Summation Techniques, Padé Approximants, and Continued Fractions

Transcription

1 Chapter 5 Summation Techniques, Padé Approximants, and Continued Fractions 5. Accelerated Convergence Conditionally convergent series, such as = ( ) n+ = ln2, (5.) n converge very slowly. The same is true for absolutely convergent series, such as n= If we call the partial sum for the latter n= π2 = ζ(2) = n2 6. (5.2) N n= n 2 = S N, (5.3) the difference between the limit S and the Nth partial sum is S S N = n=n+ n 2 N dn n 2 = N, (5.4) which means that it takes 0 6 terms to get 6-figure accuracy. Thus,toevaluateaconvergentseries,thelast thingyouwanttodoisactually literally carry out the sum. We need a method to accelerate the convergence, andgetgoodaccuracyfromafew termsin the series. Thereareseveralstandard methods. 4 Version of September 27, 206

2 42 Version of September 27, 206 CHAPTER 5. APPROXIMANTS 5.. Shanks Transformation The Shanks transformation is good for alternating series, or oscillating partial sums, such as Eq. (5.). For the series consider the Nth partial sum S = S N = a n, (5.5) n= N a n. (5.6) n= Let us suppose that, for sufficiently large N, S N = S +Ab N, (5.7) where 0 < b <, so that as N, S N S. We will take this as an ansatz for all N, to obtain an estimate for the limit S. Then, successive partial sums satisfy S N = S +Ab N, S N = S +Ab N, S N+ = S +Ab N+, (5.8a) (5.8b) (5.8c) so that which may be immediately solved for S, b = S N+ S S N S = S N S S N S, (5.9) S (N) = S N+S N S 2 N S N+ +S N 2S N, (5.0) 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 +, and N partial sums. For the series (5.) the first 5 partial sums are S =, S 2 = 2 = 0.5, S 3 = 5 6 = 0.833, S 4 = 7 2 = , S 5 = 47 = , (5.) 60 which oscillate around the correct limit ln2 = , but are not good approximations. Using the Shanks transformation (5.0) we obtain much better approximants: S () = 7 0 = 0.700, S (2) = = 0.690, S (3) = 25 = , (5.2) 36

3 5.. ACCELERATED CONVERGENCE 43 Version of September 27, 206 which use only the first 3, 4, and 5 terms in the original series. We can do even better by iterating the Shanks transformation, S [2] (N) = S (N+)S (N ) S 2 (N) S (N+) +S (N ) 2S (N), (5.3) and then we find using the same data (only 5 terms in the series) S [2] (2) = 65 = , (5.4) 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 Richardson Extrapolation For monotone series, Richardson extrapolation is often very useful. In this case we are considering partial sums S N which approach their limit S monotonically. In this case we assume an asymptotic form for large N S N S + a N + b N 2 + c +... (5.5) N3 The first Richardson extrapolation consists of keeping only the first correction term, S N = S + a N, S N+ = S + a N +, (5.6) which may be solved for the limit S [] (N) = (N +)S N+ NS N, (5.7) where again we ve inserted on the left a superscript [] indicating the first Richardson extrapolation, and a subscript (N) to indicate the approximant comes from the Nth and N +st partial sums. We consider as an example Eq. (5.2). Here, the first 4 partial sums are S =, S 2 = 5 4 =.25, S 3 = =.36, S 4 = 205 =.424, (5.8) 44 to be compared with π 2 /6 = The first three Richardson extrapolants are much better: S [] () = 3 =.5, S[] 2 (2) = 9 =.58, S[] 2 (3) = 29 =.6. (5.9) 8 Iteration of these results by inserting S [] (N) in (5.7) yields further improvement: 5/3 =.667, but this iteration improves only slowly with N. To do better we keep the first two terms in (5.5). This gives the second Richardson extrapolant, S [2] (N) = [ (N +2) 2 S N+2 2(N +) 2 S N+ +N 2 ] S N. (5.20) 2

4 44 Version of September 27, 206 CHAPTER 5. APPROXIMANTS When applied to the series (5.2) the first three terms in the series yields nearly % accuracy: S [2] () = 3 =.625. (5.2) 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 , (5.22a) (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 5.2. Euler Summation Suppose a n. (5.23) a n x n = f(x) (5.24) converges if x <. Then we define the limit of the series (5.23) by Thus, for the series (5.22b), f(x) is f(x) = S = lim x f(x). (5.25) S = ( ) n, (5.26) ( ) n x n = +x, (5.27) sos = /2. Tosupplymorecredencetothis result, wenote thatit isreproduced by the Shanks transformation. The partial sums of the series are S 0 =, S = 0, S 2 =, S 3 = 0,..., (5.28)

5 5.2. SUMMING DIVERGENT SERIES 45 Version of September 27, 206 so S = S N+S N S 2 n S N+ +S N 2S n = 2 for all N. What if we apply Euler summation to the series (5.29) ? (5.30) Now f(x) = x 2 +x 3 x 5 +x 6 x 8 +x 9... = x 3n x 2 x 3n = x2 x 3 = +x +x+x2, (5.3) so the sum of (5.30) is S = f() = 2 3. (5.32) Thus the process of summation is not (infinitely) associative. In this case the Shanks transformation does not work Borel Summation Now we use the Euler representation of the Gamma function, or the factorial, n! = Then we formally interchange summation and integration: S = a n n! 0 0 dtt n e t = dtt n e t. (5.33) 0 dte t n! a nt n, (5.34) which defines the sum if exists. Thus for (5.22b), and so g(t) = g(t) = S = n! a nt n (5.35) ( ) ntn n! = e t, (5.36) 0 dte 2t = 2, (5.37)

6 46 Version of September 27, 206 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. linear, meaning that if a n = A, b n = B, (5.38a) then and (αa n +βb n ) = αa+βb, (5.38b) 2. satisfies a n = a 0 + a n, (5.39) n= is unique. In fact, from these two properties alone (which are satisfied by both EulerandBorelsummation)wecanfindthe valueofthesum. Thusforexample, = S = ( ) = S, (5.40) implies S = /2. Slightly more complicated is S = ( ) = +( ) = +0+( ), (5.4) where adding the three lines gives 3S = 2+( ) = 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 = +(+++...) = +S, (5.43) which is only satisfied by S =. Yet such a series can be summed Zeta-function Summation Recall that the zeta function is defined by ζ(s) = n= ns, Res >. (5.44)

7 5.2. SUMMING DIVERGENT SERIES 47 Version of September 27, 206 In fact, ζ(s) exists for all s, so we can use that function to define the sum almost everywhere in the complex s plane. In particular, for s = 0: = ζ(0) = 2. (5.45) Even a more divergent sum can be evaluated this way: n = ζ( ) = 2. (5.46) n= Note the remarkable fact that these sums are not only finite, but negative, even though each term in the sum is positive! Casimir Effect Here we give a physical example of the utility of this last mode of summation. The physics is that of a pair of parallel metallic plates, separated by a distance a in the vacuum. Because the plates modify the properties of the vacuum, there is a change in the zero-point energy of the electromagnetic field, which feels the plates because they are conductors. The result is an attraction between the plates, the famous Casimir effect, predicted by Casimir in 948 (the same year that Schwinger discovered how to renormalize quantum electrodynamics), and now verified by many experiments at the percent level. The zero-point energy (per unit area) of modes confined by the plane boundaries at z = 0 and z = a is E = 2 hω = hc 2 n= d 2 k (2π) 2 k 2 + ( nπ ), (5.47) a where in the mode sum we have integrated over the two transverse wavenumbers k x andk y, andsummedoverthediscretemodes,which, say,mustvanishatz = 0 and a, that is, be given by an (unnormalized) mode function φ(z) = sin nπ z. (5.48) a Now we write the square root as integral, putting its argument in the exponential: ( nπ ) 2 k 2 + = a Γ ( ds ) 2 0 s s /2 e (k2 +(nπ/a) 2 )s, (5.49) and then interchange the two integrals: E = hc 2 n= 0 ds ) 2 s( 2 dk s s 3/2e (nπ/a) 2π e k2 2 π. (5.50) Here we have recognized that the two-dimensionalintegral over k = (k x,k y ) can be broken into the product of two one-dimensional integrals because e (k2 x +k2 y )s = e k2 x s e k2 y s. (5.5)

8 48 Version of September 27, 206 CHAPTER 5. APPROXIMANTS These one-dimensional integrals are simply Gaussians, so the squared factor in (5.50) is simply /(4πs). The remaining s-integral is again a gamma function: E = hc 6π 3/2 n= 0 = hc ( 6π 3/2Γ 3 2 ds s 5/2e (nπ/a) ) ( nπ ) 3 a = hcπ2 440a3, (5.52) where we have used the facts that ( Γ 3 ) = 4 π, ζ( 3) = , (5.53) together with the zeta-function continuation embodied in Eq. (5.44) When multiplied by 2, for the two polarization states of the photon, this is exactly Casimir s result, which implies an attractive force per unit area between the plates, P = a E = hcπ2 240a 4 = Nm 2 /a 4. (5.54) 5.3 Padé Approximants Consider a partial Taylor sum, T N+M (z) = N+M 2 s a n z n, (5.55) which is an N +Mth degree polynomial. Write this in a rational form, N PM(z) N = A nz n M m=0 B (5.56) mzm, which is called the [N,M]th Padé approximant. Here the coefficients are determined from the Taylor series coefficients as follows: We set B 0 =, and determine the (N+M+) coefficients A 0, A,..., A N and B, B 2,..., B M by requiring that when the rational function (5.56) be expanded in a Taylor series about z = 0 the first N +M + coefficients match those of the original Taylor expansion (5.55). Example Consider the exponential function e z = +z + 2 z (5.57)

9 5.3. PADÉ APPROXIMANTS 49 Version of September 27, 206 The [,] Padé of this is of the form which, when expanded in a series about z = 0 reads P (z) = A 0 +A z +B z, (5.58) P (z) A 0 +(A B A 0 )z +(B 2 A 0 A B )z 2. (5.59) Matching this with Eq. (5.57), we obtain the equations A 0 =, A B A 0 =, B (B A 0 A ) = 2, (5.60a) (5.60b) (5.60c) so we learn immediately that A 0 =, B = 2, A = 2, (5.6) so the [,] Padé is P(z) = + 2 z (5.62) 2z. How good is this? For example, at z =, P () = 3, (5.63) which is 0% larger than the exact answer e = , and is not quite as good as the result obtained from the first three terms in the Taylor series, +z + 2 z2 z= = 2.5, (5.64) about 8% low. However, in higher orders, Padé approximants rapidly outstrip Taylor approximants. Table 5. compares the numerical accuracy of P M N with T N+M. Note that typically the Padé approximant, obtained from a partial Taylor sum, is more accurate than the latter. This comes at a price, however; the Padé, being a rational expression, has poles, which are not present in the original function. Thus, e z is an entire function, while the [,] Padé approximant of this function has a pole at z = 2. Example Here s another example: z log(+z) = z 2 + z2 3 z3 4 + z4 5 z (5.65) 6

10 50 Version of September 27, 206 CHAPTER 5. APPROXIMANTS T N+M () PM N () Relative error of Padé T 3 () = P2 () = % T 4 () = P2 2 () = % T 5 () = 2.77 P3() 2 = % T 6 () = P3 3 () = % T 7 () = P4 3 () = % Table 5.: Comparison of partial Taylor series with successive Padé approximants for the exponential function, evaluated at z =. Note that precisely the same data is incorporated in T N+M and in P N M. Approximant z = 0.5 z = z = 2 Exact P P Table 5.2: Padé approximations for the function (/z) log( + z) compared with the exact values. Note that the Taylor series for this function has a radius of convergence of unity, yet the Padé approximations converge rapidly even beyond the circle of convergence. It is a simple algebraic task to expand the form of an [N,M] Padé in a Taylor series and compute the Padé coefficients by matching with the above. This can, of course, be easily implemented in a symbolic program. For example, in Mathematica, Doing so here yields PM N (z) = PadeApproximant[f[z],{z,0,{N,M}}]. (5.66) P (z) = z + 3 z z z z (5.67) 35z3 Table 5.2 shows representative numerical values for P 3 3 and P 3 4. The Padé approximants rapidly converge to the correct value even well beyond the circle of convergence of the original series. Note further in this example that PN N is larger than the function, and decreases monotonically toward it, and

11 5.3. PADÉ APPROXIMANTS 5 Version of September 27, 206 PN+ N is smaller than the function, and increases monotonically toward it. This bounding behavior is typical of a class of functions. For more detail see C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers (McGraw-Hill, New York, 978), pp. 383ff. Field Theory Examples The following function occurs in the field theory of a massless particle in zero dimensions, Z(δ) = = 2 π = 2 π Γ dx π e (x2 ) +δ (2+2δ) ( 3+2δ 2+2δ where the gamma function was defined by Euler as and satisfies the identity 0 dt t t/(2+2δ) e t = 2 ( ) π (2+2δ) Γ 2+2δ ), (5.68) Γ(z) = 0 dt t tz e t, (5.69) Γ(z +) = zγ(z). (5.70) The gamma function generalizes the factorial to complex values: Γ(n+) = n!, n = 0,, 2,... (5.7) Becausethe gamma function Γ(z) has poles when z = N, N = 0,, 2,..., this function has an infinite number of singularities between δ = 3/2 and δ =. Thus the radius of convergence of the Taylor series about δ = 0 is. Yet low order Padé s for E(δ) = log Z(δ) give an excellent approximation well outside of this radius, as Table 5.3 shows. The partition function for a zero-dimensional field theory with a mass µ is given by the function 2 Z(δ) = µ π 0 dxe µ2 2 x2 λ(x 2 ) +δ. (5.72) We consider two cases. If µ 2 > 0, the power series in δ again has radius of convergence, but the Padé approximants are accurate far beyond this radius, as shown in Table 5.4. If, ontheotherhandµ 2 < 0(whichcorrespondstothe Higgsmechanism in particle physics), the Taylor series converges nowhere, yet the Padé approximant is still quite good, as seen in Table 5.5.

12 52 Version of September 27, 206 CHAPTER 5. APPROXIMANTS δ T 0 (δ) T 20 (δ) P2 3(δ) P5 4 (δ) E(δ) Table 5.3: Approximations to the function (5.68). What is approximated is E(δ) = logz(δ). The Padé approximants based on 6 and 0 terms in the Taylor series of this function are far more accurate that the 0 and 20 term truncated Taylor series, and even are remarkably accurate far outside the circle of convergence, where the Taylor series is meaningless. δ T 8 (δ) P 4 4(δ) Z(δ) Table 5.4: Comparison of Z(δ), Eq. (5.72), µ 2 > 0, with the 8-term truncated power series, and the corresponding [4,4] Padé. Here we have taken µ 2 =, λ =. δ T 8 (δ) P 4 4(δ) Z(δ) Table 5.5: Comparison of Z(δ), Eq. (5.72), µ 2 < 0, with the 8-term truncated power series, and the corresponding [4,4] Padé. Here we have taken µ 2 =, λ =.

13 5.4. CONTINUED FRACTIONS 53 Version of September 27, Continued Fractions 5.4. Number Theory The most familiar way of representing real numbers is in terms of a decimal fraction, which is nonterminating and nonrepeating if the number is irrational. However, there are other representations which, if less familiar, can be very useful. For example, the base of the natural logarithms e can be written in the form of a continued fraction, e = (5.73a) Because this built-up form is cumbersome to write, we could write this as e = 2+/(+/(2+/(+/(+/(4+/(+/(+/(6+/(+/(+..., (5.73b) or even more compactly as e = (5.73c) The form seen here is the representation of a real number x in the form x = a 0 + a + a 2 + a 3 + a , (5.74) where the numbers a n are integers called partial quotients. The rational number formed by including only the first n+ partial quotients a 0, a,..., a n is called the n convergent of x. So the continued fraction is given by the set of a n s: e = {2,,2,,,4,,,6,,,8,,,0,,,2,...}, (5.75) and the successive convergents, which rapidly approach e = , are { 2,3, 8 3, 4, 9 7, 87 32, 06 39, 93 7, , , , , , 497 } 8089,... = {2, 3, , , , , , , , , , , , ,...}. (5.76) The partial quotients of x are determined by successively determining the unique integer that provides a bound for x for a given truncation of the partial fraction. Thus in the above example, where in each case 0 < r <, 2 < e, (5.77a) 5 2 < 2+ < 3, +r (5.77b)

14 54 Version of September 27, 206 CHAPTER 5. APPROXIMANTS 8 3 < r < 4, (5.77c) 9 7 < r < 4, (5.77d) 9 7 < r < 30, (5.77e) and so on. The successive convergents are the upper and lower bounds corresponding to r = 0. The partial fraction representation of real numbers can be generated using your favorite symbolic program. For example, in Mathematica the first n partial quotients of x are given by and the first n convergents are given by ContinuedF raction[x, n], (5.78) Convergents[x, n]. (5.79) Let us conclude this subsection with the following comments. Evidently, a rational number is represented by a terminating continued fraction. For example, exactly = {0,99908,2,,,,,3,3,2,,2} (5.80) An algebraic number, that is one which is a solution of an algebraic equation, which is not rational, is represented by a repeating pattern of partial quotients. For example, 37 = {,,2,2,,,2,2,,22,,2,2,,,2,2,,22,...}. (5.8) A trancendental number is represented by a nonrepeating pattern. That pattern is simple in the case of e, but not so for the case of π: π = {3,7,5,,292,,,,2,,3,,4,2,,,2,2,2,2,,84,2,,,5, 3,3,,4,2,6,6,99,,2,2,6,3,5,,,6,8,,7,,2,3,7,...}.(5.82) The first few convergents are π = {3, , , , , ,...}, (5.83) so ten figure accuracy requires 6 terms. However, there are other continued fraction representations for π that have simple patterns: , (5.84)

15 5.4. CONTINUED FRACTIONS 55 Version of September 27, 206 which has rather poor partial sums: π = {4, , , , ,...}; (5.85) 3 2 π = , (5.86) where the convergents are somewhat better 5 2 π = {3, , , , , 3.427,...}; (5.87) which is comparable, π = , (5.88) π = {4,3,3.6667,3.3725,3.4234,...}. (5.89) All of these are much worse than the rapid convergence of the standard convergents. But it is the existence of simple patterns that is perhaps remarkable Continued Fraction Representation of Functions If a function is represented by a power series about the origin, f(x) = a n x n, (5.90) we can also write it in a continued-fraction form. The standard approach here is to write f(x) = + = b 0 + b 0 b x b + 2 x + b 3 x + b 4 x +... b x + b 2 x + b 3 x + b 4 x (5.9) Evidently, there is a one-to-one correspondance between the Taylor-series coefficients {a n } and the continued-fraction coefficients {b n }, which may be determined by expanding the continued fraction in a power series for small x. The theory of such a representation is discussed also in the book by Bender and Orzag. Let us consider a function with the property f(0) = ; this is merely a convenient choice of normalization. Then the relation between the continued fraction coefficients and the series coefficients is easily found to be a 0 = b 0 =, a = b, a 2 = b (b +b 2 ), a 3 = b [b 2 b 3 +(b +b 2 ) 2 ], a 4 = b [b 2 b 3 (b 3 +b 4 )+2(b +b 2 )b 2 b 3 +(b +b 2 ) 3 ], (5.92a) (5.92b) (5.92c) (5.92d) (5.92e)

16 56 Version of September 27, 206 CHAPTER 5. APPROXIMANTS and so on. This constitutes a nonlinear mapping from the set of numbers {b n } to the set {a n } or vice versa. This mapping seems to be quite remarkable in that the sequence of b n s is typically much simpler than the sequence of a n s. Here are some examples: Example Let b n = n, that is, b =, b 2 = 2, etc. Then by computing the first few a n s from the above formulæ we find b n = n a n = (2n )!!. (5.93) Example 2 Let the continued-fraction sequence be {b n } = {,,2,2,3,3,4,4,...}. Then the power series coefficients are given by the factorial, a n = n!. (5.94) Example 3 What if b n = n 2? The first few a n are a =, a 2 = 5, a 3 = 6, a 4 = 385. (5.95a) (5.95b) (5.95c) (5.95d) These are recognized as the first few Euler numbers, defined by the generating function cosht = t n E n n!, (5.96) namely E 0 =, E 2 =, E 4 = 5, E 6 = 6, E 4 = 385, (5.97a) (5.97b) (5.97c) (5.97d) (5.97e) and we conclude a n = E 2n. (5.98)

17 5.4. CONTINUED FRACTIONS 57 Version of September 27, 206 Example 4 This suggests that we ask what sequence of b n s corresponds to the Bernoulli numbers. It takes a bit of playing around to find the correct normalization, which matters since the transformation is nonlinear. If we take a n = 6B 2n+2, (5.99) we find that the corresponding continued-fraction coefficients are given by b n = n(n+)2 (n+2) 4(2n+)(2n+3). (5.00) Although the latter seems a bit complicated, it is a closed algebraic expression. It further grows with n as a low power. Neither of these features hold for the Bernoulli numbers, which grow more rapidly than exponentially, and have no closed-form representation. These ideas are provocative, yet the general significance of these results remains elusive. There appears also to be some deep connection to field theory. See C. M. Bender and K. A. Milton, J. Math. Phys. 35, 364 (994) for more details.