Series Solutions. 8.1 Taylor Polynomials

Transcription

1 8 Series Solutions 8.1 Taylor Polynomials Polynomial functions, as we have seen, are well behaved. They are continuous everywhere, and have continuous derivatives of all orders everywhere. It also turns out that, given any continuous function f that has continuous derivatives of all orders, a polynomial function P can be found that approximates f on any arbitrary interval I Dom(f) to an arbitrary degree of accuracy. The precise way of going about this is to construct what is called a Taylor polynomial. Definition 8.1. Let f be a function for which f (x 0 ), f (x 0 ),..., f (n) (x 0 ) R. For n N the nth-order Taylor polynomial for f with center x 0 is the polynomial function P n given by P n (x) = n f (k) (x 0 ) (x x 0 ) k = f(x 0 ) + f (x 0 )(x x 0 ) + f (x 0 ) (x x 0 ) f (n) (x 0 ) (x x 0 ) n, 2! n! where we define f (0) (x 0 ) (x x 0 ) 0 = f(x 0 ) 0! for all x R. where More compactly we may write P n (x) = n a k (x x 0 ) k, a k = f (k) (x 0 ) is the kth coefficient of the polynomial P n (x). The remainder R n associated with the nth-order Taylor polynomial for f is the function given by R n (x) = f(x) P n (x)

2 and can be seen to be the error incurred when using P n to approximate f for any x Dom(f). The absolute error in approximating f(x) with P n (x) is given by R n (x) = f(x) P n (x). The summation notation is too convenient to pass up when working with Taylor polynomials, but using it can give the indeterminate form 0 0 occasion to arise now and again. Whenever this happens in the present context it is understood that 0 0 = 1. Summation notation and expanded form will both be employed side-by-side in any situation when it seems timely to remind the reader of this convention. Clearly P n (x 0 ) = f(x 0 ), and since for any 1 k n we find that the kth derivative of P n is given by P (k) n (x) = f (k) (x 0 ) + f (k+1) (x 0 ) (k + 1)(k)(k 1) (2)(x x 0 ) 1 (k + 1)! + f (k+2) (x 0 ) (k + 2)(k + 1)(k) (3)(x x 0 ) 2 (k + 2)!. + f (n) (x 0 ) (n)(n 1)(n 2) (n (k 1))(x x 0 ) n k n! = f (k) (x 0 ) + f (k+1) (x 0 ) 1! (x x 0 ) + f (k+2) (x 0 ) (x x 0 ) f (n) (x 0 ) 2! (n k)! (x x 0) n k, it follows that P n (k) (x 0 ) = f (k) (x 0 ). (1) We will need this fact in the proof of the following theorem. Theorem 8.2 (Taylor s Theorem). Suppose that f : [a, b] R, f (n) is continuous on [a, b] and differentiable on (a, b) for some n N, and x 0 [a, b]. Given any x [a, b] with x x 0, there exists some c between x and x 0 such that f(x) = P n (x) + f (n+1) (c) (n + 1)! (x x 0) n+1 (2) Proof. Fix a x 0 b, and let x [a, b] be such that x x 0. Letting t be a variable, we have n f (k) (x 0 ) P n (t) = (t x 0 ) k for a t b. If we let M = f(x) P n(x) (x x 0 ) n+1, which is a constant, then to show (2) is equivalent to showing f (n+1) (c) = M(n + 1)! for some c between x and x 0. Define a continuous function g : [a, b] R by g(t) = f(t) P n (t) M(t x 0 ) n+1. 2

3 The continuity of f (n) on [a, b] implies the differentiability of f (n 1) on [a, b], which in turn implies the differentiability of f (n 2) on [a, b], and so on. Thus it is clear that f (k) is differentiable on [a, b] for all 0 k n 1 (where f (0) = f). From these observations and the given hypotheses we conclude that g (n) is continuous on [a, b] and differentiable on (a, b), and g (k) is differentiable (and hence continuous) on [a, b] for 0 k n 1. In particular g (n+1) (t) exists for all a < t < b, and so is defined for all t between x and x 0. What we wish to determine is whether there exists some c between x and x 0 for which g (n+1) (c) = 0. First note that g(x 0 ) = 0, and also Moreover, from g(x) = f(x) P n (x) f(x) P n(x) (x x 0 ) n+1 (x x 0) n+1 = f(x) P n (x) [f(x) P n (x)] = 0. g (k) (t) = f (k) (t) P (k) n (t) M(n + 1)(n)(n 1) (n (k 2))(t x 0 ) n+1 k (3) and equation (1) we obtain g (k) (x 0 ) = f (k) (x 0 ) P (k) n (x 0 ) = f (k) (x 0 ) f (k) (x 0 ) = 0 for all 1 k n. Since g(x) = g(x 0 ) = 0, g is continuous on the closed interval with endpoints x and x 0, and g is differentiable on the open interval with endpoints x and x 0, by the Mean Value Theorem there exists some c 1 between x and x 0 for which g (c 1 ) = 0. Next, since g (c 1 ) = g (x 0 ) = 0, g is continuous on the closed interval with endpoints c 1 and x 0, and g is differentiable on the open interval with endpoints c 1 and x 0, there exists some c 2 between c 1 and x 0 for which g (c 2 ) = 0. Repeating this argument leads to the conclusion that there exists some c n (a, b), not equal to x 0, for which g (n) (c n ) = 0. Then, since g (n) (x 0 ) = 0 also, g (n) is continuous on the closed interval I with endpoints c n and x 0, and g (n) is differentiable on Int(I), it must be that there is some c between c n and x 0 for which g (n+1) (c) = 0. Because each c k lies between x and x 0 it is easy to see that c does also. Now, from (3) and the observation that P n (n+1) = 0, g (n+1) (t) = f (n+1) (t) P (n+1) n (t) M(n + 1)! = f (n+1) (t) M(n + 1)!, and so g (n+1) (c) = 0 implies that 3 Therefore which immediately yields (2) as desired. f (n+1) (c) M(n + 1)! = 0. f (n+1) (c) = M(n + 1)! = f(x) P n(x) (n + 1)!, (x x 0 ) n+1 Proposition 8.3. Suppose that f : [a, b] R, f (n) is continuous on [a, b] and differentiable on (a, b) for some n N, a x 0, x b with x 0 x, and I is the open interval with endpoints

4 4 y P 4 P f x P 5 P 3 P 1 Figure 1 x and x 0. If P n is the nth-order Taylor polynomial for f with center x 0 and there exists some M R such that f (n+1) (t) M for all t I, then R n (x) M x x 0 n+1. (n + 1)! Proof. Recall that by definition R n = f P n. By Taylor s Theorem there exists some c between x and x 0 such that R n (x) = f(x) P n (x) = f (n+1) (c) (n + 1)! (x x 0) n+1, and since c I we obtain R n (x) = f (n+1) (c) (n + 1)! (x x 0) n+1 = f (n+1) (c) x x 0 n+1 (n + 1)! M x x 0 n+1, (n + 1)! as was to be shown. Example 8.4. Find the nth-order Taylor polynomial for centered at 0 for n = 1, 2, 3, 4, 5. f(x) = 1 (1 + x) 2

5 5 Solution. First we obtain the needed derivatives for f, along with their values at 0. Thus we have f (x) = 2(1 + x) 3 f (0) = 2 f (x) = 6(1 + x) 4 f (0) = 6 f (x) = 24(1 + x) 5 f (0) = 24 f (4) (x) = 120(1 + x) 6 f (4) (0) = 120 f (5) (x) = 720(1 + x) 7 f (5) (0) = 720 P 1 (x) = f(0) + f (0)x = 1 2x P 2 (x) = f(0) + f (0)x + f (0) x 2 = 1 2x + 3x 2 2! P 3 (x) = f(0) + f (0)x + f (0) 2! x 2 + f (0) x 3 = 1 2x + 3x 2 4x 3 3! P 4 (x) = P 3 (x) + f (4) (0) = 1 2x + 3x 2 4x 3 + 5x 4 4! P 5 (x) = P 4 (x) + f (5) (0) = 1 2x + 3x 2 4x 3 + 5x 4 6x 5 5! Figure 1 shows the graphs of these Taylor polynomials. It can be seen that P n provides a better approximation for f in the neighborhood of 0 as n increases. Example 8.5. Let f(x) = (1 + x) 2 as in the previous example. Find an upper bound on the absolute error that may be incurred by approximating f(0.1) using the 4th-order Taylor polynomial for f with center 0. Solution. The absolute error in question is R 4 (0.1) = f(0.1) P 4 (0.1). From Example 8.4 we found that f (5) (x) = 720(1 + x) 7, which certainly is continuous on [0, 0.1] and differentiable on (0, 0.1). Now, for any 0 < t < 0.1, f (5) (t) 720 = 1 + t = 720, 7 and so by Proposition 8.3 we obtain an upper bound on R 4 (x) : R 4 (x) (720) (4 + 1)! = 720(0.1)5 5! = Thus, if we use P 4 (0.1) to estimate f(0.1), the absolute error will be no greater than Of course, nothing here prevents us from actually calculating the absolute error in this case. Since f(0.1) = ( ) and P 4 (0.1) = 1 2(0.1) + 3(0.1) 2 4(0.1) 3 + 5(0.1) 4 = ,

6 it can be seen that the absolute error is about This is indeed less than the upper bound Example 8.6. Find an upper bound on the absolute error in approximating f(x) = (1 + x) 2 on the interval [ 0.3, 0.3] using the 4th-order Taylor polynomial for f with center 0. Solution. The goal here is to find some number N such that R 4 (x) N for all x [ 0.3, 0.3]. Example 8.4 gives an expression for f (5), which is seen to be continuous on [ 0.3, 0.3] and differentiable on ( 0.3, 0.3), and so Proposition 8.3 can be used with x 0 = 0 in order to determine a value for N. Fix x [ 0.3, 0.3], and let I be the open interval with endpoints 0 and x. For any t I we have f (5) (t) 720 = 1 + t < , ( 0.3) 7 since t I implies that 0.3 t 0.3, and so by Proposition 8.3 and the fact that x 0.3 we obtain R 4 (x) (8742.8) x 5 (8742.8) < ! 5! Now, since x [ 0.3, 0.3] is arbitrary, it follows that R 4 (x) for all 0.3 x 0.3. That is, serves as an upper bound on the absolute error in approximating f on [ 0.3, 0.3] using P 4. In the case when x = 0.3 we have f( 0.3) = (1 0.3) and P 4 ( 0.3) = , and so the absolute error is well less than In the next example we see at last how Taylor polynomials may be used to obtain approximate solutions to initial value problems encountered in the theory of differential equations. Example 8.7. Determine the first four nonzero terms in the Taylor polynomial approximation of the solution to the initial value problem y + 2y y 2 = t 2, y(0) = 1, y (0) = 1. 6 Solution. We have y (t) = y 2 (t) 2y (t) + t 2, and so using the initial conditions we obtain y (0) = y 2 (0) 2y (0) = 1 2 2(1) = 1. Next, y = 2yy 2y + 2t, so y (0) = 2(1)(1) 2( 1) + 2(0) = 4. Finally, from y (4) = 2yy + 2(y ) 2 2y + 2 we have y (4) (0) = 2(1)( 1) + 2(1) 2 2(4) + 2 = 6.

7 7 The nth-order Taylor polynomial for y with center 0 is n y (k) (0) P n (t) = and so That is, t k = y(0) + y (0)t + y (0) 2! P 4 (t) = y(0) + y (0)t + y (0) 2! = 1 + t 1 2 t t3 1 4 t4. t 2 + y (0) 3! t y(n) (0) t n, n! t 3 + y(4) (0) t 4 4! for all t near 0. y(t) 1 + t 1 2 t t3 1 4 t4

8 8 8.2 Power Series The basic definition of an infinite series, found in any calculus book, is assumed to be familiar to the reader and so is not included here. Rather, we begin with the definition of a special kind of infinite series known as a power series. Definition 8.8. An infinite series of the form c k (x x 0 ) k (4) is a power series with center x 0, and the c k values are the coefficients of the power series. Just as an infinite series need not have its index k start at 1, the index of a power series as defined here does not need to start at 0. However having the initial value of k be 0 is the most common scenario. Power series may be used to define functions. That is, we can define a function f by f(x) = c k (x x 0 ) k, with the understanding that the domain of f consists of the set of all x R for which the series converges. Define { } S = x R : c k (x x 0 ) k R to be the set of convergence for the series (4). As the next theorem makes clear, given any power series (4) the set S can only ever be {x 0 }, (, ), or an interval with endpoints x 0 R and x 0 + R for some R > 0. Here R is called the radius of convergence of the power series. We define R = 0 if S = {x 0 }, and R = if S = (, ). Theorem 8.9. A power series c k (x x 0 ) exhibits one of three behaviors: 1. The series converges absolutely for all x R, so that S = (, ) and R =. 2. The series converges only for x = x 0, so that S = {x 0 } and R = For some 0 < R < the series converges absolutely for all x (x 0 R, x + R) and diverges for all x (, x 0 R) (x 0 + R, ). In part (3) of the theorem notice that nothing is said about whether the power series converges at x = x 0 ± R, and that is because nothing can be said in general. If part (3) applies to a particular power series, then the set of convergence S of the series will be an interval of convergence that may be of the form (x 0 R, x 0 + R), [x 0 R, x 0 + R), (x 0 R, x 0 + R], or [x 0 R, x 0 + R]. The endpoints x 0 R and x 0 +R will have to be investigated individually to determine whether the series converges or diverges there.

9 To determine for what values of x a power series converges, the best tools are the Ratio Test and Root Test, stated here without proof. 1 Theorem 8.10 (Root Test). Given the series a k, let ρ = lim k k a k. 1. If ρ [0, 1), then a k converges. 2. If ρ (1, ], then a k diverges. Theorem 8.11 (Ratio Test). Given the series a k for which a k = 0 for at most a finite number of k values, let ρ = lim k a k+1 /a k. 1. If ρ [0, 1), then a k converges. 2. If ρ (1, ], then a k diverges. If {b k } is a sequence such that b k > 0 for all k, then a series of the form ( 1) k b k or ( 1) k+1 b k is called alternating. Thus, the terms of an alternating series alternate between positive and negative values. An easy example is the series ( 1) k+1 1 k, which it known as the alternating harmonic series. To determine whether an alternating series converges or diverges, there is the following test which will soon prove useful. Theorem 8.12 (Alternating Series Test). If {b k } is such that 0 < b k+1 b k for all k and lim k b k = 0, then the series ( 1) k+1 b k converges. Example Find the interval of convergence of the power series k 1 xk ( 1) k, (5) 3 and state the radius of convergence. Solution. Clearly the series converges when x = 0. Assuming x 0, we can employ the Ratio Test with a k = ( 1) k 1 x k /k 3 : lim a k+1 k a k = lim ( 1) k x k+1 k 3 k (k + 1) 3 ( 1) k 1 x k = lim k 3 x k (k + 1) 3 = lim k 3 x = x, k (k + 1) 3 since k 3 lim k (k + 1) = The proofs for these tests, as well as for most of the other results in this section, can be found in Chapters 9 and 10 of my Calculus Notes. 9

10 Thus the series converges if x < 1, or equivalently 1 < x < 1. The Ratio Test is inconclusive when x = 1 or x = 1, so we analyze these endpoints separately. When x = 1 the series becomes ( 1) k 1 ( 1) k k 3 = ( 1) 2k 1 k 3 = Recall that 1/k 3 is a convergent p-series, and thus 1/k 3 = s for some s R. Now Proposition 9.12 implies that = 1 k 3 which shows that 1/k 3 also converges. When x = 1 the series becomes ( 1) k 1 k 3, 1 k 3 = s, 1 k 3. which is an alternating series ( 1) k 1 b k with b k = 1/k 3. Since lim k b k = 0 and b k+1 = 1 (k + 1) 3 < 1 k 3 = b k for all k, by the Alternating Series Test this series converges. We conclude that the series (5) converges on the interval [ 1, 1], and the radius of convergence is R = 1 1 ( 1) = 1. 2 Example Find the interval of convergence of the power series k (x + 2)k ( 1), (6) k 2 k and state the radius of convergence. Solution. Clearly the series converges when x = 2. Assuming x 2, we can employ the Ratio Test with to obtain k (x + 2)k a k = ( 1) k 2 k lim a k+1 k a k = lim ( 1) k+1 (x + 2) k+1 k 2 k k (k + 1) 2 k+1 ( 1) k (x + 2) k = lim ( 1)(x + 2) k k 2(k + 1) 1 = lim k x + 2 k 2k + 2 = 1 x Thus the series converges if 1 2 x + 2 < 1, which implies x + 2 < 2 and thus 4 < x < 0. The Ratio Test is inconclusive when x = 4 or x = 0, so we analyze these endpoint separately. 10

11 When x = 4 the series becomes ( 1) k ( 2) k = k 2 k 2 k k 2 = k which is the harmonic series and therefore diverges. When x = 0 the series becomes ( 1) k 2 k = ( 1) k 1 k 2 k k, which is an alternating series ( 1) k b k with b k = 1/k. Since lim k b k = 0 and b k+1 = 1 k + 1 < 1 k = b k for all k, by the Alternating Series Test this series converges. Therefore the series (6) converges on the interval ( 4, 0], and the radius of convergence is R = 1 0 ( 4) = 2. 2 Example Find the interval of convergence of the power series x k (ln k), (7) k and state the radius of convergence. Solution. In this case it would be easier to use the Root Test with a k = x k /(ln k) k ; so, for any x R, we obtain k k x lim ak = lim k k k (ln k) k = lim k x k k ln k = lim x k k ln k = 0, since ln k as k. Thus, the series (7) converges for all real x, which implies that the interval of convergence is (, ) and the radius of convergence is R =. Theorem Suppose the series c k(x x 0 ) k converges on an interval I, and define f : I R by f(x) = c k (x x 0 ) k. 1 k, f is continuous on I. 2. f is differentiable on Int(I), with for all x Int(I). f (x) = kc k (x x 0 ) k 1.

12 12 3. f is integrable on Int(I), with for arbitrary constant c. f = c k k + 1 (x x 0) k+1 + c In Theorem 8.16(2) we could have written f (x) = kc k (x x 0 ) k 1, with the understanding that the term corresponding to k = 0, which is 0c 0 (x x 0 ) 1, is equal to 0 even if x = x 0 and the form 0 1 results! We will avoid this here, although we will continue to take 0 0 to be 1 wherever it arises in the power series notation. The first part of Theorem 8.16 states that, for any x Int(I), lim t x c k (t x 0 ) k = lim t x f(t) = f(x) = c k (x x 0 ) k = lim c k(t x 0 ) k, (8) t x which is to say the limit of a convergent series can be carried out termwise so long as the limit operates in the interior of the interval of convergence I of the series. If x is an endpoint of I then the appropriate one-sided limit is executed in (8) instead. The second and third parts of the theorem state that a convergent power series can be differentiated and integrated termwise on the interior of I, meaning the differentiation or integration operator can be brought inside the series: and d dx c k (x x 0 ) k = [ c k (x x 0 ) ]dx k = d [ ck (x x 0 ) k] = dx [ kc k (x x 0 ) k 1 = ] c k (x x 0 ) k dx + c = kc k (x x 0 ) k 1, c k k + 1 (x x 0) k+1 + c. Moreover the new series that results from differentiating or integrating the old series will be convergent on Int(I) as mentioned in the theorem, although nothing can be said in general about the behavior of the new series at the endpoints of I. The following proposition is proven in more advanced texts using Theorem 8.16, among other things. It will be needed in later sections. Proposition If c k (x x 0 ) k = 0 for all x in some open interval, then c k = 0 for all k 0.

13 Definition A function f is said to be analytic at x 0 if there exists R > 0 and coefficients c k R such that f(x) = c k (x x 0 ) k (9) for all x (x 0 R, x 0 + R). If a function f is such that (9) holds for all x in an open interval I, then f is said to be analytic on I. An immediate consequence of Theorem 8.16 is that a function f for which (9) holds for all x (x 0 R, x 0 + R) must have derivatives of all orders at x 0. Given a function f whose derivatives can be explicitly determined at x 0 by the usual rules of differentiation, the customary way of going about finding a power series expression for f (i.e. one that satisfies (9) on some open interval containing x 0 ) is to construct the Taylor series for f with center at x 0. Definition Let f be a function that has derivatives of all orders on an open interval I containing x 0. Then the power series of the form f (k) (x 0 ) (x x 0 ) k is the Taylor series for f centered at x 0. Maclaurin series. 13 A Taylor series centered at 0 is called a Recalling the definition for the nth-order Taylor polynomial for f with center x 0 given in Section 8.1, it can be seen that f (k) (x 0 ) (x x 0 ) k = lim P n (x). n

14 Series Solutions Near an Ordinary Point Our objective is to devise a method to express the general solution of an ordinary differential equation in terms of a power series that converges in some open interval I containing a point x 0. To start, we will focus first on second-order linear differential equations of the form Dividing (10) by a 2 (x) puts the equation in the standard form a 2 (x)y + a 1 (x)y + a 0 (x)y = 0. (10) y + p(x)y + q(x)y = 0, (11) where of course p(x) = a 1(x) a 2 (x) and q(x) = a 0(x) a 2 (x). Definition A point x 0 in the interior of Dom(p) Dom(q) is called an ordinary point for the ODE (11) if p and q are both analytic at x 0. Otherwise x 0 is called a singular point. The theory that underlies our stated objective is best expressed in the following theorem. Theorem Let p and q in (11) be analytic at an ordinary point x 0, so that there exist p k, q k R and R 1, R 2 > 0 such that p(x) = p k (x x 0 ) k on (x 0 R 1, x 0 + R 1 ), and q(x) = qk (x x 0 ) k on (x 0 R 2, x 0 + R 2 ). If R = min{r 1, R 2 }, then the IVP y + p(x)y + q(x)y = 0, y(x 0 ) = b 0, y (x 0 ) = b 1 (12) has a unique solution that is analytic on (x 0 R, x 0 + R). According to the developments in Section 4.7, the general solution to an equation of the form (11) is a two-parameter family of functions of the form d 1 y 1 (x) + d 2 y 2 (x), where y 1 and y 2 are two particular linearly independent solutions to (11), and d 1 and d 2 are arbitrary constants. According to Theorem 8.21 the IVP (12) has a solution y(x) expressible as a power series on I = (x 0 R, x 0 + R), which is to say there exist a k R such that y(x) = c k (x x 0 ) k = c 0 + c 1 (x x 0 ) + c 2 (x x 0 ) 2 + (13) for all x I, and the initial conditions y(x 0 ) = b 0 and y (x 0 ) = b 1 are satisfied. Indeed from (13) we find that c 0 = y(x 0 ) = b 0, and from y (x) = kc k (x x 0 ) k 1 = c 1 + 2c 2 (x x 0 ) + 3c 3 (x x 0 ) 2 + we find that c 1 = y (x 0 ) = b 1. Thus if the initial conditions in (12) were not given, then we would find that c 0 and c 1 would be left as arbitrary constants and so (13) would in fact be a two-parameter family of

15 solutions. Reconciling the developments of Section 4.7 with the consequences of Theorem 8.21, we conclude that the general solution to (11) is y(x) = c k (x x 0 ) k = d 1 y 1 (x) + d 2 y 2 (x), where again c 0, c 1, d 1, d 2 are arbitrary constants. What we have done is concoct two different ways of expressing the general solution to an ODE of the form (11). The following example should help illustrate how this is all put into practice. Example Find the general solution to in the form of a power series about 0. (x 2 4)y + 3xy + y = 0 Solution. Since x = 0 is an ordinary point for the ODE, we expect to find a general solution of the form y(x) = c k x k, (14) with the power series converging on some open interval I containing 0. The task is to determine all coefficients c k that can be determined. By the discussion above we expect that c 0 and c 1 will be left as arbitrary, and all c k for k 2 will not. Now, substituting (14), y (x) = kc k x k 1, and y (x) = k(k 1)c k x k 2 into the ODE yields (x 2 4) k(k 1)c k x k 2 + 3x kc k x k 1 + Assuming that x I, this equation becomes k(k 1)c k x k 4 k(k 1)c k x k kc k x k + c k x k = c k x k = 0 (15) Now we will shift indexes so that all the power series in (15) begin at k = 0. This can be done effortlessly for the first and third series, since it merely results in adding terms that equal 0. As for the second series, we need only replace k with k + 2 to obtain (k + 2)[(k + 2) 1]c k+2 x (k+2) 2, which becomes k+2=2 (k + 1)(k + 2)c k+2 x k.

16 16 Putting this into (15) gives k(k 1)c k x k 4 (k + 1)(k + 2)c k+2 x k + 3 kc k x k + whence we obtain and finally c k x k = 0, [k(k 1)c k 4(k + 1)(k + 2)c k+2 + 3kc k + c k ] x k = 0, [ ] (k + 1) 2 c k 4(k + 1)(k + 2)c k+2 x k = 0 for all x I. Therefore by Proposition 8.17 we have for all k 0. Solving for c k+2 gives (k + 1) 2 c k 4(k + 1)(k + 2)c k+2 = 0 c k+2 = k + 1 4(k + 2) c k, a recurrence relation which will now be used to obtain an explicit expression for c k. We have c 2 = (0 + 2) c 0 = c 0 c 3 = (1 + 2) c 1 = 2 4(1 3) c 1 c 4 = (2 + 2) c 2 = c 5 = (3 + 2) c 3 = c 6 = (4 + 2) c 4 = c 7 = (5 + 2) c 5 = c 8 = (6 + 2) c 6 = c 0 = (2 4) c c = 4 2 (1 3 5) c (2 4) c 0 = (3 5) c 1 = 4 3 (2 4 6) c ( ) c (2 4 6) c 0 = ( ) c 0 Two patterns are emerging here: one for c k when k is even, and another for c k when k is odd. For n any whole number we have and c 2n = c 2n+1 = (2n 1) 4 n [2 4 6 (2n)] c 0 = (2n) 4 n [3 5 7 (2n + 1)] c 1 = (2n 1) c 4 n (2 n 0 = n!) (2n)! 4 n (2 n n!) c 2 0 = (2n)! 4 2n (n!) c 2 0 [2 4 6 (2n)]2 c 4 n 1 = (2n n!) 2 (2n + 1)! 4 n (2n + 1)! c 1 = (n!)2 (2n + 1)! c 1.

17 These two results can be rigorously proven by a routine use of the principle of induction, but we shall refrain in the interests of brevity. What is evident, however, is that all c k are expressible in terms of either c 0 or c 1, which are arbitrary. So, as a general solution to the ODE, we have y(x) = c kx k with 4 k [(k/2)!] c 0, if k is even 2 c k = [((k 1)/2)!] 2 c 1, if k is odd There is another way to express the ODE s solution which is nicer. If we set c 0 = 0 and c 1 = 1, then c k = 0 if k is even and [((k 1)/2)!]2 c k = if k is odd, and so only the odd-indexed terms in the series (14) are nonzero. Now, k odd means here that k = 2n + 1 for some n = 0, 1, 2,..., and so (14) can be written as y 1 (x) = c 2n+1 x 2n+1 = which is in fact a particular solution to the ODE. If we set c 0 = 1 and c 1 = 0, then c k = 0 if k is odd and c k = 4 k [(k/2)!] 2 (n!) 2 (2n + 1)! x2n+1, if k is even, and so only the even-indexed terms in (14) are nonzero. Now, k even implies that k = 2n for n = 0, 1, 2,..., and so (14) can be written as y 2 (x) = c 2n x 2n (2n)! = 4 2n (n!) 2 x2n, which is another particular solution to the ODE. The particular solutions y 1 (x) and y 2 (x) are linearly independent, and so the general solution to the ODE can be expressed as y(x) = d 1 y 1 (x) + d 2 y 2 (x) = d 1 where d 1 and d 2 are arbitrary constants. Example Find the general solution to in the form of a power series about x 0 = 0. (n!) 2 (2n + 1)! x2n+1 + d 2 y x 2 y xy = 0 (2n)! 4 2n (n!) 2 x2n, Solution. Since 0 is an ordinary point for the ODE we expect to find a general solution of the form y(x) = c k x k, 17

18 with the series converging on some open interval I containing 0. Substituting this into the ODE yields k(k 1)c k x k 2 x 2 kc k x k 1 x c k x k = 0, and thus k(k 1)c k x k 2 kc k x k+1 Reindexing so that all series feature x k, we have (k + 1)(k + 2)c k+2 x k (k 1)c k 1 x k c k x k+1 = 0. c k 1 x k = 0. Finally we contrive to have the index of each series start at 2 by removing the first two terms of the leftmost series and the first term of the rightmost series: [ ] [ ] 2c 2 + 6c 3 x + (k + 1)(k + 2)c k+2 x k (k 1)c k 1 x k c 0 x + c k 1 x k = 0. Hence 2c 2 + (6c 3 c 0 )x + which simplifies to become [(k + 1)(k + 2)c k+2 (k 1)c k 1 c k 1 ] x k = 0, 2c 2 + (6c 3 c 0 )x + [(k + 1)(k + 2)c k+2 kc k 1 ] x k = 0. Now, Proposition 8.17 implies that 2c 2 = 0, 6c 3 c 0 = 0, and (k + 1)(k + 2)c k+2 kc k 1 = 0 for all k 2. That is, c 2 = 0, c 3 = c 0 /6 = c 0 /(2 3), and c k+2 = k (k + 1)(k + 2) c k 1 for k 2. The recursion relation enables us to express all c k exclusively in terms of c 0 and c 1 : c 4 = c 1 c 5 = c 2 = 0 c 6 = c 4 3 = c 0 c 7 = c = c 1 c 8 = c 5 = 0 c 9 = c 6 = c 10 = c = c 1 c 11 = c 8 = 0 c 12 = c = c 0 So we have c 0 18

19 y(x) = c 0 + c 1 x + c x3 + 2c 1 4c x x c 1 4 7c x x c c x x12 + Setting c 0 = 0 and c 1 = 1 yields the particular solution y 1 (x) = x x x x10 + = x ! x x x ! 10! (3k 1) 2 = x + x 3k+1, (3k + 1)! and setting c 0 = 1 and c 1 = 0 yields the particular solution y 2 (x) = x x x9 + = ! x ! x x x ! 12! (3k 2) 2 = 1 + x 3k. (3k)! x12 + Since y 1 (x) and y 2 (x) are linearly independent, the general solution to the ODE may be expressed as ] ] 2 y(x) = d 1 [x (3k 1) 2 + x 3k d 2 [ (3k 2) 2 + x 3k (3k + 1)! (3k)! for all x I, where d 1 and d 2 are arbitrary constants. Example Find the first four nonzero terms in a power series expansion about x 0 = 2 for a general solution to x 2 y y + y = 0. Solution. Since x = 2 is an ordinary point for the ODE, we expect to find a general solution of the form y(x) = c k (x 2) k, (16) with the power series converging on some open interval I containing 2. From (16) comes y (x) = kc k (x 2) k 1 and y (x) = k(k 1)c k (x 2) k 2, 19

20 which when substituted into the ODE yields k(k 1)c k (x 2) k 2 kc k (x 2) k 1 + x 2 20 c k (x 2) k = 0. (17) It will be expedient to express x 2 in terms of x 2. Since (x 2) 2 = x 2 4x + 4 we have x 2 = (x 2) 2 + 4x 4 = (x 2) 2 + 4(x 2) + 4, and so (17) becomes [(x 2) 2 + 4(x 2) + 4] k(k 1)c k (x 2) k 2 kc k (x 2) k 1 + c k (x 2) k = 0, and thus k(k 1)c k (x 2) k + 4 k(k 1)c k (x 2) k k(k 1)c k (x 2) k 2 kc k (x 2) k 1 + c k (x 2) k = 0. Adding zero terms and reindexing where needed, we obtain k(k 1)c k (x 2) k + 4 k(k + 1)c k+1 (x 2) k + 4 (k + 1)(k + 2)c k+2 (x 2) k (k + 1)c k+1 (x 2) k + c k (x 2) k = 0, or equivalently [k(k 1)c k + 4k(k + 1)c k+1 + 4(k + 1)(k + 2)c k+2 (k + 1)c k+1 + c k ] (x 2) k = 0 for all x I. Therefore by Proposition 8.17 we have k(k 1)c k + 4k(k + 1)c k+1 + 4(k + 1)(k + 2)c k+2 (k + 1)c k+1 + c k = 0 for all k 0, which rearranges to become c k+2 = (4k2 + 3k 1)c k+1 + (k 2 k + 1)c k. (18) 4k k + 8 Using the recursion relation (18), we obtain and Hence c 3 = 6c 2 + c 1 24 = 1 4 c 2 = c 1 c 0 8 ( ) c1 c c 1 = 3c 0 7c y(x) = c 0 + c 1 (x 2) + c 2 (x 2) 2 + c 3 (x 2) 3 +

21 21 = c 0 + c 1 (x 2) + c 1 c 0 8 (x 2) 2 + 3c 0 7c 1 (x 2) is a power series expansion about 2 for a general solution to the ODE. Alternatively, setting c 0 = 1 and c 1 = 0 yields the particular solution y 1 (x) = (x 2) (x 2)3 +, while setting c 0 = 0 and c 1 = 1 yields the particular solution y 2 (x) = (x 2) (x 2) (x 2)3 + ; and since y 1 (x) and y 2 (x) are linearly independent it follows that y(x) = a 0 y 1 (x) + a 1 y 2 (x) is a general solution to the ODE. That is, [ y(x) = d (x 8 2)2 + 1 (x 32 2)3 + ]+ [ d 2 (x 2) + 1 (x 8 2)2 7 (x 96 2)3 + ] for arbitrary constants d 1 and d 2. Example Find the first four nonzero terms in a power series expansion about x 0 = 0 for a solution to the initial value problem y + ty + e t y = 0, y(0) = 0, y (0) = 1. Solution. Since t = 0 is an ordinary point for the ODE we expect to find a general solution of the form y(t) = c k t k, with the power series converging on some open interval I containing 0. Substituting this power series into the ODE, as well as the power series representation for e t centered at 0, yields ( )( k(k 1)c k t k 2 + t kc k t k 1 t k ) + c k t k = 0, (19) We take the Cauchy product of the series for e t and y(t): ( )( e t t k ) y(t) = c k t k ( ) = 1 + t + t2 2 + t3 6 + (c0 + c 1 t + c 2 t 2 + c 3 t 3 + c 4 t 4 + ) so = c 0 + (c 0 + c 1 ) t + }{{} d 1 ( c0 ) 2 + c 1 + c 2 }{{} d 2 e t y(t) = ( t 2 c0 + d k t k, 6 + c ) c 2 + c 3 }{{} d 3 t 3 +,

22 where d 0 = c 0, and d 1, d 2, and d 3 are as defined above. Putting this in (19) and reindexing, we have (k + 1)(k + 2)c k+2 t k + kc k t k + d k t k = 0, or By Proposition 8.17 [(k + 1)(k + 2)c k+2 + kc k + d k ] t k = 0. (k + 1)(k + 2)c k+2 + kc k + d k = 0 for all k 0, which solved for c k+2 becomes c k+2 = kc k + d k (k + 1)(k + 2). (20) The recursion relation (20) may now be used to determine c k for any k 2, at least in terms of c 0 and c 1 : c 2 = d 0 2 = c 0 2 c 3 = c 1 + d 1 6 c 4 = 2c 2 + d 2 12 c 5 = 3c 3 + d 3 20 Thus we have = c 1 + (c 0 + c 1 ) 6 = c 0 + 2c 1 6 = 2( c 0/2) + (c 0 /2 + c 1 + c 2 ) 12 = c 0 (c 0 /2 + c 1 c 0 /2) 12 = 3( c 0/6 c 1 /3) + (c 0 /6 + c 1 /2 + c 2 + c 3 ) 20 y(t) = c 0 + c 1 t + c 2 t 2 + c 3 t 3 + c 4 t 4 + c 5 t 5 + = 6c 0 + 5c = c 0 c 1 12 = c 0 + c 1 t c 0 2 t2 c 0 + 2c 1 t 3 + c 0 c 1 t 4 + 6c 0 + 5c 1 t 5 + (21) From (21) and the initial condition y(0) = 0 we readily find that c 0 = 0. Putting this into (21) gives From y(t) = c 1 t 1 3 c 1t c 1t c 1t 5 + y (t) = c 1 c 1 t c 1t c 1t 4 + and the initial condition y (0) = 1 we obtain c 1 = 1, and so y(t) = t t t t is the solution to the IVP on I.