INTRODUCTION TO PADÉ APPROXIMANTS E. B. Saff Center for Constructive Approximation
H. Padé (1863-1953) Student of Hermite Histhesiswon French Academy of Sciences Prize C. Hermite (1822-1901) Used Padé approximants to prove that e is trancendental But origins of subject go back to Cauchy, Jacobi, Frobenius. Historical Reference: C. Brezinski, History of Continued Fractions and Padé Approximants, Springer-Verlag, (Berlin, 1991)
Why Padé? 1) Convergence Acceleration [e.g. ɛ-algorithm] 2) Numerical Solutions to Partial Differential Equations [exp(at) Q(At) 1 P (At)] 3) Analytic Continuation of Power Series [regions of convergence beyond a disk] 4) Includes Study of Orthogonal Polys on Interval [Padé denominators for Markov functions are orthogonal] 5) Finding Zeros/Roots, Poles/Singularities [usezerosandpolesofpadé approximants to predict - e.g. QD algorithm]
Padé Approximants (PA) generalize Taylor Polynomials Given f(z) = k=0 c k z k Taylor poly P m (z) = m k=0 c k z k Then f(z) P m (z) = k=m+1 c k z k Equivalently, f(z) P m (z) =O ( z m+1) P m (0) = f(0) P m(0) = f (0) P (m) (0) = f (m) (0).
Idea of PA: Given m, n Rational function R = P/Q deg P m, deg Q n Choose P, Q so that l as large as possible. (f R)(z) =O ( z l), How large can we expect l to be? P has m +1 parameters Q has n +1 parameters P/Q has 1 parameter So total of m + n +1 parameters Expect: ( ) f P (z) =O ( z m+n+1). Q
NOT ALWAYS POSSIBLE Ex: m = n =1, f(z) =1+z 2 + z 4 +. Want R(z) = P (z) Q(z) = az + b cz + d. (1) R(z) =1+z 2 + O ( z 3). But R is either identically constant or one-to-one. From (1), neither is possible [ R (0) = 0 ]. Idea: Linearize by requiring Qf P = O ( z m+n+1) P k := all polynomials of degree k.
DEF Let f(z) = 0 c k z k be a formal power series, and m, n nonnegative integers. A Padé form (PF) of type (m, n) isapair(p, Q) such that P = Q 0and m k=0 p k z k P m, Q = n k=0 q k z k P n, (2) Qf P = O ( z m+n+1) as z 0. Proposition Padé formsoftype(m, n) always exist. Proof. (2) is a system of m+n+1 homogeneous equations in m + n + 2 unknowns: (3) n j=0 c k j q j p k =0, 0 k m (4) n j=0 c k j q j =0, k = m +1,...,m+ n. c m,n := ( c m+i j ) n i,j=1 Toeplitz matrix
THM Every PF of type (m, n) forf(z) yields the same rational function. Proof. (P, Q) and ( ˆP, ˆQ ) are PF s. Qf P = O ( z m+n+1) ˆQf ˆP = O ( z m+n+1) so ˆQP + ˆPQ = O ( z m+n+1) P m+n. Thus ˆPQ ˆQP ˆP/ˆQ P/Q. DEF The uniquely determined rational P/Q is called the Padé Approximant (PA) of type (m, n) forf(z), and is denoted by [m/n] f (z) or r m,n (f; z).
Remark In reduced form [m/n] f (z) =p m,n (z)/q m,n (z), where we (often) normalize so that q m,n (0) = 1, p m,n (0) = c 0, p m,n and q m,n relatively prime. Padé Table for f [0/0] [0/1] [0/2] [1/0] [1/1] [1/2] Taylor polys [2/0] [2/1] [2/2] Equal entries occur in square blocks.
Ex: f(z) =1+z 2 + z 4 + z 6 + ( = 1 ) 1 z 2 Block structure [0/0] = [0/1] [0/2] = [1/0] = [1/1] [1/2] = [2/0] = [2/1] [ ] = [3/0] = [3/1] [ ] = [4/0] = [4/1] [ ] = THM Let p/q be a reduced PA for f(z), with c 0 0. Suppose and m = exact deg of p n = exact deg of q qf p = O ( z m+n+k+1) exactly.
Then (a) k 0 (b) [µ/ν] f = p/q iff m µ m + k, n ν n + k. See: W. B. Gragg, The Padé Table and its Relation to Certain Algorithms of Numerical Analysis, SIAM Review (1972), 1-62. DEF APadé approximant is said to be normal if it appears exactly once in table. We say f is normal if every entry in its Padé table is normal. Ex: f(z) =e z is normal.
Determinant Representations and Frobenius Identities. f(z) = k=0 c k z k, u m,n (z) := v m,n (z) := f m (z) := m k=0 c k z k P m f m (z) zf m 1 (z) z n f m n (z) c m+1 c m c m n+1 c m+2 c m+1 c m n+2 c m+n c m+n 1 c m 1 z z n c m+1 c m c m n+1 c m+2 c m+1 c m n+2 c m+n c m+n 1 c m Note: u m,n (z) P m, v m,n (z) P n. THM f(z)v m,n (z) u m,n (z) = O ( z m+n+1).
DEF For arbitrary, but fixed polys g, h, let w m,n (z) :=g(z)u m,n (z) + h(z)v m,n (z) c m,n := det ( c m+i j ) n i,j=1 THM Between any 3 entries in the table of w m,n functions, there is a homogeneous linear relation with poly coefficients which can be computed from the coefficients c k of f. c m,n+1 w m+1,n c m+1,n w m,n+1 = c m+1,n+1 zw m,n c m+1,n w m 1,n + c m,n+1 w m,n 1 = c m,n w m,n c m,n c m+1,n w m,n+1 c m,n+1 c m+1,n+1 zw m,n 1 =(c m+1,n c m,n+1 c m,n c m+1,n+1 z)w m,n Proof. Use Sylvester s identity on determinant representation for Padé denominator v m,n. deta det A i,j;k,l =deta i;k deta j;l deta i;l deta j;k
Padé Approximants for the Exponential f(z) =e z = k=0 Want to find p m,n P m,q m,n P n such that (5) q m,n (z)e z p m,n (z) =O ( z m+n+1). Let D := d/dz. Then z k k! D [qe z ] = qe z + q e z = e z (I + D) q Apply D m+1 to (5) e z (I + D) m+1 q m,n +0=O (z n ) (I + D) m+1 q m,n = k m,n z n q m,n = k m,n (I + D) (m+1) z n. Recall (1 + x) (m+1) = j=0 ( 1) j( m + j m ) x j.
So q m,n (z)=k m,n n j=0 = k m,n n j=0 ( 1) j( m + j ) D j z n m ( 1) j( m + j m ) n! (n j)! zn j. q m,n (z) = n k=0 (m + n k)!n! ( z) k (m + n)!(n k)! k! q m,n e z p m,n = O ( z m+n+1), q m,n p m,n e z = O ( z m+n+1). So p m,n ( z) =q n,m (z), Also from p m,n (z) = m k=0 (m + n k)!m! (m + n)!(m k)! z k k!. D m+1 [q m,n e z p m,n ]=k m,n z n e z, and integration by-parts we get
q m,n (z)e z p m,n (z) = ( 1)n 1 (m + n)! zm+n+1 0 sn (1 s) m e sz ds. Remark For z ρ, q m,n (z) 1+ρ + ρ2 2! + ρ3 3! + = eρ. So q m,n form a normal family in C. Further, if m + n, m/n λ, Hence... coeff of z k ( 1)k (1 + λ) k k!.
THM (Padé) Let m j, n j Z + satisfy m j + n j, m j /n j λ as j. Then and lim j q m j,n j (z) =e z/(1+λ), lim j p m j,n j (z) =e λz/(1+λ), lim [m j/n j ](z) =e z, j locally uniformly in C. More precisely (m = m j,n= n j ) [m/n](z) e z = m!n! z m+n+1 e 2Re(z)/(1+λ) (m + n)!(m + n +1)! (1 + o(1)). COR All zeros and poles of PA s to e z go to infinity as m + n. But where are they located?
Zeros of p m,0 (z) = m k=0 z k /k!,m=1, 2,...,40 THM (S+Varga) For every m, n 0, the normalized Padé numerator p m,n ((n +1)z) for e z is zero-free in the parabolic region Result is sharp! P : y 2 4(x +1), x > 1.
THM (S+Varga) Consider any ray sequence [m j /n j ](z) wheren j /m j σ (0 σ< ). S σ := { z : arg z > cos 1 [(1 σ)/(1 + σ)] } C σ ze g(z) w σ (z) = 2 [1 + z + g(z)] (1+σ) [1 z + g(z)] where g(z) := 1+z 2 2z ( ) 1 σ 1+σ. Then 2σ (1+σ), (i) ẑ is a lim. pt. of zeros of [m j /n j ]((m j + n j )z) iff ẑ D σ := { z S σ : w σ (z) =1, z 1 }. (ii) ẑ is a lim. pt. of poles of [m j /n j ]((m j + n j )z) iff ẑ E σ := { z C \ S σ : w σ (z) =1, z 1 }.
More recent variations: Multi-point Padé Approx. Let B (m+n) = { x (m+n) } m+n k k=0 R, R m,n = P m,n /Q m,n, deg P m,n = m, deg Q m,n = n, interpolates e z in B (m+n). THM (Baratchart+S+Wielonsky) If B (m+n) [ ρ, ρ], m = m ν, n = n ν (m + n ), then R m,n (z) e z z C. Moreover, the zeros and poles of R m,n lie within ρ of the zeros and poles, respectively, of the Padé approximants [m/n](z) toe z. COR Conclusion holds for best uniform rational approx. to e x on any compact subinterval of R. Analogous results for best L 2 -rational approximants to e z on unit circle.
Introduction to Convergence Theory f(z) = k=0 c k z k [m/0] f (z) = m k=0 c k z k converges in largest open disk centered at z = 0 in which f is analytic: 1 z <R, where = lim sup R m c m 1/m. Next simplest case: [m/1] f. v m,1 (z) =det ( Assume c m+1 0. c m /c m+1. lim inf m c m+1 c m 1 z c m+1 c m 1 R ) = c m zc m+1. Then v m,1 has zero at lim sup m c m+1 c. m It s possible for ratios to have many limit points different from 1/R.
ALL IS NOT ROSES - There can be spurious poles. Perron s Example: f entire (R = ) such that every point in C is a limit point of poles of some subsequence of [m/1] f. THM (de Montessus de Ballore, 1902) Let f be meromorphic with precisely ν poles (counting multiplicity) in the disk : z <ρ, with no poles at z =0. Then lim m [m/ν] f(z) =f(z) uniformly on compact subsets of \{ν poles of f}. Furthermore, as m, the poles of [m/ν] f tend, respectively, to the ν poles of f in. Ex: f(z) =zγ(z) haspolesatz = 1, 2,... The n-th column of Padé table will converge to zγ(z) in{ z <n+1}\{ 1,..., n}.
ProofofdeMontessusdeBalloreTheorem: Hermite s Formula Suppose g is analytic inside and on Γ, a simple closed contour. Let z 1,z 2,...,z µ be points interior to Γ, regarded with multiplicities n 1,n 2,...,n µ. Set N := n 1 + n 2 + + n µ. Then a unique poly p P N 1 such that p (j) (z k )=g (j) (z k ), j =0, 1,...,n k 1, k =1,...,µ. Moreover, p(z) = 1 2πi Γ g(z) p(z) = 1 2πi where ω(ζ) ω(z) ω(ζ)(ζ z) Γ ω(z) := ω(z)g(ζ) ω(ζ)(ζ z) µ j=1 g(ζ) dζ z C, (z z j ) n j. dζ z inside Γ,
IdeaofProofofdeM.deBalloreThm f meromorphic with ν poles in z <ρ. u m,ν, v m,ν PF of type (m, ν) forf. (6) fv m,ν u m,ν = O ( z m+ν+1). Let Q ν P ν have zeros at poles of f with same multiplicity. Q ν fv m,ν Q ν u m,ν = O ( z m+ν+1) = 1 2πi ζ =ρ ɛ for z <ρ ɛ. z m+ν+1 (Q ν fv m,ν ) (ζ) ζ m+ν+1 (ζ z) dζ, For v m,ν suitably normalized, integral 0 for z <ρ ɛ. Method extends to multi-point Padé.
What about other sequences from Padé Table, such as rows, diagonals, ray sequences? THM (Wallin) There exists f entire such that the diagonal sequence [n/n] f (z), n =0, 1, 2,..., is unbounded at every point in C except z =0. Baker-Gammel-Wills Conjecture: If f is analytic in z < 1 except for m poles ( 0), then there exists a subsequence of diagonal PAs [n/n] f (z) that converges to f locally uniformly in { z < 1}\{m poles of f}. Conjecture is FALSE! D. S. Lubinsky, Rogers-Ramanujan and..., Annals of Math, 157 (2003), 847-889.
Next step: Consider a weaker form of convergence, such as convergence in measure or convergence in capacity. Nuttall-Pommerenke Near-diagonal PAs will be inaccurate approximations to f only on sets of small capacity (transfinite diameter). THM f analytic at andinadomaind C with cap(c \ D) =0. LetR m,n (z) denote the PA to f at. Fix r > 1, λ>1. Then for ɛ, η > 0 there exists an m 0 such that R m,n (z) f(z) <ɛ m for all m>m 0,1/λ m/n λ, andforallz in z <r, z E m,n,cap(e m,n ) <η.
THM (Stahl) Let f(z) be analytic at infinity. There exists a unique compact set K 0 C such that (i) D 0 := C \K 0 is a domain in which f(z) has a single-valued analytic continuation, (ii) cap(k 0 )=inf K cap(k), where the infimum is over all compact sets K C satisfying (i), (iii) K 0 K for all compact sets K C satisfying (i) and (ii). The set K 0 is called minimal set (for singlevalued analytical continuation of f(z)) and the domain D 0 C is called extremal domain.
THM (Stahl) Let the function f(z) bedefined by f(z) = j=0 f j z j and have all its singularities in a compact set E C of capacity zero. Then any close to diagonal sequence of Padé approximants [m/n](z) to the function f(z) converges in capacity to f(z) in the extremal domain D 0.