16.5. Maclaurin and Taylor Series. Introduction. Prerequisites. Learning Outcomes

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1 Maclaurin and Talor Series 6.5 Introduction In this Section we eamine how functions ma be epressed in terms of power series. This is an etremel useful wa of epressing a function since (as we shall see) we can then replace complicated functions in terms of simple polnomials. The onl requirement (of an 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 ou should... Learning Outcomes On completion ou 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 Talor series epansions of given functions find Maclaurin epansions of functions b combining known power series together find Maclaurin epansions b using differentiation and integration 40 HELM (2008): Workbook 6: Sequences and Series

2 . Maclaurin and Talor series As we shall see, man functions can be represented b power series. In fact we have alread seen in earlier Sections eamples of such a representation: = < ln( + ) = < e = ! + 3 3! + all The first two eamples show that, as long as we constrain to lie within the domain < (or, equivalentl, < < ), then in the first case has the same numerical value as and in the second case ln(+) has the same numerical value as In the third eample we see that e has the same numerical value as but in this 2! case there is no restriction to be placed on the value of since this power series converges for all values of. Figure 5 shows this situation geometricall. As more and more terms are used from the series ! + 3 3! the curve representing e is a better and better approimation. In (a) we show the linear approimation to e. In (b) and (c) we show, respectivel, the quadratic and cubic approimations ! ! + 3 3! e e e (a) (b) (c) Figure 5: Linear, quadratic and cubic approimations to e These power series representations are etremel important, from man points of view. Numericall, we can simpl replace the function b the quadratic epression as long as is so small that powers of greater than or equal to 3 can be ignored in comparison to quadratic terms. This approach can be used to approimate more complicated functions in terms of simpler polnomials. Our aim now is to see how these power series epansions are obtained. HELM (2008): Section 6.5: Maclaurin and Talor Series 4

3 2. The Maclaurin series Consider a function f() which can be differentiated at = 0 as often as we please. For eample e, cos, sin would fit into this categor but would not. Let us assume that f() can be represented b a power series in : f() = b 0 + b + b b b = b p p where b 0, b, b 2,... are constants to be determined. If we substitute = 0 then, clearl f(0) = b 0 The other constants can be determined b further differentiating and, on each differentiation, substituting = 0. For eample, differentiating once: f () = 0 + b + 2b 2 + 3b b so, putting = 0, we have f (0) = b. Continuing to differentiate: so Further: f () = 0 + 2b 2 + 3(2)b 3 + 4(3)b f (0) = 2b 2 or b 2 = 2 f (0) p=0 f () = 3(2)b 3 +4(3)(2)b 4 + so f (0) = 3(2)b 3 impling b 3 = 3(2) f (0) Continuing in this wa we easil find that (remembering that 0! = ) b n = n! f (n) (0) n = 0,, 2,... where f (n) (0) means the value of the n th derivative at = 0 and f (0) (0) means f(0). Bringing all these results together we have: If f() can be differentiated as often as required: Ke Point 4 Maclaurin Series f() = f(0) + f (0) + 2 2! f (0) + 3 3! f (0) + = This is called the Maclaurin epansion of f(). p=0 p p! f (p) (0) 42 HELM (2008): Workbook 6: Sequences and Series

4 Eample 4 Find the Maclaurin epansion of cos. Solution Here f() = cos and, differentiating a number of times: f() = cos, f () = sin, f () = cos, f () = sin etc. Evaluating each of these at = 0: f(0) =, f (0) = 0, f (0) =, f (0) = 0 etc. Substituting into f() = f(0) + f (0) + 2 2! f (0) + 3 3! f (0) +, gives: cos = 2 2! + 4 4! 6 6! + The reader should confirm (b finding the radius of convergence) that this series is convergent for all values of. The geometrical approimation to cos b the first few terms of its Maclaurin series are shown in Figure ! + 4 4! cos cos 2 2! cos Figure 6: Linear, quadratic and cubic approimations to cos Task Find the Maclaurin epansion of ln( + ). (Note that we cannot find a Maclaurin epansion of the function ln since ln does not eist at = 0 and so cannot be differentiated at = 0.) Find the first four derivatives of f() = ln( + ): f () = f () = f () = f () = HELM (2008): Section 6.5: Maclaurin and Talor Series 43

5 f () = +, f () = ( + ) 2, f () = generall: f (n) () = ( )n+ (n )! ( + ) n 2 ( + ) 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) = 2, generall: f (n) (0) = ( ) n+ (n )! Hence, obtain the Maclaurin epansion of ln( + ): ln( + ) = ln( + ) = ( )n+ n + (This was obtained in Section 6.4, page 37.) n Now obtain the radius of convergence and consider the situation at the boundar values: Radius of convergence R = R =. Also at = the series is convergent (alternating harmonic series) and at = the series is divergent. Hence this Maclaurin epansion is onl valid if <. The geometrical closeness of the polnomial terms with the function ln( + ) for < is displaed in Figure 7: ln( + ) ln( + ) ln( + ) 2 2 Figure 7: Linear, quadratic and cubic approimations to ln( + ) 44 HELM (2008): Workbook 6: Sequences and Series

6 Note that when = ln 2 = so the alternating harmonic series converges to 4 ln , as stated in Section 6.2, page 7. The Maclaurin epansion of a product of two functions: f()g() is obtained b multipling together the Maclaurin epansions of f() and of g() 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. Eample 5 Find the Maclaurin epansion of e ln( + ). Solution Here, instead of finding the derivatives of f() = e ln(+), we can more simpl multipl together the Maclaurin epansions for e and ln( + ) which we alread know: and e = ! + 3 3! + all ln( + ) = < The resulting power series will onl be convergent if <. Multipling: ) ) e ln( + ) = ( + + ( 2 2! + 3 3! = = < (You must take care not to miss relevant terms when carring through the multiplication.) HELM (2008): Section 6.5: Maclaurin and Talor Series 45

7 Task Find the Maclaurin epansion of cos 2 up to powers of 4. Hence write down the epansion of sin 2 to powers of 6. First, write down the epansion of cos : cos = cos = 2 2! + 4 4! + Now, b multiplication, find the epansion of cos 2 : cos 2 = cos 2 = ) ) ( ( 2 2! + 4 4! 2 2! + 4 4! = ( 2 2! + 4 4! ) + ( 2 2! ) + (4 4! ) + = Now obtain the epansion of sin 2 using a suitable trigonometric identit: sin 2 = sin 2 = cos 2 = ) ( = As an alternative approach the reader could obtain the power series epansion for cos 2 b using the trigonometric identit cos 2 ( + cos 2) HELM (2008): Workbook 6: Sequences and Series

8 Eample 6 Find the Maclaurin epansion of tanh up to powers of 5. Solution The first two derivatives of f() = tanh are f () = sech 2 f () = 2sech 2 tanh f () = 4sech 2 tanh 2 2sech 4 giving f(0) = 0, f (0), f (0) = 0 f (0) = 2 This leads directl to the Maclaurin epansion as tanh = Eample 7 The relationship between the wavelength, L, the wave period, T, and the water depth, d, for a surface wave in water is given b: L = gt ( ) 2 2πd 2π tanh L In a particular case the wave period was 0 s and the water depth was 6. m. Taking the acceleration due to gravit, g, as 9.8 m s 2 determine the wave length. [Hint: Use the series epansion for tanh developed in Eample 6.] Solution Substituting for the wave period, water depth and g we get ( ) π 6. L = tanh = ( ) 2.2π tanh 2π L π L The series epansion of tanh is given b tanh = Using the series epansion of tanh we can approimate the equation as { (2.2π L = ) ( ) } 3 2.2π + π L 3 L Multipling through b πl 3 the equation becomes πl 4 = πL (2.2π) 3 This equation can be rewritten as L L = 0 Solving this as a quadratic in L 2 we get L = 74 m. Using Newton-Raphson iteration this can be further refined to give a wave length of 73.9 m. HELM (2008): Section 6.5: Maclaurin and Talor Series 47

9 3. Differentiation of Maclaurin series We have alread noted that, b the binomial series, = < Thus, with replaced b + = < We have previousl obtained the Maclaurin epansion of ln( + ): ln( + ) = < Now, we differentiate both sides with respect to : + = This result matches that found from the binomial series and demonstrates that the Maclaurin epansion of a function f() ma be differentiated term b term to give a series which will be the Maclaurin epansion of df d. 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 epansion of ( ) 3 and state its radius of convergence. First write down the epansion of ( ) : = < Now, b differentiation, obtain the epansion of ( ) = d ( ) = 2 d ( ) 2 : ( ) 2 = d d ( ) = HELM (2008): Workbook 6: Sequences and Series

10 Differentiate again to obtain the epansion of ( ) 3 : ( ) = 3 2 = ( d d ( ) 2 ) = 2 [ ] ( ) = ( ) d = 3 2 d ( ) 2 2 [ ] = Finall state its radius of convergence: The final series: has radius of convergence R = since the original series has this radius of convergence. This can also be found directl using the formula R = lim b n and using the fact that the coefficient of the n th term is b n = n(n + ). 2 n b n+ 4. The Talor series The Talor series is a generalisation of the Maclaurin series being a power series developed in powers of ( 0 ) rather than in powers of. Thus Ke Point 5 Talor Series If the function f() can be differentiated as often as required at = 0 then: f() = f( 0 ) + ( 0 )f ( 0 ) + ( 0) 2 f ( 0 ) + 2! This is called the Talor series of f() about the point 0. The reader will see that the Maclaurin epansion is the Talor epansion obtained if 0 is chosen to be zero. HELM (2008): Section 6.5: Maclaurin and Talor Series 49

11 Task Obtain the Talor series epansion of series in powers of ( 2).) about = 2. (That is, find a power First, obtain the first three derivatives and the n th derivative of f() = : f () = f () = f () = f (n) () = f () = ( ) 2, f () = Now evaluate these derivatives at 0 = 2: 2 ( ) 3, f () = 6 ( ) 4, f (n) () = f (2) = f (2) = f (2) = f (n) (2) = n! ( ) n+ f (2) =, f (2) = 2, f (2) = 6, f (n) (2) = ( ) n+ n! Hence, write down the Talor epansion of f() = = about = 2: = + ( 2) ( 2)2 + ( 2) ( ) n+ ( 2) n + 50 HELM (2008): Workbook 6: Sequences and Series

12 Eercises. Show that the series obtained in the last Task is convergent if 2 <. 2. Sketch the linear, quadratic and cubic approimations to last task and compare to. obtained from the series in the 2. In the following diagrams some of the terms from the Talor series are plotted to compare with ( ). 2 + ( 2) 2 + ( 2) ( 2) 2 + ( 2) ( 2) 2 + ( 2) 3 2 HELM (2008): Section 6.5: Maclaurin and Talor Series 5