Chapter 5 Summation Techniques, Pad´eApproximants, and Continued Fractions 5.1 Accelerated Convergence Conditionally convergent series, such as ∞ 1 1 1 1 1 1 1 + + ... = ( 1)n+1 = ln 2, (5.1) − 2 3 − 4 5 − 6 − n nX=1 converge very slowly. The same is true for absolutely convergent series, such as ∞ 1 π2 = ζ(2) = . (5.2) n2 6 nX=1 If we call the partial sum for the latter N 1 = S , (5.3) n2 N nX=1 the difference between the limit S and the Nth partial sum is ∞ ∞ 1 dn 1 S SN = = , (5.4) − n2 ≈ Z n2 N n=XN+1 N which means that it takes 106 terms to get 6-figure accuracy. Thus, to evaluate a convergent series, the last thing you want to do is actually literally carry out the sum. We need a method to accelerate the convergence, and get good accuracy from a few terms in the series. There are several standard methods. 41 Version of September 27, 2016 42 Version of September 27, 2016 CHAPTER 5. APPROXIMANTS 5.1.1 Shanks’ Transformation The Shanks transformation is good for alternating series, or oscillating partial sums, such as Eq. (5.1). For the series ∞ S = an, (5.5) nX=1 consider the Nth partial sum N SN = an. (5.6) nX=1 Let us suppose that, for sufficiently large N, N SN = S + Ab , (5.7) where 0 <b< 1, so that as N , SN S. We will take this as an ansatz for all N, to obtain an estimate→ for ∞ the limit→ S. Then, successive partial sums satisfy N−1 SN−1 = S + Ab , (5.8a) N SN = S + Ab , (5.8b) N+1 SN+1 = S + Ab , (5.8c) so that S S S S b = N+1 − = N − , (5.9) S S S − S N − N 1 − which may be immediately solved for S, 2 SN+1SN−1 SN S(N) = − , (5.10) S + S − 2S N+1 N 1 − N where now we’ve inserted the (N) subscript on the left to indicate this is an estimate for the limit, based on the N, N + 1, and N 1 partial sums. For the series (5.1) the first 5 partial sums are − 1 5 7 S =1, S = =0.5, S = =0.833, S = =0.5833, 1 2 2 3 6 4 12 47 S = =0.7833, (5.11) 5 60 which oscillate around the correct limit ln 2 = 0.693147, but are not good ap- proximations. Using the Shanks transformation (5.10) we obtain much better approximants: 7 29 25 S = =0.700, S = =0.690, S = =0.6944, (5.12) (1) 10 (2) 42 (3) 36 5.1. ACCELERATED CONVERGENCE 43 Version of September 27, 2016 which use only the first 3, 4, and 5 terms in the original series. We can do even better by iterating the Shanks transformation, 2 S S − S [2] (N+1) (N 1) − (N) S(N) = , (5.13) S + S − 2S (N+1) (N 1) − (N) and then we find using the same data (only 5 terms in the series) 165 S[2] = =0.693277, (5.14) (2) 238 an error of only 0.02%! For more detailed comparison of Shanks estimates for this series, see Table 8.2 on page 373 of Bender and Orzag. 5.1.2 Richardson Extrapolation For monotone series, Richardson extrapolation is often very useful. In this case we are considering partial sums SN which approach their limit S monotonically. In this case we assume an asymptotic form for large N a b c S S + + + + .... (5.15) N ∼ N N 2 N 3 The first Richardson extrapolation consists of keeping only the first correction term, a a S = S + , S = S + , (5.16) N N N+1 N +1 which may be solved for the limit S[1] = (N + 1)S NS , (5.17) (N) N+1 − N where again we’ve inserted on the left a superscript [1] indicating the first Richardson extrapolation, and a subscript (N) to indicate the approximant comes from the Nth and N + 1st partial sums. We consider as an example Eq. (5.2). Here, the first 4 partial sums are 5 49 205 S =1, S = =1.25, S = =1.361, S = =1.424, (5.18) 1 2 4 3 36 4 144 to be compared with π2/6=1.644934. The first three Richardson extrapolants are much better: 3 19 29 S[1] = =1.5, S[1] = =1.58, S[1] = =1.611. (5.19) (1) 2 (2) 12 (3) 18 [1] Iteration of these results by inserting S(N) in (5.17) yields further improvement: 5/3=1.667, but this iteration improves only slowly with N. To do better we keep the first two terms in (5.15). This gives the second Richardson extrapolant, 1 S[2] = (N + 2)2S 2(N + 1)2S + N 2S . (5.20) (N) 2 N+2 − N+1 N 44 Version of September 27, 2016 CHAPTER 5. APPROXIMANTS When applied to the series (5.2) the first three terms in the series yields nearly 1% accuracy: 13 S[2] = =1.625. (5.21) (1) 8 For further numerical details, see Table 8.4 on page 377 of Bender and Orzag. 5.2 Summing Divergent Series The series encountered in physics, typically perturbation expansions, are usually divergent. How can one extract a meaningful number from such series, which represent physical processes and so reflect real processes? On the surface, it would seem impossible to attach any meaning to such obviously divergent series as 1+1+1+1+1+ ..., (5.22a) 1 1+1 1+1 .... (5.22b) − − − However, as we will now see, perfectly finite numbers can be associated with these series. Again there are various procedures, of which we give a sampling. Throughout, we are considering a divergent series of the form ∞ an. (5.23) nX=0 5.2.1 Euler Summation Suppose ∞ n anx = f(x) (5.24) nX=0 converges if x < 1. Then we define the limit of the series (5.23) by | | S = lim f(x). (5.25) x→1 Thus, for the series (5.22b), ∞ S = ( 1)n, (5.26) − nX=0 f(x) is ∞ 1 f(x)= ( 1)nxn = , (5.27) − 1+ x nX=0 so S =1/2. To supply more credence to this result, we note that it is reproduced by the Shanks transformation. The partial sums of the series are S0 =1, S1 =0, S2 =1, S3 =0,..., (5.28) 5.2. SUMMING DIVERGENT SERIES 45 Version of September 27, 2016 so S S − S2 1 S = N+1 N 1 − n = (5.29) S + S − 2S 2 N+1 N 1 − n for all N. What if we apply Euler summation to the series 1+0 1+1+0 1+1+0 1+1+0 1+ ...? (5.30) − − − − Now f(x)=1 x2 + x3 x5 + x6 x8 + x9 ... ∞− −∞ − − = x3n x2 x3n − nX=0 nX=0 1 x2 1+ x = − = , (5.31) 1 x3 1+ x + x2 − so the sum of (5.30) is 2 S = f(1) = . (5.32) 3 Thus the process of summation is not (infinitely) associative. In this case the Shanks transformation does not work. 5.2.2 Borel Summation Now we use the Euler representation of the Gamma function, or the factorial, ∞ − n!= dt tne t. (5.33) Z0 Then we formally interchange summation and integration: ∞ ∞ ∞ ∞ 1 n −t −t 1 n S = an dt t e = dt e ant , (5.34) n! Z Z n! nX=0 0 0 nX=0 which defines the sum if ∞ 1 g(t)= a tn (5.35) n! n nX=0 exists. Thus for (5.22b), ∞ n t − g(t)= ( 1)n = e t, (5.36) − n! nX=0 and so ∞ − 1 S = dt e 2t = , (5.37) Z0 2 46 Version of September 27, 2016 CHAPTER 5. APPROXIMANTS which coincides with the result found by Euler summation. In general, Borel summation is more powerful than Euler summation, but if both Euler and Borel sums exist, they are equal. In fact, we can prove that any summation that is both 1. linear, meaning that if ∞ ∞ an = A, bn = B, (5.38a) nX=0 nX=0 then ∞ (αan + βbn)= αA + βB, (5.38b) nX=0 and 2. satisfies ∞ ∞ an = a0 + an, (5.39) nX=0 nX=1 is unique. In fact, from these two properties alone (which are satisfied by both Euler and Borel summation) we can find the value of the sum. Thus for example, 1 1+1 1+1 1+ ... = S =1 (1 1+1 1+1 1+ ...)=1 S, (5.40) − − − − − − − − implies S =1/2. Slightly more complicated is S = (1+0 1+1+0 1+1+0 1+ ...) − − − = 1+(0 1+1+0 1+1+0 1+ ...) − − − = 1+0+( 1+1+0 1+1+0 1+1+0 ...), (5.41) − − − − where adding the three lines gives 3S =2+(0+0+0+0+0+ ...)=2, (5.42) or S =2/3 as before. But there are sums resistant to such schemes. An example is (5.22a), because the above process leads to S =1+(1+1+1+ ...)=1+ S, (5.43) which is only satisfied by S = . Yet such a series can be summed. ∞ 5.2.3 Zeta-function Summation Recall that the zeta function is defined by ∞ 1 ζ(s)= , Re s> 1. (5.44) ns nX=1 5.2. SUMMING DIVERGENT SERIES 47 Version of September 27, 2016 In fact, ζ(s) exists for all s = 1, so we can use that function to define the sum almost everywhere in the complex6 s plane.