146 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 When t = 0, let x = A ⇒ C = ln A \ lnx = –kt + ln A ln x A = –kt x A = e–kt \ The solution of the differential equation is x = Ae–kt. (b) When t = 0, A = D ⇒ x = De–kt When t = 1, x = 3 4 D ⇒ 3 4 D = De–k e–k = 3 4 The solution of the differential equation x = De–kt can be written as x = D(e–k ) t . \ x = D1 3 4 2 t (c) After 2 hours: The first dose of drug administered at time t = 0 will reduce to the amount of D1 3 4 2 2 . The second dose of drug administered at time t = 1 will reduce to the amount of D1 3 4 2 1 . The third dose of drug administered at time t = 2 has amount D. \ Immediately after 2 hours the total amount of drug remaining is D + D1 3 4 2 + D1 3 4 2 2 = 2 5 16 D (d) Let the amount of drug remaining after an indefinite period be S. S = D + D1 3 4 2 + D1 3 4 2 2 + D1 3 4 2 3 + D1 3 4 2 4 + … S = D11 + 3 4 + 1 3 4 2 2 + 1 3 4 2 3 + …2 S = D1 1 1 – 3 4 2 \ S = 4D
147 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 Exercise 4.5 1. The height, h of a type of tree increases proportionately to the difference between its height at that time, t and its final height, H. Find the differential equation connecting the variables h and t. If h = 1 10H when t = 0 and h = 1 5 H at t = T, solve the differential equation. If time t is measured in days and T = 3, find the number of days passed before the height is more than 9 10H. 2. The manufacturer of a branded shampoo launches an advertisement program which results in the number of consumers, n at time t to increase at a rate proportional to the square root of n. Write down the differential equation which describes this relationship between n and t. If n = N when t = 0 and n = 9 4 N when t = T, solve the differential equation. Find t when n = 4N. 3. Under certain conditions, the rate of cooling of a liquid is proportional to the difference between its temperature and its surrounding temperature. The liquid is placed in a room of temperature 30°C and the temperature of the liquid at time t (minutes) is x. Form a differential equation to describe this phenomena. If the liquid cools from 100°C to 70°C in 8 minutes, find the further time taken for the liquid to reach a temperature of 31°C. Find also the temperature of the liquid after it has been in the room for 10 minutes. 4. In a community, the number of people, n infected by a certain disease at a certain time increases at a rate of λn. Form a differential equation connecting n and t. Initially, the number of patients is N. Show that at day t, n = Neλt . Find the value of λ if n = 2N when t = 10. Find also (a) the value of n when t = 20, (b) the value of t when n = 3N. 5. In a certain chemical process, substance X is continuously changed to form substance Y. The total mass of X and Y at any time is the same and equals to M. The rate of increase of Y at time t is proportional to the mass of X at that time. If the mass of Y at time t is m, form a differential equation that describes this chemical reaction. If M = 100 grams and 60 grams of substance X remains after 2 minutes, find the mass of Y formed in 6 minutes. 6. At time t minutes, the number of micro-organisms in a liquid of a certain colony is x. The rate of increase of x resulting from natural growth is αx and the rate of decrease due to death is β. Write down the differential equation that describes the change in the size of the colony. If the number of organisms is n0 at the beginning, (a) find the time taken for all the micro-organisms in the colony to extinct if n0 = 200, α = 2 and β = 500, (b) find the number of micro-organisms in the colony after half a minute if n0 = 200, α = 4 and β = 500. 7. At time t, the volume of water in a container is v. Due to leakage, water flows out at a rate of kv where k is a positive constant. The rate at which water is lost due to evaporation is c. Obtain a differential equation that describes the decrease in the volume of water. Given that v = V when t = 0, find the volume of water in the container after t minutes. Find also the time taken before the container is empty.
148 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 8. The rate of decomposition of a radioactive substance is proportional to the mass of the remaining substance at that time. If two-thirds of the initial mass remains after 100 days, find the time taken for one-half of the initial mass to remain. Find the percentage of the mass that remains after 120 days. 9. Heat is supplied to an electric kettle at a constant rate of 2 000 watts whereas heat is lost to the surroundings at a rate of 20 watts for every Celsius degree difference between the temperature of the kettle and the surrounding temperature. One watt of heat results in an increase in the temperature of the kettle at a rate of 1 50°C per minute. If the surrounding temperature is 15°C and θ°C is the temperature of the kettle after t minutes, show that dθ dt = 40 – 2 5 (θ – 15). How long does it take for the temperature of the kettle to increase from 15°C to 100°C? 10. In a chemical reaction, two substances A and B react together to form another substance C. At time t, the amount of A and B are a – x and b – x respectively, where a and b are constants and x is a function of t. The value of x is zero when t = 0. At any time, the rate of decrease of A is proportional to the product of the amount of A and B at that time. Find a differential equation involving x and t and solve x as a function of t in each case of (i) a = b, (ii) a b. 11. A water tank is in the shape of a cylinder with a vertical axis. The base area is A m2 . Initially, the tank is empty. Starting at time t = 0, water is poured into the tank at a constant rate of w m3 s–1, and water flows out from a small hole at the base at a rate of kx m3 s–1, where k is a constant and x m is the depth of water in the tank. Form a differential equation and solve it to obtain a function of x in terms of t. Show that no matter how long this process is repeated, the depth of the water will not exceed w k m and if the time taken to reach one-half of the depth of the cylinder is T s, then k = A T ln 2. 12. An object moves along a straight line and passes a fixed point O with velocity u in the positive direction of the x-axis. At time t the object is at a displacement x from O and the velocity of the object is v. The rate of change of velocity has magnitude c v , where c is a constant, and is directed to towards the fixed point O. (a) Write down a differential equation for the motion of the object involving the velocity v and the time t. Hence, find the velocity v as a function of the time t. (b) Show that dv dt = v dv dx and hence, write down the differential equation for the motion of the object that involves the change of velocity v with respect to the displacement x and the velocity v. Hence, find the velocity v as function of the displacement x. Hence show that after a time t and the object has moved a distance x, 3cx = u3 – (u2 – 2ct) —2 3 and deduce that 2ct , u2 . 13. A research has been set up on an island to study a particular species of turtle. Initially there are 25 turtles on the island. After t years the number of turtles x satisfies the differential equation dx dt = 1 20k x(k – x) where k is a constant. (a) Show that k = 100 if it is known that the rate of growth is 0.45 turtle per year when x = 10. (b) What is the maximum rate of growth?
149 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 (c) Explain what happens as x approaches 100? (d) Obtain the solution of the differential equation and sketch the solution curve. (e) Find (i) the number of turtles after 30 years, (ii) the time for which the number of turtles is 50. 14. The rate of destruction of a particular drug is proportional to the amount of drug present in the body. (a) If x is the amount of drug at time, t, write a differential equation relating the amount of drug and the time and hence, solve the differential equation. (b) Initially an amount, Q, of the drug is injected to a body and after a time t = 1 hour the amount of the drug remaining is 2 3 Q. Obtain an expression for x in terms of Q and t. (c) The drug is injected again to the body after t = 1 hour, t = 2 hours and t = 3 hours. Find that the amount of drug remaining in the body immediately after 3 hours. (d) If the drug is administered at regular intervals of 1 hour for an indefinite period, find the amount of drug remaining in the body. Summary 1. A differential equation is an equation involving at least one derivative of y with respect to x, e.g. dy dx , d2 y dx2 , …, dn y dxn , where x is the independent variable and y the dependent variable. 2. The order of a differential equation is the order of the highest derivative found in the differential equation. 3. The solution of a differential equation that contains of an arbitrary constant is a general solution. 4. The solution of a differential equation that satisfies certain conditions (initial or boundary conditions) is a particular solution. 5. If the differential equation dy dx = f(x, y) can be expressed as dy dx = u(x) . v(y), where u(x) and v(y) are functions of x and y, it is a differential equation with separable variables. 6. The linear differential equation dy dx + f(x) y = g(x), where f(x) and g(x) are functions of x, can be solved by multiplying both sides of the equation by an integrating factor e∫f(x)dx .
150 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 STPM PRACTICE 4 1. Solve the differential equation 2xy dy dx = x2 + 2y2 given that y = 0 when x = 1. 2. Find the solution of the differential equation x2 dy dx = 4ey that satisfies the condition y = 0 when x = 2. 3. Show that the substitution u = 1 y transforms the non-linear differential equation dy dx – y x = y2 into du dx – u x = –1. Solve this linear equation given that y = 2 when x = 1. 4. Differentiate y2 ex with respect to x. Find the particular solution of the differential equation 2yex dy dx + y2 ex = e2x if y = 0 when x = 1. 5. Using the substitution z = sin y, find the general solution of the differential equation dy dx + 1 x tan y = 1 x2 sec y. 6. Using the substitution y = vx, solve the differential equation y dy dx = 2x + y given that y = 2 when x = 2. 7. Show that the substitution y = vx transforms the differential equation x · dy dx = y + x cot 1 2y x 2 into x · dv dx = cot 2v. Hence, find the particular solution of the given differential equation for which y = 1 when x = 0. Express your answer in the form y = f(x). 8. In a study on the effectiveness of a type of insect poison, it was found that the rate of decrease of the insect population, y is given by dy dt = –1 10 1 + 5t 2, where t is the time taken in hours after the poison is administered. Initially, there are 50 insects. Find (a) the number of insects left 24 hours after the administration of poison, (b) the time taken to destroy half the insect population. 9. A cultured bacteria of a species multiply at a rate that is directly proportional to the number of cultured bacteria in the culture. If x is the number of bacteria in the culture at time t seconds, write down the differential equation that describes the growth of the bacteria. At the beginning of the experiment, there were 1 000 bacteria of a certain species. It was known that the cultured bacteria multiply at a rate of 1.5 times per hour. Find the number of bacteria in the culture after (a) 3 hours, (b) 5 hours. Find the time taken for the cultured bacteria to increase to five times the original number.
151 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 10. At a chemical reaction, there is x kg of chemical reagent X and y kg of chemical reagent Y at time t. Initially, there were 1 kg of X and 2 kg of Y. The variables of x and y satisfy the following differential equations. dx dt = –x2 y and dy dt = –xy2 . Find dy dx in terms of x and y and express y in terms of x. Hence, write a differential equation relating x and t. Express x in terms of t. 11. An object moves in a medium and experiences a resistance so much so that the velocity v decreases at a rate given by dv dt = –kv where k is a positive constant. If the initial velocity of the object is 100 m s–1 and its velocity is 40 m s–1 after 2 seconds, show that k = 1 2 ln 5 2 and find its velocity as a function of t. Hence, find the distance travelled by the object in the first 2 seconds. 12. The rate of growth of a certain animal population is directly proportional to the population. If x is the animal population at time t (measured in years), write down a differential equation relating x and t. Find the animal population in year 2000 based on the information given below. Number Year Population 1 1970 3 683 thousands 2 1980 4 453 thousands 13. A firm uses a computer software to control the work of the machines that produce certain electronic components. It is known that the time T, taken by the computer software increases with the number of machines used, N at a rate given by the equation dT dN = 1 + ln N. Given that the computer software takes one second to control 50 machines, find the time taken to control 100 machines. 14. A contagious disease spreads at a rate directly proportional to the product of the number of the population infected and the remaining population that is not infected. Initially, one-half of the population is infected and if the rate of infection is kept constant, the whole population will be infected in 24 days. Find the proportion of the population that will be infected after 12 days. [Hint: If the proportion of the population infected is x, then the proportion of the population not infected will be (1 – x).] 15. A tank contains 10 kg of sodium hydroxide in 1 000 litres of water. Water is continuously added to the tank at a rate of 5 litres min–1 so that the mixture is diluted evenly. At the same time, the solution flows out at the same rate. Initially, there were 10 grams of sodium hydroxide in every litre of water. How long will it take for the concentration of the solution to drop to 3.2 grams per litre? 16. Find the solution of the differential equation 3x · dy dx – 4y = 1 which satisfies the condition y = 0 when x = 1. Give your answer in the form y = f(x).
152 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 17. In a rabbit farm there are 500 rabbbits and two rabbits are infected with Myxomatosis, a devastating viral infection, in the month of April. The farm owner has decided to cull the rabbits if 20% of the population is infected. The rate of increase of the number of infected rabbits x at t days is given by the differential equations dx dt = kx(500 – x) where k is a constant. Assuming that no rabbits leave the farm during the outbreak, (a) show that x = 1000 498e–500kt + 2 , (b) if it is found that, after two days, there are ten infected rabbits, show that k = 1 1000 ln 249 49 . (c) determine the number of days before culling will be launched. 18. Find the general solution to the differential equation ln x dy dx = tan y x . 19. Using the substitution y = vx, show that the differential equation xy3 dy dx – x4 – y4 = 0 may be reduced to v3 x dv dx = 1. Hence, find the particular solution that satisfies y = 1 and x = 1. 20. The variables x and y, x . 0 and y . 0, satisfy the differential equation x dy dx = x3 + x2 y + y3 x2 + y2 . Show that the substitution y = ux transforms the given differential equation into the differential equation x du dx = 1 1 + u2 . Hence, find the solution of the differential equation for which y = 1 and x = 1. 21. By using the substitution y = vx transform the equation x dy dx = y + x tan1 y x 2 into x dv dx = tan v. Hence, find the solution of the given differential equation satisfying the condition y = π 2 when x = 1. Give your answer in the form y = f(x). 22. Solve the differential equation x · dy dx = 2x – y with the condition y = 2 when x = 3. Express your answer in the form of y = f(x). 23. The variables x and y are related by the differential equation dy dx = 2x + y – 1 2x + y + 5 . Show that the substitution V = 2x + y transforms the differential equation to dV dx = 3V + 9 V + 5 . Hence, find the particular solution of the differential equation given that y = 1 and x = 1. 24. Find the general solution of the differential equation cos x · dy dx – y sin x = 4 sin x cos x. 25. Find the general solution of the differential equation dy dx = y 2 – 9 6x . Express your answer in the form of y = f(x). 26. Find the particular solution of the differential equation x · dy dx – y = x2 (ln x) with the condition y = 3 when x = 1.
CHAPTER Learning Outcome (a) Find the Maclaurin series for a function and the interval of convergence. (b) Use standard series to find the series expansions of the sums, differences, products, quotients and composites of functions. (c) Perform differentiation and integration of a power series. (d) Use series expansions to find the limit of a function. convergence – menumpu Maclaurin series – siri Maclaurin Maclaurin's theorem – teorem Maclaurin power series – siri kuasa 5 MACLAURIN SERIES Bilingual Keywords
154 Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 5.1 Maclaurin Series Representations of functions as power series A power series is a series of the form ∞ n Σ = 0 cnxn = c0 + c1x + c2x2 + ··· + cnxn + ··· where x is a variable and cn are constants called the coefficients of the series. If all the coefficients of the power series are equal to 1, the power series becomes ∞ n Σ = 0 xn = 1 + x + x2 + ··· + xn + ··· This is an infinite geometric series which has first term 1 and common ratio x and has sum 1 1 – x provided –1 , x , 1. Thus 1 1 – x = 1 + x + x2 + ··· + xn + ··· , provided –1 , x , 1. The set of values of the single variable, x, for which the power series converges (i.e. the expansion is valid) is known as the interval of convergence of the series. This example shows that an infinite series in ascending power of x (power series) can be used to represent a certain function. In other words, a function may be expanded as a power series in a single variable. We can replace a function with its power series expansion as long as the single variable, x, is small and within the interval of convergence. The interval of convergence is the restriction on the values of single variable, x, in the power series so that the power series always converges. However, if the power series converges for all values of x, then there is no restriction on the values of x. In general, a power series of the form ∞ n Σ = 0 cnxn = c0 + c1x + c2x2 + ··· + cnxn + ··· always converges at x = 0. Maclaurin series Some functions may be represented by power series using Maclaurin's theorem. Maclaurin Series INFO
Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 155 Maclaurin's theorem Let 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 variable and 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 – 2 + ··· f0(0) = 2c2 \ c2 = f0(0) (2)(1) = f0(0) 2! Differentiating f0(x) and substituting x with 0, f-(x) = 6c3 + 24c4x + ··· + (n)(n – 1)(n – 2)cnxn – 3 + ··· 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! A Hence, f(x) = f(0) + f9(0)x + f0(0) 2! x2 + ··· + f (n) (0) n! xn + ··· This is known as Maclaurin's theorem. This 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(x) cannot be expressed as a Maclaurin series. Notice that 1 x , ln x and cot x cannot be represented by Maclaurin series because they are undefined at x = 0.
156 Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 Example 1 Find the Maclaurin series for ex . Solution: Let f(x) = ex Then, f9(x) = ex , f0(x) = ex , ··· In general, f(n) (x) = ex , n > 1 Thus f(0) = 1, f9(0) = 1, f0(0) = 1, ··· , f (n) (0) = 1, ··· The Maclaurin series for ex is ex = f(0) + f9(0) x + f0(0) 2! x2 + f-(0) 3! x3 + ··· = 1 + x + x2 2! + x3 3! + ··· It can be shown, by a convergence test called the ratio test, that series converges for all real values of x. The interval of convergence of the series is (–∞, ∞). Example 2 Find 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 is sin x = f(0) + f9(0) x + f0(0) 2! x2 + f -(0) 3! x3 + ··· = x – x3 3! + x5 5! – x7 7! + ··· = ∞ n Σ = 0 (–1)n (2n + 1)! x2n+1 The interval of convergence is (–∞, ∞). Example 3 Find the Maclaurin series for ln (1 + x). Solution: Let f(x) = ln (1 + x) Then, f9(x) = 1 1 + x , f0(x) = – 1 (1 + x) 2 , f -(x) = 2 (1 + x) 3 , ···
Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 157 Thus f(0) = 1, f9(0) = 1, f0(0) = –1, f -(0) = 2, f(4)(0) = –6, ··· The Maclaurin series for ln (1 + x) is ln (1 + x) = f(0) + f9(0) x + f0(0) 2! x2 + f -(0) 3! x3 + f (4)(0) 4! x4 + ··· = x – x2 2 + x3 3 – x4 4 + ··· = ∞ n Σ = 1 (–1)n + 1 n xn The interval of convergence is (–1, 1]. Standard Maclaurin series Some common functions and their Maclaurin series Function Maclaurin series Sigma notation form Interval of convergence ex 1 + x + x2 2! + x3 3! + ··· ∞ n Σ = 0 1 n! xn (–∞, ∞) sin x x – x3 3! + x5 5! – x7 7! + ··· ∞ n Σ = 0 (–1)n (2n + 1)! x2n + 1 (–∞, ∞) cos x 1 – x2 2! + x4 4! – x6 6! + ··· ∞ n Σ = 0 (–1)n (2n)! x2n (–∞, ∞) ln (1 + x) x – x2 2 + x3 3 – x4 4 + ··· ∞ n Σ = 1 (–1)n + 1 n xn (–1, 1] 1 1 – x 1 + x + x2 + x3 + ··· ∞ n Σ = 0 xn (–1, 1) (1 + x) n 1 + nx – n(n – 1) 2! x2 + n(n – 1)(n – 2) 3! x3 + ··· – (–1, 1) tan–1 x x – x3 3 + x5 5 – x7 7 + ··· ∞ n Σ = 0 (–1)n x2n + 1 2n + 1 [–1, 1] Maclaurin Series VIDEO
158 Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 Geometrical approximation by Maclaurin polynomials The geometrical approximation to certain functions by the first few terms of their Maclaurin series called Maclaurin polynomial are shown in the following diagrams. 1. f(x) = ex and f(x) = 1 + x + 1 2! x2 + 1 3! x3 + 1 4! x4 + ··· (a) x 2 2 2 4 4 4 6 8 – 0 –2 – f(x) = ex f(x) = 1 + x f(x) (b) x 2 2 2 4 4 4 6 8 0 – – 2 – 1 2! x2 f(x) = ex f(x) = 1 + x + f(x) Figure 5.1(a) Figure 5.1(b) (c) (d) x 2 2 2 4 4 4 6 8 0 – –2 – f(x) = ex f(x) = 1 + x + 1 2! 1 3! x2 + x3 f(x) 2 2 2 4 4 4 6 8 0 – –2 – f(x) = 1 + x + 1 2! 1 3! 1 4! x2 + + x3 x4 x f(x) = ex f(x) Figure 5.1(c) Figure 5.1(d) 2. f(x) = sin x and f(x) = x – 1 3! x3 + 1 5! x5 – 1 7! x7 + 1 9! x9 – ··· (a) (b) Figure 5.2(a) Figure 5.2(b) f(x) = sin x f(x) = x 2 2 2 1 1 1 3 3 3 6 5 4 54 6 0 – – –1 –––– 1 3! x 3 – x f(x) f(x) = sin x f(x) = x – — x3 + — x5 – — x7 2 2 2 1 1 1 3 3 3 6 5 4 54 6 0 – – –1 –––– x f(x) 111 3! 5! 7!
Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 159 (c) f(x) = x – – 1 3! x3 1 5! 1 7! 1 9! + x + 5 x9 x7 f(x) = sin x x 2 2 2 1 1 1 3 3 3 6 5 4 54 6 0 – – –1 –––– f(x) Figure 5.2(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. Exercise 5.1 1. Using Maclaurin's theorem, obtain the Maclaurin series for (a) cos x, (b) 1 1 – x , (c) (1 + x) n . 2. Using Maclaurin's theorem, obtain the expansion for each of the following functions. State the interval of convergence in each case. (a) e–5x (b) 2 sin (–x) (c) cos x (d) 2 3 – x (e) x2 x + 1 3. Use Maclaurin's theorem to find the series expansion sin–1 x. Write the sum of n terms of the Maclaurin series using sigma notation, and state its interval of convergence. 4. Use Maclaurin's theorem to find the series expansion for ax up to the term in x5 . Series expansions of the composites of functions Based on the standard Maclaurin series, we are able to generate the Maclaurin series for functions which are the composites of common functions. Example 4 Find the first four terms of the Maclaurin series for (a) e2x , (b) ex cos x . Solution: (a) ex = 1 + x + 1 2! x2 + 1 3! x3 + ··· for –∞ , x , ∞ Replace the single variable x in ex with 2x, e2x = 1 + (2x) + 1 2! (2x) 2 + 1 3! (2x) 3 + ··· for –∞ , x , ∞ e2x = 1 + 2x + 2x2 + 4 3 x3 + ··· for –∞ , x , ∞.
160 Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 (b) ex = 1 + x + 1 2! x2 + 1 3! x3 + ··· for –∞ , x , ∞ Replace the single variable x in ex with x cos x, ex cos x = 1 + (x cos x) + 1 2! (x cos x) 2 + 1 3! (x cos x) 3 + ··· for –∞ , x , ∞ Now replace cos x with its series expansion, ex cos x = 1 + x(1 – 1 2! x2 + ···) + 1 2! x2 (1 – 1 2! x2 + ···)2 + 1 3! x3 (1 – 1 2! x2 + ···)3 + ··· = 1 + x – 1 2 x3 + 1 2 x2 + 1 6 x3 + ··· = 1 + x + 1 2 x2 – 1 3 x3 + ··· for –∞ , x , ∞. Example 5 Find the first four terms of the Maclaurin series for cos πx. Solution: cos x = 1 – 1 2! x2 + 1 4! x4 – 1 6! x6 + ··· for –∞ , x , ∞ Replace the single variable x with πx, cos πx = 1 – 1 2! (πx) 2 + 1 4! (πx) 4 – 1 6! (πx) 6 + ··· for –∞ , x , ∞ = 1 – 1 2! π2 x2 + 1 4! π4 x4 – 1 6! π6 x6 + ··· for –∞ , x , ∞. Example 6 Find the Maclaurin series for 1 1 + 5x in ascending powers of x up to the term in x3 . Solution: 1 1 – x = 1 + x + x2 + x3 + ··· for –1 , x , 1. Replace the single variable x with –5x, 1 1 – (–5x) = 1 + (–5x) + (–5x) 2 + (–5x) 3 + ··· = 1 – 5x + 25x2 – 125x3 + ··· for – 1 5 , x , 1 5 .
Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 161 Example 7 Find the Maclaurin expansion of 1 4 – x2 . Determine the interval of convergence of this series. Solution: 1 1 – x = 1 + x + x2 + x3 + ··· for –1 , x , 1. Thus 1 4 – x2 = 1 4 1 1 1 – x2 4 2 for –1 , x 2 4 , 1 –4 , x2 , 4 –2 , x , 2 = 1 4 11 + 1 x2 4 2 + 1 x2 4 2 2 + 1 x2 4 2 3 + ··· 2 = 1 4 11 + x2 4 + x4 16 + x6 64 + ··· 2 = 1 4 + 1 16 x2 + 1 64 x4 + 1 256 x6 + ··· The series is convergent for –1 , x2 4 , 1, i.e. –2 , x , 2. The interval of convergence of the series (–2, 2). Example 8 Find the power series representation of x3 x + 2 . State the set of values of x for which the series expansion is valid. Solution: x3 x + 2 = x3 1 1 x + 22 = x3 1 1 2 1 + x 2 2 = x3 2 1 1 1 + x 2 2 Maclaurin series of 1 1 – x is 1 1 – x = 1 + x + x2 + x3 + ··· for –1 , x , 1. Replace the variable x in with 1– x 2 2 , 1 1 + x 2 = 1 + 1– 1 2 x2 + 1– 1 2 x2 2 + 1– 1 2 x2 3 + ··· = 1 – 1 2 x + 1 4 x2 – 1 8 x3 + ··· \ 1 2 x3 1 1 1 + x 2 2 = 1 2 x3 11 – 1 2 x + 1 4 x2 – 1 8 x3 + ··· 2 \ x3 x + 2 = 1 2 x3 – 1 4 x4 + 1 8 x5 – 1 16 x6 + ··· The series is convergent for –1 , x 2 , 1, i.e. –2 , x , 2. The set of values of x for which the series expansion is valid is (–2, 2).
162 Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 Exercise 5.2 1. Use a standard Maclaurin series to find the first four non-zero terms of the series expansion for the following functions. State the interval of convergence in each case. (a) 1 4 + x2 (b) 1 1 – x3 (c) e–2x2 (d) sin 2x (e) tan–1 (x2 ) (f) ln (1 + 2x4 ) (g) sin (x4 ) 2. Find the Maclaurin series for 1 (1 + x) 3 . Determine the range of values of x for which the series converges. Series expansions of the sums, differences, products and quotients We may use the series expansions of functions to find the series expansions of their sums, differences, products and quotients. Example 9 Find the Maclaurin series for ex + e–3x . State the interval of convergence of the series. Solution: ex = 1 + x + 1 2! x2 + 1 3! x3 + ··· for –∞ , x , ∞ \ e–3x = 1 + (–3x) + 1 2! (–3x) 2 + 1 3! (–3x) 3 + ··· = 1 – 3x + 9 2 x2 – 9 2 x3 + ··· \ ex + e–3x = (1 + x + 1 2 x2 + 1 6 x3 + ···) + (1 – 3x + 9 2 x2 – 9 2 x3 + ···) = 2 – 2x + 5x2 – 13 3 x3 + ··· The interval of convergence of the series is (–∞, ∞) Example 10 Expand ln 1 1 – 2x 1 + 2x 2 as a series of ascending powers of x. Find the range of values of x for which the expansion is valid. Solution: Since ln (1 + x) = x – x2 2 + x3 3 – x4 4 + ··· for –1 , x < 1, We have ln (1 – 2x) = (–2x) – 1 2 (–2x) 2 + 1 3 (–2x) 3 – 1 4 (–2x) 4 + ···, –1 < 2x , 1 = –2x – 2x2 – 8 3 x3 – 4x4 – ··· for – 1 2 < x , 1 2 We also have ln (1 + 2x) = 2x – 1 2 (2x) 2 + 1 3 (2x) 3 – 1 4 (2x) 4 + ··· for –1 , 2x < 1 = 2x – 2x2 + 8 3 x3 – 4x4 + ··· for – 1 2 , x < 1 2
Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 163 Hence, ln 1 1 – 2x 1 + 2x 2 = ln(1 – 2x) – ln (1 + 2x) = 1–2x – 2x2 – 8 3 x3 – 4x4 – ···2 – 12x – 2x2 + 8 3 x3 – 4x4 + ···2 = –4x – 16 3 x3 – ··· provided that – 1 2 < x , 1 2 and – 1 2 , x < 1 2 ⇒ i.e. – 1 2 , x , 1 2 The range of values of x for which the expansion is valid is 1– 1 2 , 1 2 2. Example 11 Find the series expansion of e2x sin 2x in ascending powers of x up to the term in x3 . Solution: ex = 1 + x + 1 2! x2 + 1 3! x3 + ··· \ e2x = 1 + 2x + 1 2! (2x) 2 + 1 3! (2x) 3 + ··· = 1 + 2x + 2x2 + 4 3 x3 + ··· Since sin x = x – 1 3! x3 + ··· sin 2x = 2x – 1 3! (2x) 3 + ··· = 2x – 4 3 x3 + ··· Therefore, e2x sin 2x = 11 + 2x + 2x2 + 4 3 x3 + ···212x – 4 3 x3 + ··· 2 = 2x – 4 3 x3 + 4x2 + 4x3 + ··· = 2x + 4x2 – 8 3 x3 + ··· Example 12 Express (1 – x) 2 cos 2x as a series of ascending powers of x up to the term in x5 . Solution: cos x = 1 – 1 2 x2 + 1 24 x4 – ··· for –∞ , x , ∞ \ cos 2x = 1 – 1 2 (2x) 2 + 1 24 (2x) 4 – ··· = 1 – 2x2 + 2 3 x4 – ···
164 Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 \ (1 – x) 2 cos 2x = 11 – 2x + x2 211 – 2x2 + 2 3 x4 – ···2 = 1 – 2x2 + 2 3 x4 – 2x + 4x3 – 4 3 x5 + x2 – 2x4 + … = 1 – 2x – x2 + 4x3 – 4 3 x4 – 4 3 x5 + ··· Example 13 Find the Maclaurin series for tan x up to the term in x5 . Solution: tan x = sin x cos x = x – 1 3! x3 + 1 5! x5 – ··· 1 – 1 2! x2 + 1 4! x4 – ··· x + 1 3 x3 + 2 15 x5 + ··· 1 – 1 2 x2 + 1 24 x4 – ··· x – 1 6 x3 + 1 120 x5 – ··· x – 1 2 x3 + 1 24 x5 – ··· 1 3 x3 – 1 30 x5 + ··· 1 3 x3 – 1 6 x5 + ··· 2 15 x5 + ··· Therefore, tan x = x + 1 3 x3 + 2 15 x5 + ··· Exercise 5.3 1. Use addition or subtraction of power series to find the first three non-zero terms in the Maclaurin series for the following functions. (a) ln (1 + x)(1 – x) (b) ln 1 1 + x 1 – x 2 (c) ln (1 + 2x) 2 (1 – 3x) (d) sinh x = 1 2 (ex – e–x ) (e) cosh x = 1 2 (ex + e–x )
Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 165 2. Use multiplication or division of power series to find the first four non-zero terms in the Maclaurin series for the following functions. (a) e–x2 cos x (b) x3 e–x (c) (x + 1) sin x (d) x3 x + 2 (e) sin x ex 3. Expand sin2 x as a series of ascending powers of x as far as the term in x8 . 4. Find the Maclaurin series of 3 cos x + 1 cos x + 3 . State the range of values of x for which the series converges. 5. Find the Maclaurin series for eax cos bx, where a and b are two constants. State the interval of convergence. Differentiation and integration of a power series Using Maclaurin's theorem or standard Maclaurin series, we may express many functions in the form of convergent power series. By differentiating and integrating a convergent power series of a function, we can find the Maclaurin series for other functions. Given that f(x) = c0 + c1x + c2x2 + ··· + cnxn + ··· Differentiating f(x) with respect to x, f9(x) = c1 + 2c2x + 3c3x2 + ··· + ncnxn – 1 + ··· Integrating f(x) with respect to x, from 0 to x, where x lies inside the interval of convergent of the Maclaurin series. ∫ 0 x f(x) dx = c0x + 1 2 c1x2 + 1 3 c2x3 + ··· + 1 n + 1 cnxn + 1 + ··· Notice that we have to perform term by term differentiation and integration of f(x). Example 14 Write the Maclaurin series for 1 1 – x . Hence find the series expansion for 1 (1 – x) 2 . Solution: 1 1 – x = 1 + x + x2 + ··· + xn + ··· = ∞ n Σ = 0 xn Differentiating both sides of the equation with respect to x 1 (1 – x) 2 = 1 + 2x + 3x2 + ··· + nxn – 1 + ··· = ∞ n Σ = 1 nxn – 1 The interval of convergence is (–1, 1).
166 Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 Example 15 Using the result that ∫ 0 x 1 1 + t 2 dx = tan–1 x, find the series expansion for tan–1 x. Solution: Let y = tan–1 x \ tan y = x Differentiating with respect to x, sec2 y · dy dx = 1 dy dx = 1 1 + tan2 y dy dx = 1 1 + x2 We have 1 1 – x = 1 + x + x2 + x3 + x4 + ··· for –1 , x , 1 \ 1 1 + x2 = 1 + (–x2 ) + (–x2 ) 2 + (–x2 ) 3 + ··· = 1 – x2 + x4 – x6 + ··· Integrating both sides between 0 and x, where x lies inside the interval of convergence. ∫ 0 x tan–1 · dx = ∫ 0 x (1 – x2 + x4 – x6 + …) dx = 3x – 1 3 x3 + 1 5 x5 – 1 7 x7 + ···4 0 x = x – 1 3 x3 + 1 5 x5 – 1 7 x7 + ··· 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] Example 16 Using the Maclaurin series for 1 1 – x , express ln (1 + x) as a series up to the term in x4 . Solution: 1 1 – x = 1 + x + x2 + x3 + x4 + ··· \ 1 1 + x = 1 + (–x) + (–x) 2 + (–x) 3 + (–x) 4 + ··· = 1 – x + x2 – x3 + x4 – ··· Integrating, ln(1 + x) = ∫ 0 x (1 – x + x2 – x3 + x4 – ···) dx = x – 1 2 x2 + 1 3 x3 – 1 4 x4 + ···
Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 167 5.2 Applications of Maclaurin Series Estimating the value of a definite integral Example 17 Using the first three non-zero terms of the series expansion for sin2 x x , estimate the value of the integral ∫ 0 —π 6 sin2 x x dx. Solution: sin x = x – 1 6 x3 + 1 120 x5 – ··· \ sin2 x = 1x – 1 6 x3 + 1 120 x5 – ···21x – 1 6 x3 + 1 120 x5 – ···2 = x2 – 1 6 x4 + 1 120 x6 – 1 6 x4 + 1 36 x6 + 1 120 x6 – ··· = x2 – 1 3 x4 + 2 45 x6 – ··· \ sin2 x x = x – 1 3 x3 + 2 45 x5 – ··· ∫ 0 —π 6 sin2 x x dx ≈ ∫ 0 —π 6 1x – 1 3 x3 + 2 45 x5 2 dx ≈ 3 1 2 x2 – 1 12 x4 + 1 135 x6 4 0 —π 6 ≈ 1 2 1 π 6 2 2 – 1 12 1 π 6 2 4 + 1 135 1 π 6 2 6 ≈ 0.1310 Finding the limit of a function By expressing functions as Maclaurin series, we are able to evaluate the limit of a function as x → 0. Example 18 Evaluate lim x → 0 1 – cos x 2 sin2 x Solution: lim x → 0 1 – cos x 2 sin2 x = lim x → 0 1 – 11 – 1 2! x2 + 1 4! x4 – ···2 21x – 1 3! x3 + 1 5! x5 – ···2 2 = lim x → 0 1 2! x2 – 1 4! x4 + ··· 2x2 – 4 3! x4 + ··· = lim x → 0 1 2! + higher powers of x 2 + higher powers of x = 1 4
168 Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 Example 19 Find the first three terms of the series expansion for sin(sin x). Hence, evaluate lim x → 0 x – sin (sin x) 3x3 . Solution: sin x = x – 1 3! x3 + 1 5! x5 – ··· \ sin (sin x) = 1x – 1 3! x3 + 1 5! x5 – ···2 – 1x – 1 3! x3 + 1 5! x5 – ···2 3 3! + 1x – 1 3! x3 + 1 5! x5 – ···2 5 5! = x – 1 3! x3 + 1 5! x5 – 1 3! x3 + 3 3!3! x5 + 1 5! x5 + ··· = x – 1 3 x3 + 1 10 x5 + ··· lim x → 0 x – sin (sin x) 3x3 = lim x → 0 x – 1x – 1 3 x3 + 1 10 x5 + …2 3x3 = lim x → 0 1 3 x3 – 1 10 x5 + higher powers of x 3x3 = lim x → 0 1 3 – 1 10 x2 + higher powers of x 3 = 1 9 Example 20 By using series expansion, evaluate lim x → 0 1 ex – e–x 2 sin x 2. Solution: lim x → 0 1 ex – e–x 2 sin x 2 = lim x → 0 1 11 + x + 1 2! x2 + 1 3! x3 + ···2 – 11 – x + 1 2! x2 – 1 3! x3 + ···2 21x – 1 3! x3 + 1 5! x5 – ···2 2 = lim x → 0 2x + 2 3! x3 + higher powers of x 2x – 2 3! x3 + higher powers of x = lim x → 0 1 + 1 3! x2 + higher powers of x 1 – 1 3! x2 + higher powers of x = 1
Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 169 Exercise 5.4 1. Use series expansions to evaluate the following limits. (a) lim x → 0 1 e2x – ex x 2 (b) lim x → 0 1 1 – ex3 x 3 2 (c) lim x → 0 1 ex – esin x x3 2 (d) lim x → 0 1 1 sin x – 1 x 2 (e) lim x → 0 1 3 sin x e2x – 1 2 (f) lim x → 0 1 ex + e–x – 2 2 cos 2x – 2 2 (g) lim x → 0 1 tan x – x x3 2 (h) lim x → 0 1 x2 + 2 ln(cos x) x4 2 (i) lim x → 0 1 (5 + x) ln(x + 1) ex – 1 2 Summary 1. A power series is an infinite series in ascending powers of x in the form ∞ n Σ = 0 cnxn = 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 ∞ n Σ = 0 cnxn = c0 + c1x + c2x2 + c3x3 + … + cnxn + ··· always converges at x = 0. 3. Maclaurin series for the function f is given by f(x) = ∞ n Σ = 0 f (n) (0) n! xn = f(0) + f9(0)x + f 0(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 interval of convergence. Function Maclaurin series Interval of convergence ex 1 + x + x2 2! + x3 3! + ··· (–∞, ∞) sin x x – x3 3! + x5 5! – x7 7! + ··· (–∞, ∞) cos x 1 – x2 2! + x4 4! – x6 6! + ··· (–∞, ∞) ln (1 + x) x – x2 2 + x3 3 – x4 4 + ··· (–1, 1] 1 1 – x 1 + x + x2 + x3 + ··· (–1, 1) (1 + x) n 1 + nx + n(n – 1) 2! x2 + n(n – 1)(n – 2) 3! x3 + ··· (–1, 1) tan–1 x x – x3 3 + x5 5 – x7 7 + ··· [–1, 1] 6. The limit of a function as x → 0 can be evaluated by first expressing functions involved as series expansions.
170 Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 STPM PRACTICE 5 1. Using Maclaurin's theorem, obtain the first three non-zero terms of the Maclaurin series for (a) 4 + sin2 x (b) 1 + 8ex (c) tan2 x (d) (sin–1 x) 2 2. Using the Maclaurin theorem, find the expansion of ln(1 + tan x) up to the term in x3 . Hence, evaluate lim x → 0 3x – ln(1 + tan x) 3 6x2 – x4 . 3. If y = etan–1 x , show that dy dx = y 1 + x2 . Hence, find the Maclaurin series for etan–1 x up to and including the term in x2 . 4. Find the first three non-zero terms of the expansion, in ascending powers of x, of (a) ln(1 + xex ) (b) ln(1 + x + x2 + x3 ) 5. State the Maclaurin series for e–3x up to the term in x3 . If y = tan1 1 4 π + x2, show that d2 y dx2 = 2y dy dx . Obtain the series expansion of y in ascending powers of x up to and including the term in x3 . Hence, find the first three non-zero terms of the Maclaurin series of e –3x tan1 1 4 π + x2. 6. Find the Maclaurin series for tan–1 x up to the term in x. Hence, find the series expansion of e x2 (tan–1 x) up to the term in x5 . 7. Given that y = e–sin 2x . Determine the values of dy dx and d4 y dx4 when x = 0. Hence, obtain the Maclaurin series for y up to the term in x4 . 8. It y = (sin–1 x) 2 , show that (1 – x2 ) d2 y dx2 = 2 + x dy dx . Hence, deduce the Maclaurin series for (sin–1 x) 2 up to and including the term in x4 . 9. Given that y = cos—1 2 x. Show that d2 y dx2 = – 1 4 1y + 1 y3 2. By further differentiation, obtain the series expansion of y in ascending powers of x up to the term in x4 . 10. The function f is defined by f(x) = e –2x sin 3x. (a) Show that f9(x) = 3e–2x cos 3x – 2e–2x sin 3x and f0(x) + 4f9(x) + 13f(x) = 0. (b) Find the Maclaurin series of f(x) up to and including the term in x3 . (c) Use the Maclaurin series obtained in (b) to (i) find the expansion of f(x) 1 + x in ascending powers of x up to the term in x3 . (ii) find the value of lim x → 0 f(x) x . 11. Find the Maclaurin series for 1 (1 + x) 2 up to the term in x3 . Hence, (a) find the series expansion of ln(1 – x) (1 + 2x) 2 up to the term in x3 . State the range of values of x for which the expansion is valid. (b) using the series expansion in (a), deduce the approximate value of ∫ 0 0.2 ln(1 – x) (1 + 2x) 2 dx correct to three significant figures. 12. The first two terms in the series expansion in ascending powers of x, of (1 + 4 3 x) n are equal to the first two terms of the Maclaurin series for ex (1 + sin 2x). Find the value of n.
Mathematics Semester 2 STPM Chapter 5 Maclaurin Series 5 171 13. State the Maclaurin series for e –3x up to the term in x3 . If y = tan x, show that d2 y dx2 = 2ydy dx . Obtain the Maclaurin series for tan x up to the term in x3 . Hence, find the series expansion of e –3x tan x up to and including the term in x3 . 14. State the first four non-zero terms of series expansion cos 4x. Hence, find the value of lim x → 0 cos 4x + 8x2 – 1 32x4 . 15. Using the Maclaurin series for ln(1 + x), find the series expansion of ln 1 1 + x 1 – x 2 up to the term in x3 and state its interval of convergence. Hence, (a) obtain the Maclaurin series for e ln 1 + x 1 – x 1 2 up to the term in x3 . (b) deduce the approximate value of ∫ 0 0.3 e ln 1 + x 1 – x 1 2 dx, correct to three significant figures. 16. Use a Maclaurin polynomial to evaluate 31.05 correct to four decimal places. 17. (a) If y = ln(sec x), show that d2 y dx2 = sec2 x. (b) Obtain the Maclaurin series for ln(sec x) up to the term in x4 . (c) Using the Maclaurin series in part (b), deduce the approximate value of ∫ 0 0.4 ln(sec x) dx correct to three decimal places. 18. Evaluate the following limits by using appropriate Maclaurin series (a) lim x → 0 1 x – tan–1 x 3x3 2 (b) lim x → 0 (3 + x) ln(x + 1) ex – 1 19. Use appropriate Maclaurin series to evaluate (a) lim x → 0 1 sin2 x – x2 cos x x4 2 (b) lim x → 0 1 –xe2x e2x – 1 2 20. Using the Maclaurin series for ln(1 + x), evaluate lim x → 0 ln(1 + x) – x 3x2 . 21. Evaluate lim x → 0 1 3 sin x – sin 3x x (cos x – cos 3x) 2 . 22. Given that y = e sin–1 x + 1. Show that y = (1 – x2 ) d2 y dx2 – x dy dx . Find the Maclaurin series for esin–1 x + 1 in ascending powers of x up to the term in x2 . Hence, (a) determine lim x → 0 e sin–1 x + 1 3 (b) approximate the value of ∫ 0 0.1 esin–1 x + 1 dx, correct to three decimal places. 23. Using the Maclaurin series for ln (1 + x), deduce the first four terms of the series expansion for ln (1 – x). Hence, evaluate lim x → 0 x + ln (1 – x) x2 24. Given that y = (sin–1 x) 2 . (a) Show that (1 – x2 ) d2 y dx2 – x dy dx = 2. (b) Using Maclaurin theorem, 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. (c) Using the series expansion in (b) where x = 1 2 , estimate the value of π correct to three decimal places.
172 CHAPTER Subtopic Learning Outcome 6.1 Numerical solution of equations (a) Locate a root of an equation approximately by means of graphical considerations and by searching for a sign change. (b) Use an iterative formula of the form xn + 1 = f(xn) to find a root of an equation to a prescribed degree of accuracy. (c) Identify an iteration which converges or diverges. (d) Use the Newton-Raphson method. 6.2 Numerical integration (a) Use the trapezium rule. (b) Use sketch graphs to determine whether the trapezium rule gives an over-estimate or an under-estimate in simple cases. converge – menumpu initial approximation – penghampiran awal iteration – lelaran Newton-Raphson method – kaedah Newton-Raphson numerical method – kaedah berangka trapezium rule – petua trapezium 6 NUMERICAL METHODS Bilingual Keywords
Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 173 6.1 Numerical Solution of Equations In some cases, it is possible to solve equations using algebraic methods. But in some instances it may be impossible. As such we have to use numerical techniques which give good approximations to the solutions. One such approach is the graphical method. Locating an approximate root To locate approximate root of the equation f(x) = 0 graphically, we sketch or draw the graph y = f(x) and read from it the values of x for which f(x) = 0, i.e. the x-coordinates of the points where the graph y = f(x) cuts the x-axis. Alternatively, we could rearrange f(x) = 0 in the form g(x) = h(x) and find the x-coordinates of the points where the two graphs y = g(x) and y = h(x) intersect. Example 1 Find the approximate root of the equation x3 + x = 1 graphically. Solution: Rearrange the equation x3 + x = 1 we have x(x2 + 1) = 1 x2 + 1 = 1 x Sketch the graphs of y = x2 + 1 and y = 1 x . y = y x 2 0.7 0 1 2 3 4 4 6 8 x 10 1 y = x 2 +1 The x-coordinate of the point of intersection of the graphs y = 1 x and y = x2 + 1 is a root of the equation x3 + x – 1 = 0. Hence, the approximate root is x = 0.7.
174 Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 We can also locate an approximate root of an equation by finding two values a and b such that the signs of f(a) and f(b) are opposite, i.e. f(a) · f(b) , 0. At least one of the roots must lie between a and b if f(x) is continuous. Example 2 Consider the equation x3 + ln x = 0 Solution: Let f(x) = x3 + ln x f(0.5) = –0.568 , 0 f(1) = 1 . 0 Since f(x) is continuous on the interval [0.5, 1] with f(0.5) , 0 and f(1) . 0, there is a root of the equation between x = 0.5 and x = 1. Iterative methods If we want to solve an equation f(x) = 0 by an iterative method, we need a relationship xn + 1 = F(xn) where xn + 1 is a better approximation to the solution of f(x) = 0 than is xn. Suppose we can rearrange the equation f(x) = 0 into the form x = F(x). Consider the graphs of y = x and y = F(x) as shown below. x x2 x1 x0 y 0 Root y = F(x) y = x Figure 6.1 If our initial guess for the root is x0 , then x1 = F(x0 ) gives a better approximation and x2 = F(x1 ) gives a better approximation than x1 and so on. By repeating the process, we can obtain a value of x, which is as close to the root as is required. Theorem: If x = a is an approximation to a root of the equation x = F(x), then a sequence defined by xn + 1 = F(xn) with a starting value, close to a will converge to a so long as |F9(a)| , 1.
Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 175 Example 3 Show that the equation x2 + 4x – 1 = 0 has a root between x = 0 and x = 1. Using an iterative method to find this root correct to two decimal places. Solution: Let f(x) = x2 + 4x – 1 f(0) = –1 , 0 f(1) = 4 . 0 Since f(0) · f(1) , 0, there is a root between x = 0 and x = 1. The equation x2 + 4x – 1 = 0 can be rearranged as follows: (a) x(x + 4) – 1 = 0 (b) x2 = 1 – 4x which gives x = 1 x + 4 which gives x = 1 x – 4 The iterative formulae are (a) xn + 1 = 1 xn + 4 and (b) xn + 1 = 1 xn – 4 Consider the iterative formula xn + 1 = 1 xn + 4 with an initial guess of x0 = 1, x1 = 1 1 + 4 = 0.2 x2 = 1 0.2 + 4 = 0.2381 x3 = 1 0.2381 + 4 = 0.2359 x4 = 1 0.2359 + 4 = 0.2361 x5 = 1 0.2361 + 4 = 0.2361 Thus the required root is x = 0.24 correct to two decimal places. The graphs of y = x and y = 1 x + 4 below show how xn is converging to the required root. y x x3 x4 x2 x 0 x1 0 = 1 y = x y = x + 4 1 Root 0.24
176 Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 Using the iterative formula xn + 1 = 1 xn – 4 with an initial guess of x0 = 1 gives x1 = 1 1 – 4 = –3 x2 = 1 –3 – 4 = –4.333 x3 = 1 –4.333 – 4 = –4.231 x4 = 1 –4.231 – 4 = –4.236 x5 = 1 –4.236 – 4 = –4.236 This gives the other root –4.24 (two decimal places) The graph below shows how xn is converging to the required root. y –4 = 1 x Root –4.24 x2 x4 x3 x1 x0 0 The above example shows two possible arrangements of the given equation. Both the iterative formulae converged to a root of the equation. However, this is not always the case.
Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 177 Example 4 Use an iterative formula to show that the equation x3 + 2x – 4 = 0 has a root near x = 1. Solution: The equation can be rearranged as (a) x = 1 2 (4 – x3 ) (b) x = 34 – 2x Using the respective iterative formulae we have (a) xn + 1 = 1 2 [4 – (xn) 3 ] If x0 = 1 x1 = 1 2 (4 – 1) = 1.5 x2 = 1 2 (4 – 1.53 ) = 0.313 x3 = 1.194 x4 = –1.909 x5 = 5.477 i.e. x0 , x1 , x2 , x3 , … do not converge. (b) xn + 1 = 34 – 2xn If x0 = 1 x1 = 32 = 1.259 x2 = 34 – 2(1.259) = 1.140 x3 = 1.198 x4 = 1.170 x5 = 1.184 x6 = 1.177 \ x = 1.18 (2 decimal places) Example 5 Show that the equation x3 + 7x – 1 = 0 has a root between x = 0 and x = 1. Show that the equation x3 + 7x – 1 = 0 can be rearranged in the form x = 1 x2 + 7 . Hence, use an iterative method to find this root correct to 3 decimal places. Solution: Let f(x) = x3 + 7x – 1 f(0) = –1 , 0 f(1) = 7 . 0 Since f(0) · f(1) , 0 and f(x) is continuous, the equation has a root between x = 0 and x = 1. x3 + 7x – 1 = 0 can be rearranged as x(x2 + 7) = 1 x = 1 x2 + 7
178 Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 Use the iterative formula xn + 1 = 1 xn 2 + 7 Let x0 = 1 x1 = 1 1 + 7 = 0.125 x2 = 1 0.1252 + 7 = 0.1426 x3 = 0.1424 x4 = 0.1424 The root is 0.142 (3 decimal places). Exercise 6.1 1. For each of the following, (i) show how the iterative formula is derived from the given equation, (ii) determine whether the iterative formula converges to a root of the equation, (iii) state the root of the equation correct to 3 decimal places if the formula converges. (a) 3x3 – 2x – 6 = 0 ; xn + 1 = 2 x + 2 3 ; x ≠ 0, x0 = 1.43 (b) x + e–x – 2 = 0 ; xn + 1 = 2 – e–x ; x0 = 1.84 (c) 2x3 – x2 + 1 = 0 ; xn + 1 = 2xn 3 + 1 xn ; x0 = 0.5 2. (a) Find the equation which is solved by the iterative formula xn + 1 = 2 + 1 xn 2 . (b) Starting with x0 = 2.5, use the iterative formula above to find x1 , x2 , x3 and x4 correct to 3 decimal places. 3. (a) Find the equation which is solved by the iterative formula xn + 1 = – 1 5 (x3 + 2x2 – 1). (b) Starting with x0 = 0, use the iterative formula to find a solution of this equation correct to 3 decimal places. 4. The equation x2 + 4x – 1 = 0 has two roots, one near x = 0 and the other near x = –4. (a) Using xn + 1 = 1 xn + 4 with x0 = 0, find the root near x = 0, correct to 3 decimal places. (b) Why could we not use the formula xn + 1 = 1 xn + 4 with x0 = –4? (c) Using xn + 1 = 1 xn – 4 with x0 = –4, find the root near x = –4 correct to 3 decimal places. 5. Show that the equation x3 – x + 3 = 0 has a root in the range [–3, 3]. Use the rearranged form x = 3(x – 3) to obtain this root correct to two decimal places.
Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 179 The Newton-Raphson method When exact formulae for solving an equation f(x) = 0 are not available, we can turn to numerical techniques from calculus to approximate the roots we seek. One of these techniques is the Newton-Raphson method. It is based on the idea of using tangent lines to replace the graph of y = f(x) near the points where f(x) is zero. The aim of Newton-Raphson method for estimating a root of an equation f(x) = 0 is to produce a sequence of approximations that approach the root. We pick the first number x0 of the sequence. Then, under favourable circumstances, the method does the rest by moving step by step towards a point where the graph of f crosses the x-axis. (Figure 6.2) 0 Root sought Fourth Third y x (x1, f(x1)) (x2, f(x2)) (x 3, f(x 3)) x 1 x 2 x 3 x 4 y = f(x) Second First approximation Figure 6.2 The initial estimate, x0 may be found by sketching the graph or just plain guessing. The method uses the tangent to the curve y = f(x) at [x0 , f(x0 )] to approximate the curve, naming the point where the tangent meets the x-axis x1 . The number x1 is usually a better approximation to the root than x0 . The point x2 where the tangent to the curve at [x1 , f(x1 )] crosses the x-axis is the next approximation in the sequence. We continue using each approximation to generate the next until we are close to the root to a prescribed accuracy. We can derive a formula for generating successive approximations in the following way. Given the approximation xn, the equation of the tangent to the curve at [xn, f(xn)] (Figure 6.3) is y – f(xn) = f9(xn)(x – xn) ……………… Where the tangent crosses the x-axis, we equate y to 0 in this equation and solve for x, giving 0 – f(xn) = f9(xn)(x – xn) f’(xn)x = f9(xn)xn – f(xn) x = xn – f(xn) f9(xn) , f9(xn) ≠ 0. Newton Raphson VIDEO Newton Raphson INFO
180 Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 The value of x is the next approximation, xn + 1. 0 Tangent y x (xn, f(xn)) x n x n + 1 = xn – f(xn) _ f'(xn) y = f(x) Root sought Figure 6.3 The strategy for the Newton-Raphson method 1. Locate a first approximation to a root of the equation f(x) = 0, using a sketch graph of y = f(x) or searching for a sign change of f(x). 2. Use the first approximation to get a second, the second to get a third, and so on, using the formula xn + 1 = xn – f(xn) f9(xn) , f9(xn) ≠ 0 where f9(xn) is the derivative of f at xn. Convergence of the Newton-Raphson method The success of the Newton-Raphson method depends on 1. the initial approximation and 2. the shape of the curve in the neighbourhood of the root. In extreme cases, the Newton-Raphson method may fail. Case 1: x0 , the first approximation, is too far from the root. (example beyond a turning point) x1 x0 α(root) Turning point y = f(x) Case 1
Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 181 Case 2: f9(x0 ) = 0 or is too small. x1 x0 (Root) y = f(x) α Case 2 Thus, the Newton-Raphson method works best if the initial approximation x0 is sufficiently close to the root and f(x), f9(x), and f0(x) are continuous and bounded in a small interval containing the root. Example 6 Show that the equation x + ex = 0 has only one real root and find its value correct to three decimal places. Solution: Rewrite the equation as ex = –x. Sketch the graphs of y = ex and y = –x. y x y = ex y = –x 0 The sketch shows that there is only one point of intersection and so the equation x + ex = 0 has only one real root which is approximately x = –0.5. Using the Newton-Raphson method, with f(x) = x + ex f9(x) = 1 + ex Taking x0 = –0.5, x1 = x0 – f(x0) f9(x0 ) = –0.5 – (–0.5 + e–0.5) 1 + e–0.5 = –0.5663
182 Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 Repeating the process, x2 = –0.5663 – (–0.5663 + e– 0.5663) 1 + e– 0.5663 = –0.5671 x3 = –0.5671 – (–0.5671 + e– 0.5671) 1 + e– 0.5671 = –0.5671 Hence, the approximate root is –0.567 (correct to 3 decimal places) Example 7 Show that the equation x3 – 2x – 17 = 0 has a root between 2 and 3. Using the Newton-Raphson method, with x = 2.8 as a first approximation, determine, by means of two iterations, an approximation to this root, giving your answer to 2 decimal places. Solution: Let f(x) = x3 – 2x – 17 f9(x) = 3x2 – 2 f(2) = –13 f(3) = 4 Since f(2) and f(3) have opposite signs, f(x) has a root between 2 and 3. Let the first approximation be x1 = 2.8. f(2.8) = –0.648, f9(2.8) = 21.52 Using the Newton-Raphson method, x2 = 2.8 – 1 –0.648 21.52 2 = 2.8301 = 2.830 (correct to 3 decimal places) f(2.830) = 0.005, f9(2.830) = 22.027 x3 = 2.830 – 0.005 22.027 = 2.8298 = 2.830 (correct to 3 decimal places) Hence, an approximation to the root is 2.83 (correct to 2 decimal places). Example 8 If a1 is the first approximation to a root of the equation x4 = b, where b is given, show by using the NewtonRaphson method that there is a second approximation, a2 , given by a2 = 3a1 4 + b 4a1 3 (a) Use this result to find two further approximations to 2—1 4 by taking a1 = 1 as a first approximation. Give your results in the appropriate decimal places. (b) Sketch the graph of y = x4 – 2 and show the geometrical relationship between the approximations on the curve. (c) Would you expect the next approximation to be smaller or greater than the value of 2—1 4 ? Give reasons for your answer.
Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 183 Solution: f(x) = x4 – b f9(x) = 4x3 Let the first approximation = a1 Using the Newton-Raphson method, the second approximation, a2 = a1 – f(a1) f9(a1 ) = a1 – 1 a1 4 – b 4a1 3 2 = a1 – a1 4 + b 4a1 3 = 3a1 4 + b 4a1 3 ……… (a) x = 2 —1 4 x4 = 2 b = 2 1st approximation, a1 = 1 2nd approximation, a2 = 3 4 + 2 4 From 1 = 1.25 3rd approximation, a3 = 3(1.25) 4 + 2 4(1.25)3 = 1.1935 (b) 0 1 The root (4th approximation) y x 1.25 f(x) = x 4 – 2 a3 a4 a1 a2 1.1935 (c) From the graph, it is observed that the tangents at the points x = a2 , a3 , a4 , … approach the tangent at x = 2—1 4 (root of the equation x4 – 2 = 0) from the right. Hence, the next approximation is greater that 2—1 4 .
184 Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 Example 9 Show that the equation 2 sin x = x has a root between x = 1 and x = 2. Using x = 2 as the first approximation, find, by using the Newton-Raphson method, two more approximations to the root. Give your answer correct to 3 significant figures. Solution: Let f(x) = 2 sin x – x f9(x) = 2 cos x – 1 f(1) = 0.683 . 0 f(2) = –0.181 , 0 Since f(x) changes sign between x = 1 and x = 2, the equation 2 sin x = x has a root between x = 1 and x = 2. Let the first approximation, x1 = 2 Using Newton-Raphson method, 2nd approximation, x2 = x1 – f(x1) f9(x1 ) = 2 – 1 –0.181 –1.832 2 = 1.901 3rd approximation, x3 = 1.901 – f(1.901) f9(1.901) = 1.901 – 1 –0.009 –1.648 2 = 1.896 = 1.90 (correct to 3 significant figures) To illustrate a case where Newton-Raphson method fails. Let the first approximation x1 = 1. f(1) = 2 sin 1 – 1 = 0.683 f9(1) = 2 cos 1 – 1 = 0.081 Using Newton-Raphson method, 2nd approximation, x2 = 1 – 0.683 0.081 = –7.432 Clearly this is absurd. The failure of the method is due to the fact that the approximation x1 = 1 is close to the root of f9(x) = 0. This problem does not arise when x1 is 2.
Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 185 Exercise 6.2 1. Show that each of the following equations has a root between x = 0 and x = 1. Using the Newton-Raphson method, find the root correct to 2 decimal places. (a) x3 – x2 + 10x – 2 = 0 (b) 3x3 – 2x2 – 9x + 2 = 0 (c) 2x – 1 = e–2x 2. Show that the equation x4 = 2x3 + 8 has a root such that 2 , , 3. Obtain the value of this root correct to 2 decimal places. 3. Starting with x = 1 as a first approximation, use the Newton-Raphson method to find a second approximation to 2—1 20 . 4. By investigating the turning points of the function f(x) = x3 + 3x2 + 6x – 38, show that the equation f(x) = 0 has only one real root. Find two consecutive integers n and n + 1 between which the root lies. Starting with the value n as the first approximation, find, by the Newton-Raphson method, two more approximations to the root. Give your answer correct to 3 significant figures. 5. Sketch, on the same coordinate axes, the graphs of y = 1 x and y = sin πx for 0 , x , 3. (It will be sufficient, in each case, to determine the values of y for x = 1 2 , 1, 1 1 2 , 2, 2 1 2 , 3). Using 2 1 6 as a first approximation, calculate a second approximation to the smallest positive root of the equation sin πx = 1 x by the Newton-Raphson method. Give your answer correct to 2 decimal places. 6. Show that the equation x3 + x + 3 = 0 has only one real root. Show also that this root lies between –2 and –1. Using x = –1.2 as a first approximation, calculate, by the Newton-Raphson method, a second approximation to the root, correct to 3 significant figures. 7. Given that 37 = 2 187 and that 3 can be used as a first approximation to the root of the equation x7 + x3 = 2 210, find a second approximation to the root. 8. Show that x3 – 3x – 1 = 0 has only one real root. Starting with x = 3, use two iterations of the Newton-Raphson method to find an approximate value for the root, correct to 3 decimal places. 9. Sketch, on the same coordinate axes, the graphs of y = e–—1 2 x and y = 3 – x2 . Verify that the negative root, a, of the equation x2 + e–—1 2 x = 4 is such that –2 , a , –1, and state the integer which is closest to the positive root, . Use the Newton-Raphson method to find the value of correct to 2 decimal places.
186 Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 10. Sketch the graphs of y = (x – 2)4 and y = (x – 1)3 on the same coordinate axes. Given that the equation (x – 2)4 = (x – 1)3 has two roots a and , where a , , use the graphs to find an integer n such that n , x , n + 1. Find two consecutive integers between which lies. With your value of n as a first approximation and taking x = n + 1 2 , find a second approximation to by the Newton-Raphson method. 11. Sketch the graphs of y = x3 and y = 5x + 11 on the same coordinate axes. Deduce from your graph that the equation x3 = 5x + 11 has a positive root. Using the Newton-Raphson method, obtain an approximation to the root. 12. Find the equations of the asymptotes of the graph y = 3 – 2x x – 2 and sketch the graph. On the same coordinate axes, sketch the graph of y = 1 – e–2x . Show that where the graphs intersect, (3x – 5)e2x = x – 2, and hence state the number of real roots of this equation. Taking x = 1.7 as a first approximation, use the Newton-Raphson method once to obtain a second approximation to one of the roots of the equation correct to 3 significant figures. 13. On the same coordinate axes, sketch the curves with equations (i) y2 = 9x, (ii) y = x2 (1 + x), Label each curve clearly. Deduce that the equation x3 (1 + x) 2 – 9 = 0 has exactly one real root. Denoting this root by , find an integer n such that n , a , n + 1, and taking n as a first approximation, use the Newton-Raphson method to find a second approximation to , correct to 2 decimal places. 6.2 Numerical Integration Approximate integration There are times when we need to evaluate the integral ∫ a b f(x) dx exactly but cannot do so by the integration methods we have learnt. The function may be one that cannot be integrated algebraically, for example, ∫ ex2 dx. y x 0 y = f(x) Since this integral ∫ a b f(x) dx gives the area bounded by the curve y = f(x), the x-axis and the ordinates x = a and x = b, an approximate value for the integration can be found by estimating this area by another method. A common method is the trapezium rule.
Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 187 The trapezium rule y x 0 y = f(x) a b d d d d d A1 y0 y1 y2 y3 A2 A3 An – 1 yn – 2 yn – 1 yn An This method divides the area required into a number of parallel strips, each of equal width. Each strip is then regarded as a trapezium. The area of a trapezium is given by the product of one half the sum of its parallel sides and the perpendicular distance between them. Thus, if the area required is divided into n strips by (n + 1) equidistant ordinates of length y0 , y1 , y2 , …, yn – 1, yn, each strip has a width of b – a n . Area under the curve ≈ Sum of areas of trapezium ≈ A1 + A2 + A3 + … + An – 1 + An ≈ b – a n 3 1 2 (y0 + y1 ) + 1 2 (y1 + y2 ) + 1 2 (y2 + y3 ) + … + 1 2 (yn – 1 + yn)4 ≈ 1 2 d [y0 + 2y1 + 2y2 + 2y3 + … + 2yn – 1 + yn], where d = b – a n . Hence ∫ a b f(x) dx ≈ 1 2 d [y0 + 2(y1 + y2 + y3 + … + yn – 1) + yn] This is known as the trapezium rule. Example 10 Find ∫ 1 2 (1 + x2 ) —1 2 dx using the trapezium rule with five ordinates. Solution: We divide the area into four strips (n = 4), each of width d = 1 4 = 0.25. x 1 1.25 1.5 1.75 2 (1 + x2 ) —1 2 1.4142 y0 1.6008 y1 1.8024 y2 2.0156 y3 2.2361 y4 ∫ 1 2 (1 + x2 ) —1 2 dx ≈ 1 2 d (y0 + 2y1 + 2y2 + 2y3 + y4 ) = 0.25 2 [1.4142 + 2(1.6008 + 1.8024 + 2.0156) + 2.2361] = 1.8111
188 Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 Example 11 Evaluate ∫ 0 —π 3 tan x dx using the trapezium rule with ordinates at x = 0, π 12 , π 6 , π 4 , π 3 , giving your answer correct to two significant figures. Explain, with the aid of a sketch, why the trapezium rule gives an estimate which is greater than the exact answer in this case. Solution: x 0 π 12 π 6 π 4 π 3 tan x 0 0.2679 0.5773 1 1.732 Thus ∫ 0 —π 3 tan x dx ≈ 1 2 1 π 122[0 + 2(0.2679 + 0.5773 + 1) + 1.732] = 0.7097 = 0.71 (2 significant figures) y x 0 From the sketch, the trapezium are each higher than the curve. Hence this total area is greater than the area under the curve. Thus, the trapezium rule gives an estimate which is greater than the exact value. Example 12 Using the trapezium rule with three ordinates, find an estimate for the area under the curve y = sin x between x = 0 and x = π 2 . Use a sketch graph of y = sin x, determine whether the trapezium rule gives an over-estimate or an underestimate of the area. Solution: d = π 2 – 0 2 = π 4 x 0 π 4 π 2 sin x 0 1.414 1
Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 189 A π 4 2 (0 + 1.414 + 1) 0.95 An estimate of the area = 0.95 y y = sin x x 1 0 4 π 2 π From the sketch graph, the trapeziums are lower than the curve. Hence, the trapizium rule gives an under-estimate of the area. Exercise 6.3 1. Use the trapezium rule with 4 intervals of equal width to estimate the areas shown. State with reasons, whether your answer is an over-estimate or under-estimate. (a) y = 3 x + 1 , x . 1 (b) y = x + 1 x (c) y = 1 x2 , x . 0 y x 0 2 6 y x 0 1 3 y x 0 1 1 5 2. Use the trapezium rule, with the given number of ordinates for each case, to estimate the following definite integrals. (a) ∫ –1 1 ln(2 + x) dx, 5 ordinates, correct to 2 significant figures. (b) ∫ 0 0.5 2e–x2 dx, 6 ordinates, correct to 2 decimal places. (c) ∫ 1 3 40 – x3 dx, 3 ordinates, correct to 3 significant figures. (d) ∫ 1 2.5 1 1 + ln x dx, 4 ordinates, correct to 2 decimal places. (e) ∫ 0 2 x ex dx, 5 ordinates, correct to 4 significant figures.
190 Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 3. The definite integral ∫ 0 1 1 1 + x2 dx is known to be equal to π 4 . Using the trapezium rule for 5 strips, find an approximation for π. 4. The graph of the function y = 2 + x is given below. y x A B C D 0 2 1 2 3 7 The area of the shaded region ABCD is to be found. (a) Make a table of values of y for integer values of x from x = 2 to x = 7, giving each value of y correct to 4 decimal places. (b) Use the trapezium rule with 5 strips, each of 1 unit wide, to calculate an estimate for the area ABCD. State, giving a reason, whether your estimate is too large or too small. 5. Tabulate the values of the function f(x) = 1 + x2 to 3 decimal places for values of x from 0 to 0.8 at intervals of 0.1. Use these values to estimate ∫ 0 0.8 f(x) dx by the trapezium rule, using all the ordinates. Summary 1. If a polynomial function f(x) is such that f(a) · f(b) , 0, then there is at least one real root of the equation f(x) = 0 between x = a and x = b. 2. If xn is a first approximation to the root of the equation f(x) = 0, then the iterative formula xn + 1 = F(xn) is a better approximation to the root of f(x) = 0 than xn. 3. The approximate root of a non-linear equation can be found by the Newton-Raphson method, xn + 1 = xn – f(xn) f9(xn) , f9(xn) ≠ 0 4. The trapezium rule, with n strips of width h ∫ a b f(x) dx = 1 2 h [y0 + 2(y1 + y2 + ··· + yn – 1) + yn] y x 0 y = f(x) a b h y0 y1 y2 yn – 1 yn
Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 191 STPM PRACTICE 6 1. (a) Using the trapezium rule with six ordinates, find the approximate value of ∫ 0 1 x —1 2 e–2x dx correct to five decimal places. (b) Given that ∫ 0 1 x – —1 2 e–2x = Ae–2 + 4 ∫ 0 1 x —1 2 e–2x dx. Determine the value of the constant A. Hence, deduce the approximate value of ∫ 0 1 x –—1 2 e–2x dx correct to four decimal places. (c) Explain briefly why trapezium value cannot be applied to approximate the value of ∫ 0 1 x –—1 2 e–2x dx. 2. Sketch the graphs of y = ln x and y = 3 – x for 1 < x < 3 on the same coordinate axes. Show that the equation ln x = 3 – x has a root between x = 2 and x = 3. Use an iterative method to find this root correct to 3 decimal places. 3. On the same coordinates axes, sketch the curve y = e—1 2 x and y = 3 sin 2x for 0 < x < π. Deduce that e —1 2 x – 3 sin 2x has two positive real roots. Using Newton-Raphson method, with initial approximation x0 = 1.2, find the largest positive real roots correct to three decimal places. 4. Show that the cubic equation x3 – 6x + 3 = 0 has a root a, such that 2 , a , 3. Taking x0 = 2 as a first approximation to a, use the Newton-Raphson method to obtain a correct to three decimal places. 5. Sketch on the same coordinate axes, the graphs of y = ln x and y = 4 – x. Hence, show that the equation ln x = 4 – x has only one real root and prove that this root lies between 2.9 and 3. Taking x0 = 2.9 as a first approximation to this root, use the Newton-Raphson method to find a second approximation to the root correct to 3 decimal places. 6. Show that the equation (x + 1)5 = (x + 2)3 + 4 has a root between 0.8 and 1. Given that x1 is the first approximation to the root, use the Newton-Raphson method to find an expression for x2 , the second approximation to the root, in terms of x1 . By taking x1 as 1, find the root correct to 3 significant figures. 7. (a) Given that y = 3–x , show that dy dx = –3–x 1n3. (b) On the same coordinate axes, sketch the curves y = 4 – x2 and y = 3–x . (c) Verify that the curves intersect at the point A(–1, 3). The curve also intersect at point B in the first quadrant whose x-coordinate is the positive root of the equation x2 + 3–x – 4 = 0. (d) Verify that 1 , a , 2 by calculation. (e) By taking 2 as the first approximation to a, use the Newton-Raphson method once to find the second approximation, giving your answer to 2 decimal places. 8. Show that the equation x3 + 8x – 3 = 0 has a real root in the interval [0, 1]. Determine which of the iterations Xn + 1 = 1 8 (3 – Xn 3 ) and Xn + 1 = 3 – 8x xn 2 is more likely to give a convergent sequence of approximation to a root in the interval [0, 1]. Use your choice with initial approximation x0 = 1 to determine the root correct to three decimal places.
192 Mathematics Semester 2 STPM Chapter 6 Numerical Methods 6 9. Sketch, on the same coordinate axes, the graphs y = 2ex and y = 4 – x2 . Show that the equation 2ex = 4 – x2 has exactly one root, a, in the interval [0, 1]. Taking a1 = 0.45 as a first approximation, use the Newton-Raphson method once to obtain a second approximation, a2 to this root. 10. By using the trapezium rule with five ordinates, estimate the value of ∫ 0 4 1 1 + x dx correct to four decimal places. 11. Use the trapezium rule with ordinates at π 6 , π 4 , π 3 , 5π 12 and π 2 to estimate the value of ∫ —π 6 —π 2 sin q dq. 12. (a) Using the trapezium rule with 3 ordinates, show that ∫ –1 1 1 1 + e–x dx is approximately 1. (b) By means of the substitution u = ex , show that the estimate obtained in (i) is correct. 13. Given that f(x) = 2x3 – 7x2 + x + k and (x – 2) is a factor of f(x), find the value of k and factorise f(x) completely. Sketch the curve y = f(x). (You are not required to find the coordinates of the stationary points). Use the trapezium rule with 4 ordinates to find an approximation to ∫ 2 –1 f(x) dx. 14. (a) If I = ∫ 0 1 (x2 + 1)–—3 2 dx, use the trapezium rule with 3 ordinates to estimate the value of I, giving your answer correct to two significant figures. (b) By using the trapezium rule with the same ordinates as part (a), estimate the volume of solid formed when the region bounded by the curve y = (x2 + 1)–—3 2 , the axes and the line x = 1 is revolved completely about the x-axis, giving your answer correct to two significant figures. 15. Show that the equation x3 + x2 – 20 = 0 has a root between 2 and 3. Given that x0 = 2 as an initial approximation, use the Newton-Raphson method to find the root correct to three decimal places. 16. By using trapezium rule with five ordinates, estimates the value of ∫ 0 1 √ — 4 – x2 dx correct to three decimal places. 17. Sketch on the same diagram, the graphs of y = e–x and y = 3x – 4. Show that the equation (3x – 4)ex – 1 = 0 has only one real root. Use the Newton-Raphson method with initial approximation x0 = 1.5, find the root of the equation correct to three decimal places. 18. Show that the equation x3 – 6x2 + 30 = 0 has a root between 4 and 5. Given initial value x0 = 4.5, use Newton-Raphson method to find the root correct to three decimal places.
193 MATHEMATICS (T) PAPER 2 (One and a half hours) Instruction to candidates: Answer all questions in Section A and any one question in Section B. Answers may be written either in English or Bahasa Malaysia. All necessary working should be shown clearly. Scientific calculators may be used. Programmable and graphic display calculators are prohibited. Section A [45 marks] Answer all questions in this section. 1. Given the function f(x) = 1 – x 3 + x . (a) Determine the intervals for the function f(x) to be defined. [3 marks] (b) Shows algebraically that the function is continuous on that interval. [3 marks] 2. Evaluate the following definite integral: (a) ∫ 0 1 2ex ex + 2e–x dx [3 marks] (b) ∫ 0 1 x5 ex3 + 1 dx [3 marks] 3. Diagram below shows part of the graph y2 = x – 2. The region in the first quadrant bounded by the axes, the curves and the line y = 3 is denoted by R. 0 3 y R y2 = x – 2 x 2 Find the volume of the solid generated when R is rotated through one revolution (a) about the y-axis. [3 marks] (b) about the x-axis. [3 marks] STPM Model Paper (954/2)
194 Mathematics Semester 2 STPM Model Paper 4. Using the substitution u = xy, transform the differential equation x2 dy dx + xy – y2 = 0 into a differential equation relating u and x. [6 marks] Hence, find the particular solution of the differential equation given that y = x = 1. [3 marks] 5. Given that f(x) = e√ —1 + x . Show that 4(1 + x)f "(x) + 2f '(x) = f(x). [6 marks] By successive differentiation of this result, find the series expansion of f(x), in ascending power of x, up to the term in x3 . [3 marks] 6. Sketch on the same coordinates axes, the graphs of y = ln(x – 1) and y = –(x + 1). Show that the equation ln(x – 1) + x + 1 = 0 has only one real root in the interval [1.1, 2]. [3 marks] Verify your answer algebraically. [3 marks] Use the Newton-Raphson method with initial estimate x0 = 1.1 to estimate this real root, correct to three decimal places. [3 marks] Section B [15 marks] Answer any one question in this section. 7. Find the general solution of differential equation dy dx + 2xy = 2x(1 + x2 ). Express your answer in the form y = f(x). [5 marks] (a) Show that the solution curves of the differential equation have (i) one stationary point when c < 1. [3 marks] (ii) three stationary points when c . 1. [3 marks] (b) Sketch on a single diagram two solution curves, corresponding to c . 1 and c , 1. [4 marks] 8. (a) A ball is thrown vertically downwards at rest. After time t, its velocity v m s –1 is given by the differential equation dv dt = 3(5 – v). (i) Express v in terms of t. [4 marks] (ii) Determine the distance travelled by the ball after 4 s. [4 marks] (d) A solid right circular cylinder of radius r cm and height h cm has volume 432π cm3 and total surface area s cm2 . (i) Express s in terms of r. [3 marks] (ii) Determine the value of r that makes the surface area a maximum or a minimum. [4 marks]
195 1 Limits and Continuity Exercise 1.1 1. (a) 6 (b) 4 (c) 3 (d) –3 2. (a) 1 4 (b) 7 3 (c) – 1 2 (c) 6 3. (a) 1 (b) 3 (c) 1 2 (d) 1 3 4. (a) 3 (b) 1 (c) 4 (d) ∞ (e) 15 4 (f) 2 5 5. (a) 1 5 (b) –1 6. (a) –1 (b) 1 (c) does not exist 7. (a) 9 (b) –10 (c) –1 (d) 5 (e) does not exist 8. (a) 2 (b) 2 (c) 2 (d) 2 (e) –3 (f) does not exist Exercise 1.2 1. x 0 f(x ) _ 2 π– _π 2 continuous 2. 0 1 2 3 1 – – 4 f(x) x 1 2 continuous 3. (a) (b) x π y 0 2π x π y 0 continuous not continuous (c) (d) x y 0 x y –1 0 1 continuous continuous (e) (f) x y –1 0 x y 1 –2 0 continuous not continuous 4. (a) 1, –1; Not continuous (b) x y 5 –1 1 (c) does not exist 5. (a) does not exist; f is not continuous (b) 0 x y –1 5 6. Yes x y 1 0 1 2 3 4 5 2 1 2 –1 – ANSWERS