Lecture 3 Power Series (See Chapter 1 in Boas) We want to continue the discussion of the previous lecture with the focus now on power series, i.e., where the terms in the series contain powers of some parameter, typically a dynamical variable, and we are expanding about the point where that parameter vanishes. (A special case is the geometric series we discussed in Lecture 2.) The power series then defines a function of that parameter. The standard examples are the Taylor and Maclaurin series mentioned at the beginning of the previous lecture corresponding to expanding in a series about x 0 or xx 0 , n S x bn x Maclaurin n0 or (3.1) n S x bn x x0 Taylor . n0 The coefficients bn are related to the derivatives of the underlying function nn n (bn 1 n ! d S dx S 0 or x0 n!), at least when the series is convergent, i.e., xx 0 or 0 when the series actually defines the function. Some specific examples of power series are n 1 x x23 x x0 b S x 1 , 01n 2n 2 4 8 n1 1 x2 x 3 x 4 x00 b 0 0, bn S 2 x x , n1 n 2 3 4 n (3.2) 1 x3 x 5 x 7 x0 b 0, b S x x , 0 2nn 2 12n 1 ! 3 3! 5! 7! 23 12x xx22 x 2 b S x 1 . 04n n 1 2 3 4 These series uniquely define the corresponding functions, whenever (i.e., at values of x where) the series converges. When we apply the convergence tests of the previous lecture to these expressions, we will define a range for the variable x within which the Physics 227 Lecture 3 1 Autumn 2008 series converges, i.e., a range in which the series expansion makes mathematical (and physical) sense. This range for x is called the interval of convergence (for the power series expansion). For values of x outside of this interval we will need to find different expressions to define the underlying functions, i.e., the power series serves to define the function only in the interval of convergence. Applying the ratio test to Sx1 we find n1 x 2 x n , x 2n 2 (3.3) 1 x 2. So the series absolutely converges for x 2 . At the endpoints, 1, x 2 , the series does not absolutely converge since both the p and s parameters of Eq. (2.29) vanish. For x 2 all terms in the series are positive and the series diverges ( SS1122 ). At the other endpoint, , x 2, we must be more careful due to the alternating signs. We apply test 5) (from Lecture 2) to this series with nn abnn21 . Since limnn a 0 , the series again diverges. We can see both of these results explicitly by looking at the series, SS1 2 1 1 1 1 N limit cycle 1,0,1,0, , (3.4) S1 2 1 1 1 1 . Hence the function is well defined (by the series, i.e., the series converges) only on the open interval 22 x (open means excluding the end points). For Sx2 we find from the ratio test that xnn1 1x 1 1 n x 1, xn n1 1 n n2 n (3.5) limnn x . Physics 227 Lecture 3 2 Autumn 2008 So the ratio test says that the series absolutely converges for x 1 ( ps1, 0 ). At the endpoint x 1 the signs are all negative and the series diverges ( SS2211 , just the negative Harmonic series). At the other endpoint, n1 x 1, the signs alternate, test 5) is satisfied ( ann 1 , aann1 , limnn a 0) and the series converges conditionally. Hence the function Sx2 is well defined (by the power series) only on the semi-open interval 11 x . Note again that we must treat the endpoints carefully. The third series may look like it is missing every other term (the even powers) and therefore you may be confused about how to apply the ratio test. This is not an issue. The idea of the ratio test is to always consider 2 contiguous terms in the series. Alternatively we can (re)write the series as (note that now every value of n contributes to the sum) n 2 n 1 S3 x x bnn x ,. b (3.6) n0 2n 1 ! Applying the ratio test we find (recall nn! 1 2 3 , 0! 1) xn23n 2 3 ! x 2 , n x21n 2 n 1 ! 2 n 2 2 n 3 (3.7) limnn 0 any x . The series converges absolutely everywhere, x ( Sx3 is just the power series expansion of the sine function sin x ). Finally apply the ratio test to Sx4 1 1 n 11n n x 2 x 2 x 2 1 , nn 221 2 n (3.8) limnn x 2 . Physics 227 Lecture 3 3 Autumn 2008 Thus the series absolutely converges for 1 x 2 1, the open interval 31 x . Since p 1 2 1, S4 diverges at the end points. For x 1 there are no alternating n signs so S4 1 also diverges. At x 3 we have ann 11 , which satisfies test 5) and the series conditionally converges. Hence the interval of convergence for Sx4 is 31 x . Now that we have verified that we know how to determine the interval of convergence, we should restate what is important about the interval. For values of x within that interval of convergence the following statements are all true (and equivalent). n The series defines the function of x, S x bn x . n The power series bxn converges (in the sense of Lecture 2) to the function Sx . The function is (accurately) represented by the series . is a power series expansion for the function . To further develop this discussion we state the following theorems (without proof), which are true in the interval of convergence where the (infinite) power series can be treated like a polynomial. I) We can integrate or differentiate the series term-by-term to find a definition of the corresponding integral or derivative of the function defined by the original series, i.e., Sx or S x dx . The resulting series has the same interval of convergence as the original series, except perhaps for the behavior at the endpoints (see below). [This is why series are so useful. Note that the effect of integrating or taking a derivative is to introduce a factor of 1/(n+1) or n in the n±1 term bn x , which does not change the interval of convergence (you should convince yourself of this fact).] II) If we have two series defining two functions with known intervals of convergence, we can add, subtract or multiply the two series to define the corresponding functions, S12 x S x or S12 x S x. The new series Physics 227 Lecture 3 4 Autumn 2008 convergences in the overlap (or common) interval of the original 2 intervals of convergence. We can also think about the function defined by dividing the two series, which with some manipulation we can express as a corresponding power series. This function is well defined in the overlap interval of the original series except points where the series in the denominator vanishes. At these points the function may still be defined if the series in the numerator also vanishes (appropriately quickly). We are guaranteed that there is an interval where the new function and its series expansion make sense, but we must explicitly calculate it. III) We can substitute one series into another series if the value of the first series is n n within the interval of convergence of the second, S1 x bn x , S2 x cn x n yields S31 x cn S x . n IV) The power series expansibxn on of a function is UNIQUE! No matter how you determine the coefficients bn in , once you have found them you are done. [This is a really important point for the Lazy but Smart!] So let’s develop power series expansions for the functions we know and love. We know the general form is given by Eq. (3.1). If we apply the Maclaurin expansion definition to ex , we find (recall the essential property of the exponential, dexx dx e ) ddnn11 ex e x 1, b e x dxnnx0 n n!! dx n xx00 (3.9) x23 x xn exx 1. 2 6n0 n ! For this power series the interval of convergence is all values of x, x , x , 0 x . (3.10) n n 1 In a similar fashion we can quickly demonstrate that Physics 227 Lecture 3 5 Autumn 2008 35 nn11 xx 11 21n sinx x x b2nn 0, b 2 1 , 3! 5!n0 2nn 1 ! 2 1 ! (3.11) 24 nn xx 11 2n cosx 1 x b2nn , b 2 1 0 . 2! 4!n0 2nn ! 2 ! Again the interval of convergence is all values of x, . ASIDE: Note that with the definition i2 1, i 1, it follows from Eqs. (3.9) and (3.11) that ixn ix2 k ix 2 l 1 eix n0n! k 0 2 k ! l 0 2 l 1 ! 11kl x2kl i x 2 1 (3.12) kl002kl ! 2 1 ! cosx i sin x . This last result is called Euler’s formula (and is extremely useful!). We can obtain more useful relations by using the four theorems above. Recall our discussion in the last lecture of the geometric and Harmonic series. We had 1 1x x23 x xn 1 x 1 (3.13) n0 1 x or equivalently n 1 x 1x x23 x x 1 x 1 .