Taylor Series and Asymptotic Expansions The importance of power series as a convenient representation, as an approximation tool, as a tool for solving differential equations and so on, is pretty obvious. What may not be so obvious is that power series can be of some use even when they diverge! Let us start by considering Taylor series. If f :[ a,a] has infinitely many continuous derivatives, − → ∞ Taylor’s theorem says that for each n IN 0 , ∈ ∪{ } f(x)= Pn(x)+ Rn(x) n 1 (m) m with Pn(x)= m! f (0)x mX=0 1 (n+1) ′ n+1 ′ Rn(x)= (n+1)! f (x )x for some x between 0 and x Since f (n+1) is continuous, there exists a constant M such that R (x) M xn+1 for all x [ a,a]. This n | n |≤ n ∈ − says that as x gets smaller and smaller, the polynomial Pn(x) becomes a better and better approximation n+1 to f(x). The rate at which Pn(x) tends to f(x) as x tends to zero is, at worst, proportional to x . Thus ∞ 1 (m) m the Taylor series m=0 m! f (0)x is an asymptotic expansion for f(x) where P Definition. ∞ a xm is said to be an asymptotic expansion for f(x), denoted f(x) ∞ a xm m=0 m ∼ m=0 m if, for each n PIN 0 , P ∈ ∪{ } n 1 m lim n f(x) amx =0 x→0 x − h mX=0 i There are several important observations to make about this definition. n m (i) The definition says that, for each fixed n, m=0 amx becomes a better and better approximation to f(x) as x gets smaller. As x 0, n aPxm approaches f(x) faster than xn tends to zero. → m=0 m (ii) The definition says nothing aboutP what happens as n . There is no guarantee that for each fixed n m → ∞ x, m=0 amx tends to f(x) as n . ∞ m→ ∞ ∞ m (iii) InP particular, the series m=0 amx may have radius of convergence zero. In this case, m=0 amx is just a formal symbol. ItP does not have any meaning as a series. P (iv) We have seen that every infinitely differentiable function has an asymptotic expansion, regardless of whether its Taylor series converges or not. Now back to our Taylor series. There are three possibilities. ∞ 1 (m) m (i) The series m=0 m! f (0)x has radius of convergence zero. ∞ 1 (m) m (ii) The series P f (0)x has radius of convergence r> 0 and lim Rn(x) = 0 for all r<x<r. m=0 m! n→∞ − In this caseP the Taylor series converges to f(x) for all x ( r, r). Then f(x) is said to be analytic on ∈ − ( r, r). − ∞ 1 (m) m (iii) The series m=0 m! f (0)x has radius of convergence r > 0 and the Taylor series converges to something otherP than f(x). Here is an example of each of these three types of behaviour. − ∞ e t Example (i) For this example, f(x)= 0 1+x2t dt. Our work on interchanging the order of differentiation and integration yields that f(x) is infinitelyR differentiable and that the derivatives are given by ∞ − dn ∂n e t n f(x)= n dt dx Z ∂x 1+x2t 0 c Joel Feldman. 2008. All rights reserved. February 28, 2008 Taylor Series and Asymptotic Expansions 1 Instead of evaluating the derivative directly, we will use a little trick to generate the Taylor series for f(x). Recall that N N+1 rn = 1 r 1−r − 1−r nX=0 We just substitute r = x2t to give − N n 2 N+1 1 = ( x2t) + (−x t) 1+x2t − 1+x2t nX=0 and N ∞ ∞ n 2 N+1 f(x)= ( x2t) e−t dt + (−x t) e−t dt Z − Z 1+x2t nX=0 0 0 N ∞ ∞ N+1 = ( 1)nx2n tne−t dt + ( 1)N+1x2(N+1) t e−t dt − Z − Z 1+x2t nX=0 0 0 N ∞ N+1 = ( 1)nn! x2n + ( 1)N+1x2(N+1) t e−t dt − − Z 1+x2t nX=0 0 (1) ∞ n −t th (m) since 0 t e dt = n!. To compute the m derivative, f (0), we choose N such that 2(N + 1) > m and applyR dm 2n 0 if m =2n dxm x = 6 x=0 (2n)! if m =2n dm N +1 2(N+1) ∞ tN+1 −t Applying m with m < 2(N + 1) to ( 1) x e dt gives a finite sum with every term dx − 0 1+x2t containing a factor xp with p> 0. When x is set to 0, everyR such term is zero. So m 0 if m is odd d m m f(x) = m dx ( 1) 2 !m! if m is even x=0 2 − So the Taylor series expansion of f is ∞ ∞ ∞ 1 (m) m 1 m m m m=2n n 2n f (0)x = ( 1) 2 !m!x = ( 1) n!x m! m! − 2 − mX=0 mX=0 nX=0 m even The high growth rate of n! makes the radius of convergence of this series zero. For example, one formula −1 −1 ∞ n an+1 n an+1 for the radius of convergence of anx is lim . When an = ( 1) n!, lim = n=0 n→∞ an − n→∞ an −1 h i h i lim (n + 1) = 0. ReplacingP x by x2 gives that ∞ ( 1)nn!x2n converges if and only if x = 0. n→∞ n=0 − Nonetheless, the error estimates P N ∞ ∞ N+1 f(x) ( 1)nn! x2n = x2(N+1) t e−t dt x2(N+1) tN+1e−t dt = (N + 1)! x2N+2 − − Z 1+x2t ≤ Z nX=0 0 0 shows that if we fix N and make x smaller and smaller, N ( 1)nn! x2n becomes a better and better n=0 − approximation to f(x). On the other hand, if we fix x Pand let N grow, we see that the error grows dramatically for large N. Here is a graph of y = (N +1)! x2N+2 against N, for several different (fixed) values of x. ∞ ∞ (1) n −t −t n −t n−1 Denote γn = t e dt. Then γ0 = e dt = 1 and, by integration by parts with u = t , dv = e dt, du = nt dt 0 0 −t and v = −e R, γn = nγn−1 for all n ∈RIN. So, by induction, γn = n!. c Joel Feldman. 2008. All rights reserved. February 28, 2008 Taylor Series and Asymptotic Expansions 2 (N + 1)! x2N+2 x =0.75 x =0.6 x =0.5 x =0.4 3 2 1 12345678910 N Example (ii) For this example, f(x)= ex. By Taylor’s theorem, n xm f(x)= m! + Rn(x) mX=0 ′ ex n+1 ′ ∞ xm with Rn(x) = (n+1)! x for some x between 0 and x. In this case, m=0 m! has radius of convergence infinity and the error R (x) 1 e|x| x n tends to zero both P n ≤ n! | | (a) when n is fixed and x tends to zero and ¯ 1 |x| n (b) when x is fixed and n tends to infinity. Use Rn(x) = n! e x to denote our bound on the error n | | introduced by truncating the series at x . Then R¯ (x)= |x| R¯ (x). So if, for example, n 2 x , n! n+1 n+1 n ≥ | | then increasing the index n of R¯n(x) by one reduces the value of R¯n(x) by a factor of at least two. Example (iii) For this example, −1/x2 f(x)= e if x =0 0 if x =06 I shall show below that f is infinitely differentiable and f (n)(0) = 0 for every n IN 0 . So in this case, ∞ n ∈ ∪{ } the Taylor series is n=0 0x and has infinite radius of convergence. When we approximate f(x) by the n m Taylor polynomial, Pm=0 0x , of degree n, the error, Rn(x)= f(x), is independent of n. The function f(x) agrees with its TaylorP series only for x = 0. Similarly, the function −1/x2 g(x)= e if x> 0 0 if x 0 ≤ agrees with its Taylor series if and only if x 0. ≤ Here is the argument that f is infinitely differentiable and f (n)(0) = 0 for every n IN 0 . For any ∈ ∪{ } x = 0 and any integer n 0 6 ≥ 2 2 n n e1/x =1+ 1 + 1 1 + + 1 1 + 1 1 x2 2! x2 ··· n! x2 ··· ≥ n! x2 Consequently, for any polynomial P (y), say of degree p, there is a constant C such that 1 1 2 P C p P 1 e−1/x = x |x| (p + 1)! C x2+p x e1/x2 ≤ 1 1 ≤ (p+1)! x2(p+1) c Joel Feldman. 2008. All rights reserved. February 28, 2008 Taylor Series and Asymptotic Expansions 3 for all x 1. Hence, for any polynomial P (y), | |≤ 2 lim P 1 e−1/x =0. (1) x→0 x Applying (1) with P (y)= y, we have 0 if x =0 ′ f(x)−f(0) 1 −1/x2 ′ f (0) = lim = lim e =0 = f (x)= 2 x→0 x x→0 x ⇒ 2 e−1/x if x =0 x3 6 Applying (1) with P (y)=2y4, we have ′ ′ 0 if x =0 ′′ f (x)−f (0) 1 2 −1/x2 ′′ f (0) = lim = lim 3 e = f (x)= 2 x→0 x x→0 x x ⇒ 6 + 4 e−1/x if x =0 − x4 x6 6 2 Indeed, for any n 1, if f (n−1)(x)= P 1 e−1/x for all x> 0, then applying (1) with P (y)= yP (y) ≥ n−1 x n−1 gives n− n− (n) f ( 1)(x)−f ( 1)(0) 1 1 −1/x2 f (0) = lim = lim Pn−1 e =0 x→0 x x→0 x x and hence 0 if x =0 (n) f (x)= 2 1 P ′ 1 + 2 P 1 e−1/x if x =0 − x2 n−1 x x3 n−1 x 6 Example (iv) Stirling’ formula ln n! n + 1 ln n n + ln √2π actually consists of the first few terms of ≈ 2 − an asymptotic expansion ln n! n + 1 ln n n + ln √2π + 1 1 1 1 + ∼ 2 − 12 n − 360 n2 ··· If n> 10, the approximation ln n! n + 1 ln n n + ln √2π is accurate to within 0.06% and the exponen- ≈ 2 − tiated form n+ 1 −n+ 1 n! n 2 √2πe 12n ≈ is accurate to one part in 300,000.