About the HELM Project HELM (Helping Engineers Learn Mathematics) materials were the outcome of a three-year curriculum development project

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2 About the HELM Project HELM (Helping Engineers Learn Mathematics) materials were the outcome of a three-year curriculum development project undertaken by a consortium of five English universities led by Loughborough University, funded by the Higher Education Funding Council for England under the Fund for the Development of Teaching and Learning for the period October 00 September 005. HELM aims to enhance the mathematical education of engineering undergraduates through a range of flexible learning resources in the form of Workbooks and web-delivered interactive segments. HELM supports two CAA regimes: an integrated web-delivered implementation and a CD-based version. HELM learning resources have been produced primarily by teams of writers at six universities: Hull, Loughborough, Manchester, Newcastle, Reading, Sunderland. HELM gratefully acknowledges the valuable support of colleagues at the following universities and colleges involved in the critical reading, trialling, enhancement and revision of the learning materials: Aston, Bournemouth & Poole College, Cambridge, City, Glamorgan, Glasgow, Glasgow Caledonian, Glenrothes Institute of Applied Technology, Harper Adams University College, Hertfordshire, Leicester, Liverpool, London Metropolitan, Moray College, Northumbria, Nottingham, Nottingham Trent, Oxford Brookes, Plymouth, Portsmouth, Queens Belfast, Robert Gordon, Royal Forest of Dean College, Salford, Sligo Institute of Technology, Southampton, Southampton Institute, Surrey, Teesside, Ulster, University of Wales Institute Cardiff, West Kingsway College (London), West Notts College. HELM Contacts: Post: HELM, Mathematics Education Centre, Loughborough University, Loughborough, LE 3TU. helm@lboro.ac.uk Web: HELM Workbooks List Basic Algebra 6 Functions of a Complex Variable Basic Functions 7 Multiple Integration 3 Equations, Inequalities & Partial Fractions 8 Differential Vector Calculus 4 Trigonometry 9 Integral Vector Calculus 5 Functions and Modelling 30 Introduction to Numerical Methods 6 Exponential and Logarithmic Functions 3 Numerical Methods of Approximation 7 Matrices 3 Numerical Initial Value Problems 8 Matrix Solution of Equations 33 Numerical Boundary Value Problems 9 Vectors 34 Modelling Motion 0 Complex Numbers 35 Sets and Probability Differentiation 36 Descriptive Statistics Applications of Differentiation 37 Discrete Probability Distributions 3 Integration 38 Continuous Probability Distributions 4 Applications of Integration 39 The Normal Distribution 5 Applications of Integration 40 Sampling Distributions and Estimation 6 Sequences and Series 4 Hypothesis Testing 7 Conics and Polar Coordinates 4 Goodness of Fit and Contingency Tables 8 Functions of Several Variables 43 Regression and Correlation 9 Differential Equations 44 Analysis of Variance 0 Laplace Transforms 45 Non-parametric Statistics z-transforms 46 Reliability and Quality Control Eigenvalues and Eigenvectors 47 Mathematics and Physics Miscellany 3 Fourier Series 48 Engineering Case Studies 4 Fourier Transforms 49 Student s Guide 5 Partial Differential Equations 50 Tutor s Guide Copyright Loughborough University, 006

3 Contents 6 Sequences and Series 6. Sequences and Series 6. Infinite Series The Binomial Series Power Series Maclaurin and Taylor Series 40 Learning outcomes In this Workbook you will learn about sequences and series. You will learn about arithmetic and geometric series and also about infinite series. You will learn how to test the for the convergence of an infinite series. You will then learn about power series, in particular you will study the binomial series. Finally you will apply your knowledge of power series to the process of finding series expansions of functions of a single variable. You will be able to find the Maclaurin and Taylor series expansions of simple functions about a point of interest.

4 Sequences and Series 6. Introduction In this Section we develop the ground work for later Sections on infinite series and on power series. We begin with simple sequences of numbers and with finite series of numbers. We introduce the summation notation for the description of series. Finally, we consider arithmetic and geometric series and obtain expressions for the sum of n terms of both types of series. Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... understand and be able to use the basic rules of algebra be able to find limits of algebraic expressions check if a sequence of numbers is convergent use the summation notation to specify series recognise arithmetic and geometric series and find their sums HELM (008): Workbook 6: Sequences and Series

5 Infinite Series 6. Introduction We extend the concept of a finite series, met in Section 6., to the situation in which the number of terms increase without bound. We define what is meant by an infinite series being convergent by considering the partial sums of the series. As prime examples of infinite series we examine the harmonic and the alternating harmonic series and show that the former is divergent and the latter is convergent. We consider various tests for the convergence of series, in particular we introduce the ratio test which is a test applicable to series of positive terms. Finally we define the meaning of the terms absolute and conditional convergence. Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... be able to use the summation notation be familiar with the properties of limits be able to use inequalities use the alternating series test on infinite series use the ratio test on infinite series understand the terms absolute and conditional convergence HELM (008): Section 6.: Infinite Series 3

6 The Binomial Series 6.3 Introduction In this Section we examine an important example of an infinite series, the binomial series: +px + p(p ) x +! p(p )(p ) x 3 + 3! We show that this series is only convergent if x < and that in this case the series sums to the value ( + x) p. As a special case of the binomial series we consider the situation when p is a positive integer n. In this case the infinite series reduces to a finite series and we obtain, by replacing x with b, the binomial theorem: a (b + a) n = b n + nb n n(n ) a + b n a + + a n.! Finally, we use the binomial series to obtain various polynomial expressions for ( + x) p when x is small. Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... understand the factorial notation have knowledge of the ratio test for convergence of infinite series. understand the use of inequalities recognise and use the binomial series state and use the binomial theorem use the binomial series to obtain numerical approximations 6 HELM (008): Workbook 6: Sequences and Series

7 . The binomial series A very important infinite series which occurs often in applications and in algebra has the form: +px + p(p ) x +! p(p )(p ) x 3 + 3! in which p is a given number and x is a variable. By using the ratio test it can be shown that this series converges, irrespective of the value of p, as long as x <. In fact, as we shall see in Section 6.5 the given series converges to the value ( + x) p as long as x <. ( + x) p =+px + Key Point 9 The Binomial Series p(p ) x +! p(p )(p ) x 3 + x < 3! The binomial theorem can be obtained directly from the binomial series if p is chosen to be a positive integer (here we need not demand that x < as the series is now finite and so is always convergent irrespective of the value of x). For example, with p =we obtain ( + x) = +x + () x = +x + x as is well known. With p =3we get ( + x) 3 = +3x + 3() x + 3()() x ! = +3x +3x + x 3 Generally if p = n (a positive integer) then ( + x) n =+nx + n(n ) x +! n(n )(n ) x x n 3! which is a form of the binomial theorem. If x is replaced by b a then + b a n =+n b n(n ) + a! b + + a n b a Now multiplying both sides by a n we have the following Key Point: HELM (008): Section 6.3: The Binomial Series 7

8 Power Series 6.4 Introduction In this Section we consider power series. These are examples of infinite series where each term contains a variable, x, raised to a positive integer power. We use the ratio test to obtain the radius of convergence R, of the power series and state the important result that the series is absolutely convergent if x <R, divergent if x >Rand may or may not be convergent if x = ±R. Finally, we extend the work to apply to general power series when the variable x is replaced by (x x 0 ). Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... have knowledge of infinite series and of the ratio test have knowledge of inequalities and of the factorial notation. explain what a power series is obtain the radius of convergence for a power series explain what a general power series is 3 HELM (008): Workbook 6: Sequences and Series

9 Maclaurin and Taylor Series 6.5 Introduction In this Section we examine how functions may be expressed in terms of power series. This is an extremely useful way of expressing a function since (as we shall see) we can then replace complicated functions in terms of simple polynomials. The only requirement (of any significance) is that the complicated function should be smooth; this means that at a point of interest, it must be possible to differentiate the function as often as we please. Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... have knowledge of power series and of the ratio test be able to differentiate simple functions be familiar with the rules for combining power series find the Maclaurin and Taylor series expansions of given functions find Maclaurin expansions of functions by combining known power series together find Maclaurin expansions by using differentiation and integration 40 HELM (008): Workbook 6: Sequences and Series

10 . Maclaurin and Taylor series As we shall see, many functions can be represented by power series. In fact we have already seen in earlier Sections examples of such a representation: x = +x + x + x < ln( + x) = x x + x3 <x 3 e x = +x + x! + x3 + all x 3! The first two examples show that, as long as we constrain x to lie within the domain x < (or, equivalently, <x<), then in the first case has the same numerical value as x +x + x + and in the second case ln(+x) has the same numerical value as x x + x3 3. In the third example we see that e x has the same numerical value as +x + x + but in this! case there is no restriction to be placed on the value of x since this power series converges for all values of x. Figure 5 shows this situation geometrically. As more and more terms are used from the series +x + x! + x3 3! the curve representing ex is a better and better approximation. In (a) we show the linear approximation to e x. In (b) and (c) we show, respectively, the quadratic and cubic approximations. y + x + x + x! y + x + x! + x3 3! y e x e x e x x x (a) (b) (c) x Figure 5: Linear, quadratic and cubic approximations to e x These power series representations are extremely important, from many points of view. Numerically, we can simply replace the function x by the quadratic expression +x + x as long as x is so small that powers of x greater than or equal to 3 can be ignored in comparison to quadratic terms. This approach can be used to approximate more complicated functions in terms of simpler polynomials. Our aim now is to see how these power series expansions are obtained. HELM (008): Section 6.5: Maclaurin and Taylor Series 4

11 . The Maclaurin series Consider a function f(x) which can be differentiated at x =0as often as we please. For example e x, cos x, sin x would fit into this category but x would not. Let us assume that f(x) can be represented by a power series in x: f(x) =b 0 + b x + b x + b 3 x 3 + b 4 x 4 + = b p x p where b 0,b,b,... are constants to be determined. If we substitute x =0then, clearly f(0) = b 0 The other constants can be determined by further differentiating and, on each differentiation, substituting x =0. For example, differentiating once: f (x) =0+b +b x +3b 3 x +4b 4 x 3 + so, putting x =0, we have f (0) = b. Continuing to differentiate: so Further: f (x) =0+b + 3()b 3 x + 4(3)b 4 x + f (0) = b or b = f (0) p=0 f (x) = 3()b 3 +4(3)()b 4 x+ so f (0) = 3()b 3 implying b 3 = 3() f (0) Continuing in this way we easily find that (remembering that 0! = ) b n = n! f (n) (0) n =0,,,... where f (n) (0) means the value of the n th derivative at x =0and f (0) (0) means f(0). Bringing all these results together we have: If f(x) can be differentiated as often as required: Key Point 4 Maclaurin Series f(x) =f(0) + xf (0) + x! f (0) + x3 3! f (0) + = This is called the Maclaurin expansion of f(x). p=0 x p p! f (p) (0) 4 HELM (008): Workbook 6: Sequences and Series

12 Example 4 Find the Maclaurin expansion of cos x. Solution Here f(x) = cos x and, differentiating a number of times: f(x) = cos x, f (x) = sin x, f (x) = cos x, f (x) = sin x etc. Evaluating each of these at x =0: f(0) =, f (0) = 0, f (0) =, f (0) = 0 etc. Substituting into f(x) =f(0) + xf (0) + x! f (0) + x3 3! f (0) +, gives: cos x = x! + x4 4! x6 6! + The reader should confirm (by finding the radius of convergence) that this series is convergent for all values of x. The geometrical approximation to cos x by the first few terms of its Maclaurin series are shown in Figure 6. y y x! + x4 4! y cos x x cos x x! x cos x x Figure 6: Linear, quadratic and cubic approximations to cos x Task Find the Maclaurin expansion of ln( + x). (Note that we cannot find a Maclaurin expansion of the function ln x since ln x does not exist at x =0and so cannot be differentiated at x =0.) Find the first four derivatives of f(x) = ln( + x): f (x) = f (x) = f (x) = f (x) = HELM (008): Section 6.5: Maclaurin and Taylor Series 43

13 f (x) = +x, f (x) = ( + x), f (x) = generally: f (n) (x) = ( )n+ (n )! ( + x) n ( + x) 3, Now obtain f(0), f (0), f (0), f (0): f(0) = f (0) = f (0) = f (0) = f(0) = 0 f (0) =, f (0) =, f (0) =, generally: f (n) (0) = ( ) n+ (n )! Hence, obtain the Maclaurin expansion of ln( + x): ln( + x) = ln( + x) =x x + x3 3 ( )n+...+ x n + (This was obtained in Section 6.4, page 37.) n Now obtain the radius of convergence and consider the situation at the boundary values: Radius of convergence R = R =. Also at x =the series is convergent (alternating harmonic series) and at x = the series is divergent. Hence this Maclaurin expansion is only valid if <x. The geometrical closeness of the polynomial terms with the function ln( + x) for <x is displayed in Figure 7: y x y y x x + x3 3 ln( + x) ln( + x) ln( + x) x x x x x Figure 7: Linear, quadratic and cubic approximations to ln( + x) 44 HELM (008): Workbook 6: Sequences and Series

14 Note that when x = ln = + 3 so the alternating harmonic series converges to 4 ln 0.693, as stated in Section 6., page 7. The Maclaurin expansion of a product of two functions: f(x)g(x) is obtained by multiplying together the Maclaurin expansions of f(x) and of g(x) and collecting like terms together. The product series will have a radius of convergence equal to the smaller of the two separate radii of convergence. Example 5 Find the Maclaurin expansion of e x ln( + x). Solution Here, instead of finding the derivatives of f(x) =e x ln(+x), we can more simply multiply together the Maclaurin expansions for e x and ln( + x) which we already know: and e x =+x + x! + x3 + all x 3! ln( + x) =x x + x3 + <x 3 The resulting power series will only be convergent if <x. Multiplying: e x ln( + x) = +x + x x! + x3 3! + x + x3 3 + = x x + x3 3 x x x3 + x x3 x4 4 + x4 6 = x + x + x x5 + <x 40 (You must take care not to miss relevant terms when carrying through the multiplication.) HELM (008): Section 6.5: Maclaurin and Taylor Series 45

15 Task Find the Maclaurin expansion of cos x up to powers of x 4. Hence write down the expansion of sin x to powers of x 6. First, write down the expansion of cos x: cos x = cos x = x! + x4 4! + Now, by multiplication, find the expansion of cos x: cos x = cos x = x! + x4 4! x! + x4 4! = ( x! + x4 4! )+( x! + x4 4 )+(x4 4! )+ = x + x4 3 x6 45 Now obtain the expansion of sin x using a suitable trigonometric identity: sin x = sin x = cos x = x + x4 3 x = x x4 3 + x As an alternative approach the reader could obtain the power series expansion for cos x by using the trigonometric identity cos x ( + cos x). 46 HELM (008): Workbook 6: Sequences and Series

16 Example 6 Find the Maclaurin expansion of tanh x up to powers of x 5. Solution The first two derivatives of f(x) = tanh x are f (x) =sech x f (x) = sech x tanh x f (x) =4sech x tanh x sech 4 x giving f(0) = 0, f (0), f (0) = 0 f (0) = This leads directly to the Maclaurin expansion as tanh x = 3 x3 + 5 x5 Example 7 The relationship between the wavelength, L, the wave period, T, and the water depth, d, for a surface wave in water is given by: L = gt πd π tanh L In a particular case the wave period was 0 s and the water depth was 6. m. Taking the acceleration due to gravity, g, as 9.8 m s determine the wave length. [Hint: Use the series expansion for tanh x developed in Example 6.] Solution Substituting for the wave period, water depth and g we get π 6. L = tanh = π tanh π L π L The series expansion of tanh x is given by tanh x = x x3 3 + x5 5 + Using the series expansion of tanh x we can approximate the equation as.π L = π + π L 3 L Multiplying through by πl 3 the equation becomes πl 4 = πL (.π) 3 This equation can be rewritten as L L = 0 Solving this as a quadratic in L we get L = 74 m. Using Newton-Raphson iteration this can be further refined to give a wave length of 73.9 m. HELM (008): Section 6.5: Maclaurin and Taylor Series 47

17 3. Differentiation of Maclaurin series We have already noted that, by the binomial series, x =+x + x + x 3 + x < Thus, with x replaced by x +x = x + x x 3 + x < We have previously obtained the Maclaurin expansion of ln( + x): ln( + x) =x x + x3 3 x4 + <x 4 Now, we differentiate both sides with respect to x: +x = x + x x 3 + This result matches that found from the binomial series and demonstrates that the Maclaurin expansion of a function f(x) may be differentiated term by term to give a series which will be the Maclaurin expansion of df dx. As we noted in Section 6.4 the derived series will have the same radius of convergence as the original series. Task Find the Maclaurin expansion of ( x) 3 and state its radius of convergence. First write down the expansion of ( x) : x x =+x + x + x < Now, by differentiation, obtain the expansion of ( x) = d = dx x ( x) : ( x) = d dx ( + x + x + )=+x +3x +4x 3 48 HELM (008): Workbook 6: Sequences and Series

18 Differentiate again to obtain the expansion of ( x) 3 : ( x) = 3 = d dx ( x) = [ ] ( x) = d = 3 dx ( x) [ + 6x +x +0x 3 + ]=+3x +6x + 0x 3 + Finally state its radius of convergence: The final series: +3x +6x + 0x 3 + has radius of convergence R =since the original series has this radius of convergence. This can also be found directly using the formula R = lim b n and using the fact that the coefficient of the n th term is b n = n(n +). n b n+ 4. The Taylor series The Taylor series is a generalisation of the Maclaurin series being a power series developed in powers of (x x 0 ) rather than in powers of x. Thus Key Point 5 Taylor Series If the function f(x) can be differentiated as often as required at x = x 0 then: f(x) =f(x 0 )+(x x 0 )f (x 0 )+ (x x 0) f (x 0 )+! This is called the Taylor series of f(x) about the point x 0. The reader will see that the Maclaurin expansion is the Taylor expansion obtained if x 0 is chosen to be zero. HELM (008): Section 6.5: Maclaurin and Taylor Series 49

19 Task Obtain the Taylor series expansion of series in powers of (x ).) x about x =. (That is, find a power First, obtain the first three derivatives and the n th derivative of f(x) = x : f (x) = f (x) = f (x) = f (n) (x) = f (x) = ( x), f (x) = Now evaluate these derivatives at x 0 =: ( x) 3, f (x) = 6 ( x) 4, f (n) (x) = f () = f () = f () = f (n) () = n! ( x) n+ f () =, f () =, f () = 6, f (n) () = ( ) n+ n! Hence, write down the Taylor expansion of f(x) = x x = about x =: x = +(x ) (x ) +(x ) 3 + +( ) n+ (x ) n + 50 HELM (008): Workbook 6: Sequences and Series

20 Exercises. Show that the series obtained in the last Task is convergent if x <.. Sketch the linear, quadratic and cubic approximations to last task and compare to x. x obtained from the series in the. In the following diagrams some of the terms from the Taylor series are plotted to compare with ( x). y x + (x ) y x x + (x ) (x ) x y + (x ) (x ) + (x ) 3 x x HELM (008): Section 6.5: Maclaurin and Taylor Series 5

21 . Power series A power series is simply a sum of terms each of which contains a variable raised to a non-negative integer power. To illustrate: x x 3 + x 5 x 7 + +x + x! + x3 3! + are examples of power series. In the binomial series: +px + p(p ) x +! 6.3 we encountered an important example of a power series, p(p )(p ) x 3 + 3! which, as we have already noted, represents the function ( + x) p as long as the variable x satisfies x <. A power series has the general form b 0 + b x + b x + = b p x p p=0 where b 0,b,b, are constants. Note that, in the summation notation, we have chosen to start the series at p =0. This is to ensure that the power series can include a constant term b 0 since x 0 =. The convergence, or otherwise, of a power series, clearly depends upon the value of x chosen. For example, the power series + x + x 3 + x3 4 + is convergent if x = (for then it is the alternating harmonic series) and divergent if x = + (for then it is the harmonic series).. The radius of convergence The most important statement one can make about a power series is that there exists a number, R, called the radius of convergence, such that if x <Rthe power series is absolutely convergent and if x >Rthe power series is divergent. At the two points x = R and x = R the power series may be convergent or divergent. HELM (008): Section 6.4: Power Series 33

22 Key Point Convergence of Power Series For a power series b p x p with radius of convergence R then p=0 the series converges absolutely if x <R the series diverges if x >R the series may be convergent or divergent at x = ±R divergent R 0 convergent R divergent x For any particular power series know, from the ratio test that b p+ x p+ lim p b p x p = lim p b p+ b p b p x p the value of R can be obtained using the ratio test. We p=0 b p x p is absolutely convergent if p=0 x < implying x < lim p b p b p+ and so R = lim p b p b p+. Example (a) Find the radius of convergence of the series + x + x 3 + x3 4 + (b) Investigate what happens at the end-points x =, x= + of the region of absolute convergence. 34 HELM (008): Workbook 6: Sequences and Series

23 Solution (a) Here + x + x 3 + x3 4 + = x p p + so b p = p + In this case, R = lim p + p p + = p=0 b p+ = p + so the given series is absolutely convergent if x < and is divergent if x >. (b) At x = + the series is which is divergent (the harmonic series). However, at 3 x = the series is + + which is convergent (the alternating harmonic series). 3 4 Finally, therefore, the series + x + x 3 + x3 4 + is convergent if x<. Task Find the range of values of x for which the following power series converges: + x 3 + x 3 + x First find the coefficient of x p : b p = b p = 3 p Now find R, the radius of convergence: R = lim p b p b p+ = R = lim p b p b p+ = lim 3 p+ p 3 p = lim (3) = 3. p When x = ±3 the series is clearly divergent. Hence the series is convergent only if 3 <x<3. HELM (008): Section 6.4: Power Series 35

24 3. Properties of power series Let P and P represent two power series with radii of convergence R and R respectively. We can combine P and P together by addition and multiplication. We find the following properties: Key Point If P and P are power series with respective radii of convergence R and R then the sum (P +P ) and the product (P P ) are each power series with the radius of convergence being the smaller of R and R. Power series can also be differentiated and integrated on a term by term basis: Key Point 3 If P is a power series with radius of convergence R then d dx (P ) and (P ) dx are each power series with radius of convergence R Example 3 Using the known result that ( + x) p =+px + p(p ) x + x <,! choose p = and by differentiating obtain the power series expression for (+x). Solution ( + x) x =+ + Differentiating both sides: x +! 3 3! ( + x) = + Multiplying through by : ( + x) = x + x 3 + x + 3 x + 3 x + This result can, of course, be obtained directly from the expansion for ( + x) p with p =. 36 HELM (008): Workbook 6: Sequences and Series

25 Task Using the known result that +x = x + x x 3 + x <, (a) Find an expression for ln( + x) (b) Use the expression to obtain an approximation to ln(.) (a) Integrate both sides of dx +x = ( x + x ) dx = +x = x + x and so deduce an expression for ln( + x): dx = ln( + x)+c where c is a constant of integration, +x ( x + x ) dx = x x + x3 + k where k is a constant of integration. 3 So we conclude ln( + x)+c = x x + x3 + k if x < 3 Choosing x =0shows that c = k so they cancel from this equation. (b) Now choose x =0. to approximate ln( + 0.) using terms up to cubic: ln(.) = 0. (0.) + (0.)3 3 ln(.) which is easily checked by calculator. HELM (008): Section 6.4: Power Series 37

26 4. General power series A general power series has the form b 0 + b (x x 0 )+b (x x 0 ) + = b p (x x 0 ) p p=0 Exactly the same considerations apply to this general power series as apply to the special series b p x p except that the variable x is replaced by (x x 0 ). The radius of convergence of the general p=0 series is obtained in the same way: R = lim b p p b p+ and the interval of convergence is now shifted to have centre at x = x 0 (see Figure 4 below). The series is absolutely convergent if x x 0 <R, diverges if x x 0 >Rand may or may not converge if x x 0 = R. divergent 0 x 0 R x 0 x 0 +R convergent divergent Figure 4 x Task Find the radius of convergence of the general power series (x ) + (x ) (x ) 3 + First find an expression for the general term: (x ) + (x ) (x ) 3 + = p=0 (x ) p ( ) p so b p =( ) p p=0 Now obtain the radius of convergence: lim b p = R = p b p+ lim b p p b p+ = lim ( ) p p ( ) p+ =. Hence R =, so the series is absolutely convergent if x <. 38 HELM (008): Workbook 6: Sequences and Series

27 Finally, decide on the convergence at x =(i.e. at x = and x =i.e. x =0and x =): At x =0the series is +++ which diverges and at x =the series is + which also diverges. Thus the given series only converges if x < i.e. 0 <x<. divergent 0 convergent divergent x. From the result Exercises x =+x + x + x , x < (a) Find an expression for ln( x) (b) Use this expression to obtain an approximation to ln(0.9) to 4 d.p.. Find the radius of convergence of the general power series (x+)+(x+) (x+) Find the range of values of x for which the power series + x 4 + x 4 + x converges By differentiating the series for ( + x) /3 find the power series for ( + x) /3 and state its radius of convergence. 5. (a) Find the radius of convergence of the series + x 3 + x 4 + x s (b) Investigate what happens at the points x = and x = +. ln( x) = x x x3 3 x ln(0.9) (4 d.p.). R =. Series converges if 3 <x<. If x = series diverges. If x = 3 series diverges. 3. Series converges if 4 <x<4. 4. ( + x) /3 = 3 x x +... valid for x <. 5. (a) R =. (b) At x = + series diverges. At x = series converges. HELM (008): Section 6.4: Power Series 39

28 Key Point 0 The Binomial Theorem If n is a positive integer then the expansion of (a + b) raised to the power n is given by: (a + b) n = a n + na n b + This is known as the binomial theorem. n(n ) a n b + + b n! Task Use the binomial theorem to obtain (a) ( + x) 7 (b) (a + b) 4 (a) Here n =7: ( + x) 7 = ( + x) 7 =+7x +x +35x 3 +35x 4 +x 5 +7x 6 + x 7 (b) Here n =4: (a + b) 4 = (a + b) 4 = a 4 +4a 3 b +6a b +4ab 3 + b 4. Task Given that x is so small that powers of x 3 and above may be ignored in comparison to lower order terms, find a quadratic approximation of ( x) and check for accuracy your approximation for x =0.. First expand ( x) ( x) = using the binomial series with p = and with x replaced by ( x): 8 HELM (008): Workbook 6: Sequences and Series

29 ( x) = x + ( ) x ( )( 3 ) 6 x 3 + Now obtain the quadratic approximation: ( x) ( x) x 8 x Now check on the validity of the approximation by choosing x =0.: On the left-hand side we have (0.9) = to 5 d.p. obtained by calculator whereas, using the quadratic expansion: (0.9) (0.) 8 (0.) = 0.05 (0.005) = so the error is only What we have done in this last Task is to replace (or approximate) the function ( x) by the simpler (polynomial) function x 8 x which is reasonable provided x is very small. This approximation is well illustrated geometrically by drawing the curves y = ( x) curves coincide when x is small. See Figure : and y = x 8 x. The two ( x) y x 8 x x Figure HELM (008): Section 6.3: The Binomial Series 9

30 Task Obtain a cubic approximation of using appropriate values of x.. Check your approximation for accuracy ( + x) First write the term ( + x) = ( + x) +x = + x = in a form suitable for the binomial series (refer to Key Point 9): + x Now expand using the binomial series with p = and x instead of x, to include terms up to x3 : + x = + x = +( ) x + ( )( ) x ( )( )( 3) x 3 +! 3! = x 4 + x 8 x3 6 State the range of x for which the binomial series of The series is valid if + x is valid: valid as long as x < i.e. x < or <x< 30 HELM (008): Workbook 6: Sequences and Series

31 Choose x = 0. to check the accuracy of your approximation: + 0. = = = to 5 d.p = Figure 3 below illustrates the close correspondence (when x is small ) between the curves y = ( + x ) and y = x 4 + x 8 x3 6. x 4 + x 8 x3 6 y ( + x) x Figure 3 Exercises. Determine the expansion of each of the following (a) (a + b) 3, (b) ( x) 5, (c) ( + x ), (d) ( x) /3.. Obtain a cubic approximation (valid if x is small) of the function ( + x) 3/. s. (a) (a + b) 3 = a 3 +3a b +3ab + b 3 (b) ( x) 5 = 5x +0x 0x 3 +5x 4 x 5 (c) ( + x ) = x + x 4 x 6 + (d) ( x) /3 = 3 x 9 x 5 8 x3 +. ( + x) 3/ =+3x + 3 x x3 + HELM (008): Section 6.3: The Binomial Series 3

32 . Introduction Many of the series considered in Section 6. were examples of finite series in that they all involved the summation of a finite number of terms. When the number of terms in the series increases without bound we refer to the sum as an infinite series. Of particular concern with infinite series is whether they are convergent or divergent. For example, the infinite series ++++ is clearly divergent because the sum of the first n terms increases without bound as more and more terms are taken. It is less clear as to whether the harmonic and alternating harmonic series: converge or diverge. Indeed you may be surprised to find that the first is divergent and the second is convergent. What we shall do in this Section is to consider some simple convergence tests for infinite series. Although we all have an intuitive idea as to the meaning of convergence of an infinite series we must be more precise in our approach. We need a definition for convergence which we can apply rigorously. First, using an obvious extension of the notation we have used for a finite sum of terms, we denote the infinite series: a + a + a a p + by the expression where a p is an expression for the p th term in the series. So, as examples: p= a p ++3+ = = p p= since the p th term is a p p p since the p th term is a p p p= = Consider the infinite series: p= a + a + + a p + = ( ) p+ p= p a p here a p ( )p+ p We consider the sequence of partial sums, S,S,..., of this series where S = a S = a + a. S n = a + a + + a n That is, S n is the sum of the first n terms of the infinite series. If the limit of the sequence S,S,...,S n,... can be found; that is 4 HELM (008): Workbook 6: Sequences and Series

33 lim S n = S n (say) then we define the sum of the infinite series to be S: S = p= a p and we say the series converges to S. Another way of stating this is to say that n a p = lim a p n p= p= An infinite series Key Point 4 Convergence of Infinite Series a p is convergent if the sequence of partial sums p= S,S,S 3,...,S k,... in which S k = k p= a p is convergent Divergence condition for an infinite series An almost obvious requirement that an infinite series should be convergent is that the individual terms in the series should get smaller and smaller. This leads to the following Key Point: The condition: Key Point 5 a p 0 as p increases (mathematically lim p a p = 0) is a necessary condition for the convergence of the series p= a p It is not possible for an infinite series to be convergent unless this condition holds. HELM (008): Section 6.: Infinite Series 5

34 Task Which of the following series cannot be convergent? (a) (b) (c) In each case, use the condition from Key Point 5: (a) a p = a p = p p lim p p + p + = Hence series is divergent. lim a p = p (b) a p = lim a p = p a p = lim p a p =0 p So this series may be convergent. Whether it is or not requires further testing. (c) a p = a p = ( )p+ p lim a p = p lim a p =0so again this series may be convergent. p Divergence of the harmonic series The harmonic series: has a general term a n = which clearly gets smaller and smaller as n. However, surprisingly, n the series is divergent. Its divergence is demonstrated by showing that the harmonic series is greater than another series which is obviously divergent. We do this by grouping the terms of the harmonic series in a particular way: HELM (008): Workbook 6: Sequences and Series

35 Now > = > = > = and so on. Hence the harmonic series satisfies: > The right-hand side of this inequality is clearly divergent so the harmonic series is divergent. Convergence of the alternating harmonic series As with the harmonic series we shall group the terms of the alternating harmonic series, this time to display its convergence. The alternating harmonic series is: S = This series may be re-grouped in two distinct ways. st re-grouping = each term in brackets is positive since > 3, 4 > and so on. So we easily conclude that S< 5 since we are subtracting only positive numbers from. nd re-grouping = Again, each term in brackets is positive since >, 3 > 4, 5 > 6 and so on. So we can also argue that S> since we are adding only positive numbers to the value of the first term,. The conclusion that is forced upon us is that <S< so the alternating series is convergent since its sum, S, lies in the range Section 6.5 that S = ln It will be shown in HELM (008): Section 6.: Infinite Series 7

36 . General tests for convergence The techniques we have applied to analyse the harmonic and the alternating harmonic series are one-off :- they cannot be applied to infinite series in general. However, there are many tests that can be used to determine the convergence properties of infinite series. Of the large number available we shall only consider two such tests in detail. The alternating series test An alternating series is a special type of series in which the sign changes from one term to the next. They have the form a a + a 3 a 4 + (in which each a i,i=,, 3,... is a positive number) Examples are: (a) + + (b) (c) For series of this type there is a simple criterion for convergence: The alternating series Key Point 6 The Alternating Series Test a a + a 3 a 4 + (in which each a i,i=,, 3,... are positive numbers) is convergent if and only if the terms continually decrease: the terms decrease to zero: a >a >a 3 >... a p 0 as p increases (mathematically lim p a p = 0) 8 HELM (008): Workbook 6: Sequences and Series

37 Task Which of the following series are convergent? p (p ) ( ) p+ (a) ( ) (b) (p +) p p= p= (a) First, write out the series: Now examine the series for convergence: ( (p ) (p +) = p ) ( + as p increases. p ) Since the individual terms of the series do not converge to zero this is therefore a divergent series. (b) Apply the procedure used in (a) to problem (b): This series is an alternating series of the form a a + a 3 a 4 + in which a p = p. The a p sequence is a decreasing sequence since > > 3 >... Also lim p =0. Hence the series is convergent by the alternating series test. p HELM (008): Section 6.: Infinite Series 9

38 3. The ratio test This test, which is one of the most useful and widely used convergence tests, applies only to series of positive terms. Key Point 7 Let p= The Ratio Test a p be a series of positive terms such that, as p increases, the limit of a p+ a p a number λ. That is lim p a p+ a p = λ. It can be shown that: equals if λ>, then if λ<, then if λ =, then a p diverges p= a p converges p= a p may converge or diverge. p= That is, the test is inconclusive in this case. 0 HELM (008): Workbook 6: Sequences and Series

39 Example Use the ratio test to examine the convergence of the series (a) +! + 3! + 4! + (b) +x + x + x 3 + Solution (a) The general term in this series is p! i.e. +! + 3! + = p= p! a p = p! a p+ = (p + )! and the ratio a p+ p! = a p (p + )! = p(p )...(3)()() (p +)p(p )...(3)()() = (p +) a p+ lim p a p = lim p (p +) =0 Since 0 < the series is convergent. In fact, it will be easily shown, using the techniques outlined in 6.5, that +! + + = e.78 3! (b) Here we must assume that x>0 since we can only apply the ratio test to a series of positive terms. Now so that +x + x + x 3 + = p= x p a p = x p, a p+ = x p and a p+ lim p a p x p = lim = lim p x x = x p p Thus, using the ratio test we deduce that (if x is a positive number) this series will only converge if x<. We will see in Section 6.4 that +x + x + x 3 + = x provided 0 <x<. HELM (008): Section 6.: Infinite Series

40 Task Use the ratio test to examine the convergence of the series: ln (ln 3) + 7 (ln 3) + 3 First, find the general term of the series: a p = ln (ln 3) + = p 3 so a (ln 3) p p = p3 (ln 3) p Now find a p+ : a p+ = (p +)3 a p+ = (ln 3) p+ Finally, obtain lim p a p+ a p : p= a p+ a p+ = lim = a p p a p 3 a p+ p + = a p p (ln 3).Now p + a p+ lim = p a p (ln 3) < Hence this is a convergent series. p 3 = + p 3 as p increases Note that in all of these Examples and Tasks we have decided upon the convergence or divergence of various series; we have not been able to use the tests to discover what actual number the convergent series converges to. HELM (008): Workbook 6: Sequences and Series

41 4. Absolute and conditional convergence The ratio test applies to series of positive terms. Indeed this is true of many related tests for convergence. However, as we have seen, not all series are series of positive terms. To apply the ratio test such series must first be converted into series of positive terms. This is easily done. Consider two series a p and a p. The latter series, obviously directly related to the first, is a series of p= positive terms. p= Using imprecise language, it is harder for the second series to converge than it is for the first, since, in the first, some of the terms may be negative and cancel out part of the contribution from the positive terms. No such cancellations can take place in the second series since they are all positive terms. Thus it is plausible that if a p converges so does a p. This leads to the following definitions. p= p= Key Point 8 Conditional Convergence and Absolute Convergence A convergent series a p is said to be conditionally convergent if a p is divergent. p= p= A convergent series a p is said to be absolutely convergent if a p is convergent. p= p= For example, the alternating harmonic series: ( ) p+ = p p= is conditionally convergent since the series of positive terms (the harmonic series): ( ) p+ p p = p= is divergent. p= HELM (008): Section 6.: Infinite Series 3

42 Task Show that the series! + 4! + is absolutely convergent. 6! First, find the general term of the series:! + 4! 6! + = ( ) a p p=! + 4! 6! + = p= ( ) p (p)! a p ( )p (p)! Write down an expression for the related series of positive terms:! + 4! + 6! + = ( ) a p = p= (p)! p= so a p = (p)! Now use the ratio test to examine the convergence of this series: p th term = (p +) th term = p th term = (p)! Find lim p (p +) th term p th term : (p +) th term lim = p p th term (p +) th term = ((p + ))! (p)! ((p + ))! = p(p )... (p + )(p + )p(p )... = (p + )(p +) So the series of positive terms is convergent by the ratio test. convergent. 0 as p increases. Hence p= ( ) p (p)! is absolutely 4 HELM (008): Workbook 6: Sequences and Series

43 Exercises. Which of the following alternating series are convergent? (a) p= ( ) p ln(3) p (b) p= ( ) p+ p +. Use the ratio test to examine the convergence of the series: e 4 p 3 (a) (b) (c) (p +) p+ p! p (d) p= p= (0.3) p (e) p= ( ) p+ p= 3 p p= (c) p= p sin(p +) π (p + 00) 3. For what values of x are the following series absolutely convergent? ( ) p x p ( ) p x p (a) (b) p p! p= s p=. (a) convergent, (b) convergent, (c) divergent. (a) λ =0so convergent (b) λ =0so convergent (c) λ =so test is inconclusive. However, since by comparison with the harmonic series. (d) λ = 0/3 so divergent, applied. 3. (a) The related series of positive terms is p / > p then the given series is divergent (e) Not a series of positive terms so the ratio test cannot be x p. For this series, using the ratio test we find p p= λ = x so the original series is absolutely convergent if x <. x p (b) The related series of positive terms is. For this series, using the ratio test we p! p= find λ =0(irrespective of the value of x) so the original series is absolutely convergent for all values of x. HELM (008): Section 6.: Infinite Series 5

44 . Introduction A sequence is any succession of numbers. For example the sequence,,, 3, 5, 8,... which is known as the Fibonacci sequence, is formed by adding two consecutive terms together to obtain the next term. The numbers in this sequence continually increase without bound and we say this sequence diverges. An example of a convergent sequence is the harmonic sequence,, 3, 4,... Here we see the magnitude of these numbers continually decrease and it is obvious that the sequence converges to the number zero. The related alternating harmonic sequence,, 3, 4,... is also convergent to the number zero. Whether or not a sequence is convergent is often easy to deduce by graphing the individual terms. The diagrams in Figure show how the individual terms of the harmonic and alternating harmonic series behave as the number of terms increase. / /3 /4 harmonic term in sequence alternating harmonic /3 /4 / term in sequence Figure Task Graph the sequence:,,,,... Is this convergent? HELM (008): Section 6.: Sequences and Series 3

45 3 4 5 term in sequence Not convergent. The terms in the sequence do not converge to a particular value. The value oscillates. A general sequence is denoted by a,a,...,a n,... in which a is the first term, a is the second term and a n is the n th term is the sequence. For example, in the harmonic sequence a =, a =,...,a n = n whilst for the alternating harmonic sequence the n th term is: a n = ( )n+ n in which ( ) n = + if n is an even number and ( ) n = if n is an odd number. Key Point The sequence a,a,...,a n,... is said to be convergent if the limit of a n as n increases can be found. (Mathematically we say that lim n a n exists.) If the sequence is not convergent it is said to be divergent. Task Verify that the following sequence is convergent 3, 4 3, 5 3 4,... First find the expression for the n th term: a n = n + n(n +) 4 HELM (008): Workbook 6: Sequences and Series

46 Now find the limit of a n as n increases: n + n(n +) = + n n + n + 0 as n increases Hence the sequence is convergent.. Arithmetic and geometric progressions Consider the sequences:, 4, 7, 0,... and 3,,, 3,... In both, any particular term is obtained from the previous term by the addition of a constant value (3 and respectively). Each of these sequences are said to be an arithmetic sequence or arithmetic progression and has general form: a, a + d, a +d, a +3d,..., a +(n )d,... in which a, d are given numbers. In the first example above a =, d =3whereas, in the second example, a = 3, d =. The difference between any two successive terms of a given arithmetic sequence gives the value of d which is called the common difference. Two sequences which are not arithmetic sequences are:,, 4, 8,..., 3, 9, 7,... In each case a particular term is obtained from the previous term by multiplying by a constant factor ( and respectively). Each is an example of a geometric sequence or geometric progression 3 with the general form: a, ar, ar, ar 3,... where a is the first term and r is called the common ratio, being the ratio of two successive terms. In the first geometric sequence above a =, r =and in the second geometric sequence a =, r = 3. HELM (008): Section 6.: Sequences and Series 5

47 Task Find a, d for the arithmetic sequence 3, 9, 5,... a = d = a =3, d =6 Task Find a, r for the geometric sequence 8, 8 7, 8 49,... a = r = a =8, r = 7 Task Write out the first four terms of the geometric series with a =4, r =. 4, 8, 6, 3,... The reader should note that many sequences (for example the harmonic sequences) are neither arithmetic nor geometric. 6 HELM (008): Workbook 6: Sequences and Series

48 3. Series A series is the sum of the terms of a sequence. For example, the harmonic series is and the alternating harmonic series is The summation notation If we consider a general sequence a,a,...,a n,... then the sum of the first k terms a + a + a a k is concisely denoted by That is, k a + a + a a k = p= a p k p= a p. k When we encounter the expression a p we let the index p in the term a p take, in turn, the values,,...,k and then add all these terms together. So, for example p= 3 7 a p = a + a + a 3 a p = a + a 3 + a 4 + a 5 + a 6 + a 7 p= p= Note that p is a dummy index; any letter could be used as the index. For example 6 a m each represent the same collection of terms: a + a + a 3 + a 4 + a 5 + a 6. m= 6 a i, and i= In order to be able to use this summation notation we need to obtain a suitable expression for the typical term in the series. For example, the finite series k may be written as In the same way k p since the typical term is clearly p in which p =,, 3,...,k in turn. p= = p= ( ) p+ p since an expression for the typical term in this alternating harmonic series is a p = ( )p+. p HELM (008): Section 6.: Sequences and Series 7

49 Task Write in summation form the series First find an expression for the typical term, the p th term : a p = a p = p(p +) Now write the series in summation form: = = p(p +) p= Task Write out all the terms of the series 5 p= ( ) p (p +). Give p the values,, 3, 4, 5 in the typical term 5 ( ) p (p +) = p= ( )p (p +) : 8 HELM (008): Workbook 6: Sequences and Series

50 4. Summing series The arithmetic series Consider the finite arithmetic series with 4 terms A simple way of working out the value of the sum is to create a second series which is the first written in reverse order. Thus we have two series, each with the same value A: and A = A = Now, adding the terms of these series in pairs A = = 8 4 = 39 so A = 96. We can use this approach to find the sum of n terms of a general arithmetic series. If A =[a]+[a + d]+[a +d]+ +[a +(n )d]+[a +(n )d] then again simply writing the terms in reverse order: A =[a +(n )d]+[a +(n )d]+ +[a +d]+[a + d]+[a] Adding these two identical equations together we have A = [a +(n )d] + [a +(n )d]+ + [a +(n )d] That is, every one of the n terms on the right-hand side has the same value: [a +(n )d]. Hence A = n[a +(n )d] so A = n[a +(n )d]. Key Point The arithmetic series [a]+[a + d]+[a +d]+ +[a +(n )d] having n terms has sum A where: A = n[a +(n )d] HELM (008): Section 6.: Sequences and Series 9

51 As an example has a =,d=,n= 4 So A =+3+ +7= 4 [ + (3)] = 96. The geometric series We can also sum a general geometric series. Let G = a + ar + ar + + ar n be a geometric series having exactly n terms. To obtain the value of G in a more convenient form we first multiply through by the common ratio r: rg = ar + ar + ar ar n Now, writing the two series together: G = a + ar + ar + + ar n rg = ar + ar + ar 3 + ar n + ar n Subtracting the second expression from the first we see that all terms on the right-hand side cancel out, except for the first term of the first expression and the last term of the second expression so that G rg = ( r)g = a ar n Hence (assuming r = ) G = a( rn ) r (Of course, if r =the geometric series simplifies to a simple arithmetic series with d =0and has sum G = na.) The geometric series having n terms has sum G where Key Point 3 a + ar + ar + + ar n G = a( rn ), if r = and G = na, if r = r 0 HELM (008): Workbook 6: Sequences and Series

52 Task Find the sum of each of the following series: (a) (b) (a) In this arithmetic series state the values of a, d, n: a = d = n = a =, d =, n = 00. Now find the sum: = = 50( + 99) = 50(0) = (b) In this geometric series state the values of a, r, n: a = r = n = a =, r = 3, n =6 Now find the sum: = = = = HELM (008): Section 6.: Sequences and Series

53 Exercises. Which of the following sequences is convergent? (a) sin π, sin π, sin 3π, sin 4π,... (b) sin π π, sin π π, sin 3π 3π, sin 4π 4π,.... Write the following series in summation form: (a) (b) ln + ln ln 5 ln ( + (00) ) + 3 ( (00) ) 4 ( + (300) ) ( (800) ) 3. Write out the first three terms and the last term of the following series: (a) 7 p= 4. Sum the series: 3 p p!(8 p) (b) 7 p=4 (a) (b) ( p) p+ p( + p) (c) s. (a) no; this sequence is, 0,, 0,,... which does not converge. (b) yes; this sequence is. (a) 4 p= ln(p ) (p + )(p) π/, 0, 3π/, 0,,... which converges to zero. 5π/ (b) 8 p= 3. (a) 7, 3!(6), 3 3!(5),...,36 7! ( ) p (p + )( + ( ) p+ p 0 4 ) (b) 45 (4)(6), 5 6 (5)(7), 67 (6)(8),..., 7 8 (7)(9) 4. (a) This is an arithmetic series with a = 5, d=4,n=9. A = 99 (b) This is an arithmetic series with a = 5, d= 4, n=9. A = 89 (c) This is a geometric series with a =,r=,n=6. G HELM (008): Workbook 6: Sequences and Series

54 Index for Workbook 6 Alternating harmonic sequence 3 Alternating harmonic series 7, 4 7, 3 Alternating series test 8 Approximations to functions , 5 Arithmetic progression 5 Arithmetic series 9 Binomial series 6-3 Binomial theorem 8 Common difference 5 Common ratio 5 Convergence 4, 5 Convergence - absolute 3 - conditional 3 - power series 34 - radius 33 - series 4 - tests 8-5 Divergence 4, 5 Geometric progression 5 Geometric series 0 Harmonic sequence 3 Harmonic series 7, 4, 6, 3 Maclaurin series 4-49 Power series 3-39 Radius of convergence 33 Ratio test 0 Sequence 3 Series 7, 4 Summation notation 7 Taylor series EXERCISES, 5, 3, 39, 5

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