + k(k – 1)2!x2 + k(k – 1)(k – 2)3!x3 + …The interval of convergence is (–1, 1).This series is called the binomial series.If k ∈ N, then the series is a finite series with the interval of convergence (–∞, ∞)
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series152Chapter 11 Example 5If y = tan 1x + p4 2, show that d2ydx2 = 2ydydx .Hence, find the Maclaurin series for tan 1x + p4 2 up to the term in x3.Solution:y = tan 1x + p4 2dydx = sec21x + p4 2 = 1 + yxd2ydx2 = 2ydydxd3ydx3 = 2yd2ydx2 + 21dydx 22At x = 0, y = 1, dydx = 2, d2ydx2 = 4, d3ydx3 = 16.The Maclaurin series for tan 1x + p4 2 istan 1x + p4 2 = 1 + 2x + 41x22! 2 + 161x33! 2 + …= 1 + 2x + 2x2 + 83x3 + …Example 6If y = esin–1x, show that (1 – x2) d2ydx2 = xdydx + y.Hence, find esin–1x up to the term in x3.Solution:y = esin–1xdydx = esin–1x1 11 – x2 2= y1 – x2d2ydx2 = 1 – x2 dydx – y112 2 –2x1 – x21 – x2(1 – x2) d2ydx2 = y + xdydx = xdydx + y(1 – x2) d3ydx3 – 2xd2ydx2 = xd2ydx2 + dydx + dydx(1 – x2) d3ydx3 = 3xd2ydx2 + 2 dydx
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series153At x = 0, y = 1, dydx = 1, d2ydx2 = 1, d3ydx3 = 2The Maclaurin series for esin–1x isesin–1x = 1 + x + x22! + 21x33! 2 + …= 1 + x + 12x2 + 13x3 + …Standard Maclaurin series Some important and their Maclaurin series are given in the table below.Function Maclaurin series Sigma notation Interval of convergence(1 + x)n1 + nx + n(n – 1)2!x2 +n(n – 1)(n – 2)3!x3 + … – (–1, 1)ln (1 + x) x – x22+ x33 – x44+ … ∞nΣ = 1(–1)n + 1nxn (–1, 1] x 22! + … ∞nΣ = 0n! sin x x – x33!+ x55! – x77!+ … ∞nΣ = 0(–1)n(2n + 1)! x2n + 1 (–∞, ∞)cos x 1 – x22!+ x44! – x66!+ … ∞nΣ = 0(–1)n(2n)! x2n (–∞, ∞)VIDEOMaclaurin series
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series154 Geometrical approximation by Maclaurin polynomialsThe geometrical approximation to certain functions by the first few terms of their Maclaurin series called Maclaurin polynomials are shown in the following diagrams.1. f(x) = ex and f(x) = 1 + x + 12!x2 + 13!x3 + 14!x4 + …(a)x22 244 468– 0– 2–f(x) = exf(x) = 1 + xf(x) (b)x22 244 4680 ––2–12!x2f(x) = exf(x) = 1 + x +f(x) Figure 11.1(a) Figure 11.1(b)(d)x22 24 4680 ––2–f(x) = exf(x) = 1 + x + 12! x2 f(x)22 24 4680 –– 2f(x) = 1 + x + 12!13! x2 + + x3 xf(x) = exf(x) Figure 11.1(c) Figure 11.1(d)2. f(x) = sin x and f(x) = x – 13!x3 + 15!x5 – 7!x7 + 19!x9 – …(a) (b) Figure 11.2(a) Figure 11.2(b)f(x) = sin xf(x) = x222 1 113 336 5 4 54 6 0 – – –1 ––––13!x 3 –xf(x)f(x) = sin xf(x) = x – — x3 + — 222 1 113 336 5 4 54 6 0 – – –1 –––– f(x)1113! 5
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series155(c)f(x) = x – – 13!x3 15!17!19! + x + 5 x9 x7f(x) = sin xx 222 1 113 336 5 4 54 6 0 – ––1––––f(x) Figure 11.2(c)It is noticed that when more terms of a series are used, the approximation becomes better. Each series tends to provide its best approximation when x is close to 0. The set of values of x for which a Maclaurin series can be made to fit a function, provided that enough terms are taken, is called the interval of validity of the series. In general, the interval of validity is the same as the interval of convergence.Series expansions of the composites of functionsBased on the standard Maclaurin series, we are able to generate the Maclaurin series for functions which are the composites of common functions.Example 7Find the first three terms of the Maclaurin expansion of each of the following functions. State the set of values of x for which the series expansion is valid in each case.(a) 14 – x2 (b) x22 + xSolution:11 + x = (1 + x)−1 = 1 – x + (−1)(−2)2!x2 − …= 1 – x + x2 – …\\ 11 – x = 1 + x + x2 + …(a) 14 – x2 = 14 1 11 – x24 2 = 14 11 + 1x24 2 + 1x24 22 + …2 = 14 11 + x24 + x416 + …2 = 14 + 116x2 + 164x4 + …The series expansion is valid for –1 , x24, 1, i.e. –2 , x , 2.
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series156Chapter 11 (b) x32 + x = x31 12 + x2 = x31 121 + x2 2 = x32 1 11 + x2 2 11 + x = 1 – x + x2 – … \\ 11 + x2= 1 – 112x2 + 112x22 – …= 1 – 12x + 14x2 – … \\ 12x3 1 11 + x2 2 = 12x311 – 12x + 14x2 – …2 \\ x32 + x = 12x3 – 14x4 + 18x5 – …The series expansion is valid for –1 , x2, 1, i.e. –2 , x , 2.Example 8Find the first four terms of the Maclaurin series for(a) e2x, (b) ex cos x.Solution:(a) ex = 1 + x + 12!x2 + 13!x3 + …Replace x in ex with 2x,e2x = 1 + (2x) + 12! (2x)2 + 13! (2x)3 + …= 1 + 2x + 2x2 + 43x3 + … (b) ex = 1 + x + 12!x2 + 13!x3 + …Replace x in ex with x cos x,ex cos x = 1 + (x cos x) + 12! (x cos x)2 + 13! (x cos x)3 + …Now replace cos x with its series expansion,ex cos x = 1 + x(1 – 12!x2 + …) + 12!x2(1 – 12!x2 + …)2 + 13!x3(1 – 12!x2 + …)3 + …= 1 + x – 12x3 + 12x2 + 16x3 + …= 1 + x + 12x2 – 13x3 + …
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series157Exercise 11.11. Using repeated differentiation, find the Maclaurin series for(a) 11 – x(b) (1 + x)n2. Using repeated differentiation, find the Maclaurin series for cos x.3. If y = ln (x2 + 2), find dydx and show that d2ydx2 = – 2x2 – 4(x2 + 2)2 . Hence, find the first two terms of the Maclaurin series for ln (x2 + 2). 4. If y = ex (1 + sin x), find dydx and show that d2ydx2 = ex (1 + 2 cos x). Hence, find the first two terms of the Maclaurin series for ex (1 + sin x).5. Given that y = e–sin 2x, show that d2ydx2 = –2 cos 2xdydx + 4y sin 2x. By further differentiation, obtain the Maclaurin series for e–sin 2x up to the term in x4.6. Given that y = cos–—12 x, show that d2ydx2 = – 14 1y + 1y3 2. By further differentiation, obtain the series expansion of in ascending powers of x up to the term in 7. If y = tan–1 x, show that d2ydx2 = –2x1dydx 22. Hence, find the Maclaurin series for tan–1 x up to the term in 8. If y = sin–1 x, show that d2ydx2 = x1dydx . Hence, find the Maclaurin series for sin–1x up to the term in x3.9. Use the Maclaurin series for (1 + x)n to find the Maclaurin series for 11 – x3 up to the term in interval of convergence.10. Using the Maclaurin series for ln (1 + x), find the Maclaurin series for ln (1 + 2x4) up to the term in x12. State the interval of convergence.11. Using the Maclaurin series for ex, find the Maclaurin series for e–2x2 up to the term in x6.12. Using the Maclaurin series for sin x, find the Maclaurin series for sin x4 up to the term in x12.
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series158Chapter 11 Series expansions of the sums, differences, products and quotients of functionsWe may use the series expansions of functions to find the series expansions of their sums, differences, products and quotients.Example 9Find the Maclaurin series for ex + e–3 up to the term in x3. State the interval of convergence of the series.Solution:ex = 1 + x + 12!x2 + 13!x3 + …\\ e–3x = 1 + (–3x) + 12! (–3x)2 + 13! (–3x)3 + …= 1 – 3x + 92x2 – 92x3 + … \\ ex + e–3x = (1 + x + 12x2 + 16x3 + …) + (1 – 3x + 92x2 – 92x3 + …) = 2 – 2x + 5x2 – 133x3 + …The interval of convergence of the series is (–∞, ∞)Example 10Expand ln 11 – 2x1 + 2x 2 as a series of ascending powers of x up to the term in x4. Find the range of values of x for which the expansion is valid.Solution:ln (1 + x) = x – x22 + x33 – x44 + …\\ ln (1 – 2x) = (–2x) – 12 (–2x)2 + 13 (–2x)3 – 14 (–2x)4 + …= –2x – 2x2 – 83x3 – 4x4 – … for – 12< x , 12ln (1 + 2x) = 2x – 12 (2x)2 + 13 (2x)3 – 14 (2x)4 + …= 2x – 2x2 + 83x3 – 4x4 + … for – 12, x < 12\\ ln 11 – 2x1 + 2x 2 = ln(1 – 2x) – ln (1 + 2x)= 1–2x – 2x2 – 83x3 – 4x4 – …2 – 12x – 2x2 + 83x3 – 4x4 + …2 = –4x – 163x3 – ···provided that – 12< x , 12 and – 12, x < 12i.e. – 12, x , 12The range of values of x for which the expansion is valid is 1– 12 , 12 2.
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series159Example 11Find the series expansion of e2x sin 2x in ascending powers of x up to the term in x3.Solution:ex = 1 + x + 12!x2 + 13!x3 + … \\ e2x = 1 + 2x + 12! (2x)2 + 13! (2x)3 + …= 1 + 2x + 2x2 + 43x3 + …sin x = x – 13!x3 + …sin 2x = 2x – 13! (2x)3 + …= 2x – 43x3 + …Therefore, e2x sin 2x = 11 + 2x + 2x2 + 43x3 + …212x – 43x3 + …2 = 2x – 43x3 + 4x2 + 4x3 + … = 2x + 4x2 – 83x3 + …Example 12Find the Maclaurin series for tan x up to the term in x5Solution:tan x = sin xcos x= x – 13!x3 + 15!x5 – …1 – 12!x2 + 14!x4 – …x + 13x3 + 215x5 + …1 – 12x2 + 124x4 – ··· x – 16x3 + 1120x5 – …x – 12x3 + 124x5 – … 13x3 – 130x5 + … 13x3 – 16x5 + … 215x5 + …Therefore, tan x = x + 13x3 + 215x5 + …
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series160Chapter 11 Differentiation and integration of power seriesWe know that, for a finite sum, the derivative of a sum is the sum of the derivatives and the integral of a sum is the sum of the integrals. The same is true for an infinite sum, provided that the infinite sum is a power series.Example 13Write the first three terms of the Maclaurin series for 11 – x. Hence find the series expansion for 1(1 – Solution:11 + x = 1 – x + x2 – …11 – x = 1 + x + x2 + …Differentiating both sides of the equation with respect to x. 1(1 – x)2 = 1 + 2x + 3x2 + …The interval of convergence is (–1, 1).Example 14Using the result that ∫ 0x 11 + t2 dxSolution:Let y = tan–1 x\\ tan y = xDifferentiating with respect to x,sec2 y · dydx = 1dydx = 11 + tan2 ydydx = 11 + x2We have11 + x = 1 – x + x2 – x3 + … for –1 , x , 1 \\ 11 + x2 = 1 – (x2) + (x2)2 – (x2)3 + …= 1 – x2 + x4 – x6 + …INFO
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series161Integrating both sides between 0 and x, where x lies inside the interval of convergence,tan–1 x = ∫ 0x(1 – x2 + x4 – x6 + …) dx= 3x – 13x3 + 15x5 – 17x7 + …40x= x – 13x3 + 15x5 – 17x7 + …The interval of convergene of a series obtained by integration may include more values than the original series. Here, the interval of convergence of the series for tan1 x is [–1, 1]Exercise 11.21. Find the first three terms of the Maclaurin series for. (a) ln (1 + x)(1 – x) (b) ln 11 + x1 – x 22. Find the first three terms of the Maclaurin series for(a) sinh x = 12 (ex – e–x) (b) cosh x = 12 (ex + e–x)3. Find the first four terms of the Maclaurin series for(a) e–x2cos x (b) (x + 1) sin x4. Find the first four terms of the Maclaurin series for(a) x3x + 2 (b) sin xex5. Expand sin2 x as a series of ascending powers of 6. Find the Maclaurin series of 3 cos x + 1cos x + 3 . State the range of values of 7. Given that the Maclaurin series for tan x is x + x33 + 215x5 + …, find the Maclaurin series for secup to the term in x4.8. State the first four terms of the Maclaurin series for 11 + x. Hence, find the first four terms of the Maclaurin series for ln (1 + x).9. The sine integral function Si(x) is defined by Si(x) = ∫ 0x sin tt dt. Find the Maclaurin series for Si(to the term in x5.
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series162Chapter 11 Applications of Maclaurin series 11.2Learning outcomes:(a) Use Maclaurin series to evaluate lim f(x)g(x), where f(0) = g(0) = 0.(b) Use Maclaurin series to approximate definite integrals.x→0Evaluating the limit of a functionA useful application of Maclaurin series is to evaluate the limits of functions of the form f(x)g(x) as x → 0 where f(0) and g(0) are both 0.Example 15Evaluate(a) limx → 01 – ex3x3 (b) limx → 0x – sin (sin x)3x3Solution:(a) ex = 1 + x + x22! + …ex3= 1 + x3 + x62 + …limx → 01 – ex3x3 = limx → 01 – 11 + x3 + x62 + …2x6 = limx → 0–x3 – x62 – …x3 = limx → 0 –1 – x32 – …= –1(b) sin x = x – 13!x3 + 15!x5 – …\\ sin (sin x) = 1x – 13!x3 + 15!x52 – 1x – 13!x3 + 15!x5233! + 1x – 13!x3 + 15!x5255! – …= 1x – 13!x3 + 15!x52 – 113!x3 – 33!3!x52 + 15!x5 – …= x – 13x3 + 110x5 – …limx → 0x – sin (sin x)3x3 = limx → 0x – 1x – 13x3 + 110x5 – …23x3= limx → 013x3 – 110x5 + … 3x3= limx → 013 – 110x2 + …3= 19
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series163Example 16Evaluate(a) limx → 01 – cos x2 sin2 x (b) limx → 0(1 + x) In (1 + x)ex – 1Solution:(a) limx → 01 – cos x2 sin2 x = limx → 01 – 11 – 12!x2 + 14!x4 – …221x – 13!x3 + 15!x5 – …2 = limx → 012!x2 – 14!x4 + …2x2 – 43!x4 + … = limx → 012! – x24!+ …2 – 43!x2 + … = 14(b) limx → 0(1 + x) In (1 + x)ex – 1 = limx → 0(1 + x) 1x – x22+ …211 + x + x22!+ …2 – 1 = limx → 0x – x22+ x2 – …x + x22+ … = limx → 0x + x22 – …x + x22+ … = limx → 01 + x2 – …1 + x2+ … = 1Estimating the value of a definite integralThere are functions that we are unable to find their antiderivatives. If they can be represented by Maclaurin series, we can integrate the Maclaurin series to approximate integrals.
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series164Chapter 11 Example 17Using the Maclaurin series for sin x, find the first three terms of the series expansion of sin2 xx .Hence, estimate the value of the definite integral ∫ 0—π6 sin2 xx dx, correct to three decimal places.Solution:sin x = x – 16x3 + 1120x5 – …\\ sin2 x = 1x – 16x3 + 1120x5 – ···21x – 16x3 + 1120x5 – …2 = x2 – 16x4 + 1120x6 – 16x4 + 136x6 + 1120x6 – …= x2 – 13x4 + 245x6 – …\\ sin2 xx = x – 13x3 + 245x5 – …∫ 0—π6 sin2 xx dx ≈ ∫ 0—π61x – 13x3 + 245x52 dx≈ 312x2 – 112x4 + 1135x640—π6≈ 12 1π6 22 – 112 1π6 24 + 1135 1π6 26≈ 0.131Example 18Using standard Maclaurin series, find the series expansion of In (cos 2x) in ascending power of x, up to the term in x4.Hence, evaluate ∫ 00.3In (cos 2x) dx, giving your answer correct to three decimal places.Solution:cos x = 1 – x22! + x44! – …\\ cos 2x = 1 – (2x)22! + (2x)44! – …= 1 + 3–2x2 + 23x4 – …4
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series165In (1 + x) = x – x22 + x23 – x24 + …\\ In (cos 2x) = In 31 + 1–2x2 + 23x4 – …24 = 1–2x2 + 23x42 – 12 1–2x2 + 23x422 + … = –2x2 + 23x4 – 12(4x4) + … = –2x2 – 43x4∫ 00.3In (cos 2x) dx = ∫ 03 1–2x2 – 43x42 dx= 3– 23x3 – 415x5400.3= – 23(0.3)3 – 415(0.3)5= –0.018648∫ 00.3In (cos 2x) dx = –0.019Exercise 11.31. Using the Maclaurin series for ln(1 + x), evaluate limx → 0ln(1 + x) – x3x2 .2. Using the Maclaurin series for cos 4x, find the value of limx → cos 4x + 8 3. Use the Maclaurin series for ex and esin x to evaluate limx → 0 1ex – esin xx3 2.4. Use the Maclaurin series for ln (cos x) to evaluate limx → 0x2 + 2 In (cos x)x4 .5. Using the Maclaurin series for ex and cos x, evaluate limx → 0x2excos x – 1 .6. Using the Maclaurin series for sin x and e2x, evaluate limx → 0 13 sin xe2x – 1 2.7. Using the Maclaurin series for cos x up to the term in x6, evaluate ∫ 01 cos x dx, giving your answer correct to three decimal places.8. Use the Maclaurin series for ln (1 + x4) up to the term in x8 to evaluate ∫ 00.4 ln (1 + x4) dx, giving your answer correct to three decimal places.9. Using the Maclaurin series for cos x2 up to the term in x5, evaluate ∫ 01 x cos x2 dx, giving your answer correct to three decimal places.INFO
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series166Chapter 11 Summary1. A power series is an infinite series in ascending powers of x in the form ∞nΣ = 0cnxn = c0 + c1x + c2x2 + c3x3 + ··· + cnxn + … where x is a variable and cn are the coefficients of the series.2. A power series of the form ∞nΣ = 0cnxn = c0 + c1x + c2x2 + c3x3 + … + cnxn + … always converges at 3. Maclaurin series for the function f is given by f(x) = ∞nΣ = 0f (n)(0)n! xn = f(0) + f9(0)x + f0(0)2!x2 + … + f (n)(0)n!xn + …4. The ranges of values of the single variable for which a Maclaurin series converges is known as the interval of convergence of the series.5. Standard Maclaurin series and their intervals of convergence are as follows.Function Maclaurin series Interval of convergence(1 + x)n 1 + nx +n(n – 1)2! x2 +n(n – 1)(n – 2)3! x3 + … (–1, 1)ln (1 + x – x2+x33 – x4+ … ex 1 + x +x22! +x33! + … (–∞, ∞)sin x x – x33! +x55! – x77! + … (–∞, ∞)cos x 1 – x22! +x44! – x66! + … (–∞, ∞)6. Maclaurin series are useful in evaluating limx → 0f(x)g(x) where f(0) = g(0) = 0.7. Maclaurin series are useful in approximating definite integralsINFOGRAPHIC
Mathematics Semester 2 STPM Chapter 11 Maclaurin Series167STPM PRACTICE 111. Using Maclaurin's theorem, obtain the first three terms of the Maclaurin series for 4 + sin2 x .2. Using repeated differentiation, find the first three terms of the Maclaurin series for tan2 x.3. Given that y = (sin–1 x)2, show that (1 – x2) d2ydx2 – x dydx = 2.Using further differentiation, find the series expansion of (sin–1 x)2 in ascending powers of x up to the term in x4. State the range of values of x for which the expansion is valid.4. If y = etan–1 x, show that dydx = y1 + x2 . Hence, find the Maclaurin series for etan–1 x up to the term in x2.5. Using the Maclaurin series for ex and sin x, find the Maclaurin series of e−2x sin 3x up to the term in 6. The first two terms in the series expansion, in ascending powers of x, of (1 + 43x)n are equal to the first two terms of the Maclaurin series for ex (1 + sin 2x). Find the value of n.7. If y = tan x, show that d2ydx2 = 2ydydx and find the Maclaurin series for tan x up to the term in Hence, find the series expansion of e–3x tan x up to the term in x3. Using repeated differentiation, show that 1(1 + x)2 = 1 – 2x + 3x2 – 4x3 + … and state the internal of convergence. Hence, find the first three terms of the Maclaurin series for 1(1 + x)3 .9. Using the result sin–1 x = ∫ 0x 11 – t2 dt, find the first three terms of the Maclaurin series for sin–1 By taking x = 12, find the approximate value of p, giving your answer correct to three decimal places.10. Using the Maclaurin series for ln (1 – x), evaluate limx → 0x + ln (1 – x)x2 .11. Use the Maclaurin series for e2x to evaluate limx → 0 1 –xe2xe2x – 1 2. 12. Using the Maclaurin theorem, find the series expansion of ln (1 + tan x) up to the term in x3. Hence, evaluate limx → 03x – ln (1 + tan x)36x2 – x4 .
Mathematics Semester 2 STPM Answers2057. x + 1 2 x3 + … ; x – 3x2 + 29 6 x3 + …8. (–1, 1) 1 – 3x + 6x2 + …9. x2 + 1 6 x3 + 3 40 x5 + … 10. – 1 211. – 1 212. x – 1 2 x2 + 2 3 x3 + … ; 1 413. 0.74314. –x + 7 2 x2 – 31 3 x3 + … ; – 1 2, x , 1 2–0.014815. 1 2 x2 + 1 12 x4 0.011ChapterNumerical Methods 12Exercise 12.11.y = –x3yy = 16 – ORoot One intersection means x3 − x2 + 16 = 0 has one root only. Let f(x) = x3 – x2 + 16, f(–3) and f(–2) have different signs, the negative root lies between –3 and –2.2.47ORoot3.5xyRooty = 7 – 2xy = 4e–x Since there are two intersections, 4e−x + 2x − 7 = 0 has two real roots. Let f(x) = 4e−x + 2x − 7, f(–1) and f(0) have different signs, the negative root lies between –1 and 0. f(3) and f(4) have different signs, the positive root lies between 3 and 4.3.O 2 3xyy = In (x – 2) y = 3xRoot Since there is only one intersection, x In (x – 2) – 3 = 0 has one real root only. Let f(x) = x In (x – 2) – 3, f(4) and f(5) have different signs, the real root lies in the interval [4, 5]4.ππ 2π1y–1O x π2x2 y = π –y = cos xRoot Since there is only one intersection, x2 + cos x – p = 0 has only one root in the interval [0, 2p]. Let f(x) = x2 + cos x – p, f(5) and f(6) have different signs, the root lies between 5 and 6.5.–π –1 π–1xyπ 12 – π2y = x2 – 1RootORooty = sin 2x Two intersections means x2 – sin 2x – 1 = 0 has two roots, one positive root and one negative root Let f(x) = x2 – sin 2x – 1, f(1) and f(2) have different signs, the positive root lies between 1 and 2.Exercise 12.21. 2.467 ; xo = 2 is not a suitable first approximation to α because f(2) = 12(2)2 – 24(2) = 0 and f(2) f(2) = 0 is undefined.2. 0.65333. 1.5354. 0.31365. 0.8354
W.M: RM34.95 / E.M: RM36.95AAEVMM2675031A2IISBN 978-629-498-780-7Purchaseebook here!Strategic STPM Mathematics Semester 2is written based on the purified STPM Examination Syllabus by the Malaysian Examinations Council (MEC) and will be implemented starting from STPM Semester 1in 2026. This book is carefully designed and well-organised, containing the following features to enhance students’ understanding of the concepts being studied.Strategic STPM Pre-U Text Titles:› Pengajian Am Semester 2 › Bahasa Melayu Semester 2 › Biology Semester 2 › Physics Semester 2 › Chemistry Semester 2 › Mathematics Semester 2 › Sejarah Semester 2 › Geografi Semester 2 › Ekonomi Semester 2 › Pengajian Perniagaan Semester 2Also available:› MUET My Way MATHEMATICS(Calculus)Pre-U TextStrategicSTPMSEMESTER2REVISION★ Comprehensive Notes★ Learning Outcomes★ Examples & SolutionsREINFORCEMENT & ASSESSMENT★ Exercises★ STPM Practices★ STPM Model Paper Semester 2★ AnswersEXTRA FEATURES ★ Bilingual Keywords★ Summary★ Digital Resources QR CODEFEATURES