Riet, R.P. van de

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STICHTING MATHEMATISCH CENTRUM 2e BOERHAAVESTRAAT 49 AMSTERDAM AFDELING TOEGEPASTE WISKUNDE TN 32 On certain Pade polynomials in connection with the block function by R.P. van de Riet November 1963 Introduction In the theory of electronics the block-function f(t), defined by f 1 for O ~ t ~ 1 f(t)= l O elsewhere plays an important role. Functions f1 (t) approximating this block-function may be n found by approximating the Laplace transform -~ -s (1) L[f(t)] = F(s) = 5 e • The function F(s) can be approximated by the functions (2) p ( s) - p ( -s) Fn(s) = n sP (s) , n where P (s) is the (n,n) Fade-polynomial belonging to the n s exponential function e . In section 1 the definition of the Pade polyn6mia£ and their connection with continued fractions will be given. In this note we prove the following theorem:. The zeros of the (h,n) Pade-polyn~mials Pn(z) belonging to the exponential function ez all lie in the left complex half-plane.*) By aid of this theorem it can easily be proved that the .functions (3) are bounded and that we have for all n the formula: it)In a fortlt 9oming report another shorter proof of this theorem wiTl be given, by aid of Laplace-transformation theory. -2-(4) and hence the electronic system, based upon these func-tions f (t), is stable. n In order to prove the theorem we need the following lemma: If (5) P (ix)= F (x) + iGn(x), n n where Fn(x) and Gn(x) are polynomials with real coeffi-cients, then the zeros of Fn(x) and Gn(x) are real and separate one another. This lemma will be proved in section 2. The theorem will be proved in section 3, and finally we show the relation (4) in section 4. 1. Pade approximation and continued fractions Let some function F(x) be approximated by a quotient of polynomials Pn,m(x) and ~,m(x) of degree m and n respect-ively, in such a way that ( ) Q ( ) P (X) = o(xn+m+1)·, F x ~ n,m x - n,m P ( ,o) = F ( o) and Q ( o) :. 1, n,m n,m then P m(x) and Q (x) are called the (n,m) Pade polyno-n, n,m mials belonging to F(x) (see [1] p. 420) The (n,m) Pade polynomials belonging to the exponential X function e are -3-(6) P (x) n,m = 1 + ....!!L x + m(m-1) x2 n+m (n+m)(n+m-1) 2T + ••• + t + m. (m+n)(m+n-1) .•. (n+1) and o (x) = P n(-x). r1,m m, In this note we are only interested in the (n,n) Pade polynomials, and we write: P (x) d~f P (x) = o (-x). n n,n r1,n In the following we need some results from continued fraction theory. Let a continued fraction C be given n ·~a· C =b + .be\11 I no n,:;- a2 I + n;::-2 + .•. + I :n I, which is a shorter n notation for b + 0 We define theseries tions: (7) with a2 b1 + --------' ' ' ' ', an bn-1 + -S-n { Am} and { Bm} by the A0 = b 0 , B0 = 1, A1 = b0 b1 + a1 and B1 = b1 • recurrence rela-The A and B are called them t_h convergents or m th m rq, approximants ~f the continued fraction en. -4-An elementary result from the continued fraction theory is that An en= B, for all n. n The following theorem gives a connection between conti.nued fractions and Pade polynomials (see (1] p. 452) Let the continued fraction (8) C +_S:c_J+I k2x21 +-~+ ... +, knx21+ ... 0 ~ 1+12x ~ 1+lnx be associated with the function F(x), of which the (n,n) Pade polynomials are P (x) and Q (x), then P (x) and th n n n Qn (x) are the n - approximants of this continued fraction and the following relations hold: (1) 1 (m) (m-1) then 11 = q1 ' m = q1 - q1 (m ~ 2) and (m) (m-1) 1 (m-1) km= q2 - q2 - m q1 ' moreover c0 and c1 are defined by I CO = F ( 0) , c1 = F ( 0) . From this theorem it follows immediately that the Pade polynomials P (x) and Q (x) belonging to exp x and given n """nth by formula (6) are then apprcximants of 2 2 1 1 I~ I j 4.3.1 I+ ... + 4(2n-1)(2n-3) I+ 1 I 1 associated with the exponential function. -5-From formula (7) it is easily seen that the following recursive relations are valid for P (x): n 2 (9) Pn(x) = Pn-1(x) + 4(2n-1)(2n-3) Pn-2(x) .n ~ 2 · Let now Pn(ix) = Fn(x) + iGn(x) , where Fn(x) and Gn(x) are polynomials with real coeffi-cients. By virtue of formula (9) we obtain the following recursive relation:;, for F (x) and G (x): ·~ n n 2 Fn(x) = Fn-1(x) - 4(2n-1)(2n-3) Fn-2(x), 2 n ~2 Gn(x) = G 1 (x) X Gn-2(x), n- - 4{2n-1 )(2n-3) while formula (6) gives: Fo = 1 J Go = 0 F1 1 G1 X . = , = 2 Now we try to find a continued fraction of which Gn(x) and Fn(x) are then t~ approximants. A simple algebraic calculation shows that the following continued fraction is the desired one: 2 X X 2 I; I+ I- 12 j, + ... +·4(2n-1)~2n-3)1, which is equal to 1 I (10) __j_J - __1_j _ ... _ 1 ~ . r;;~, ~ l(4n-2)x-Let the functions R0 (n)(x) and R1 (n)(x) be defined by (n) n_ -1 (n) n -1 ( 11) R0 (x) = x ,tt'n (x, ) and R1 (x) = x G0 (x ) , then Ro (n) (x) R1 (n) (x) -6-= 2 x - l ax1 When we define R2 (n)(x) by then 1 I - I (4n-2)x 1 I) - J (4n-2)x ' R0 (n)(x) = 2 x R1 (n)(x) - R2 (n)(x) . If we define consecutively Ri (n)(x) by (12) R1 (n)(x) = R1 _(n1 )(x) ( 1 l l (4i-2)x -then 1 I ) - K4n-2)x ' (n ~ i ~ 3), (13) Rf~~{x) = (4i-2)x R1(n)(x) - Rf~~ (x) (n-1 ~ 1 ~ 2) and R~~~(x) = (4n-2) x Rn(n)(x). As is well known R (n)(x) is the greatest common divisor of R (n) and R1 (n)~ since R (n) has the degree n, it foll~ws that Rn(n)(x) must ~e constant. The relations (13) will be used in the following section. 2. The zeros of F (x) and Gnhl In this section we prove the following lemma. The zeros of Fn(x) and Gn(x) are real and separate one another, moreover, P (x) has no pure imaginary zeros. n We first give some examples: -7-2 R (2 )(x) 2 1 F2 (x) = 1 X - 12 , = X - 12 :, 0 . G2 (x) X R (2 )(x) x = 2 , 1 = 2 , F3 (x) = 1 x2 R (3) (x) = x3 X - 10, 0 10 ' 2 1 - X - X x3 R1 (3\x) o3 (x) -~ , =2- 120 . 120 (In the sequel of this section we omit the upper index of the functions R0 (n)(x) and R1 (n)(x).) We remark that the degrees of R0 and R1 are n and n-1 respectively. When n is even then R and R1 have the same - 0 degrees as F and G resp., so the zeros of R and R1 , n n o are the reciprocals of the zeros of F, resp. G, with n n the exception of the zeros x = O. When n is odd, R has 0 one zero more (x=O) than Fn and R1 has one zero less than G . n In both cases the following statement is true: if and only if the zeros of R0 and R1 are real and separate one another then the same is true for the zeros of F and G. n n Since Rn= G.C.D (R0 ,R1 ) is a constant, the polynomials R0 and R1 have no zeros in common. In particular it follows that Pn(x) has no pure imaginary zeros, which proves the second statement of the lemma. ;-.. , n-i Let Ri(x) = Ci x + ••• then it follows from {13) that all ci are positive. Let V { >-.1 , ••• , X.n} be the number of variations of sign in the sequence of the real numbers X1 , •.. ,An. According to the Sturm theory of zeros of polynomials we study V * (x) = V { R0 (x), R1 (x), ... , Rn (x) } . v*(x) can only change in a zero of Ri(x)t · -8-Assume R.(x) = 0 (i~1), then R. 1 (x) and R. 1(x) l O l+ 0 l - 0 have different signs_. as follows from (13), provided they are not zero~ but this is impossible since R = Go G. D ( R. , R. 4 ) = cons t . /.0 for i = 1 , ... , n. n l 1-For a sufficiently small e it holds therefore that Ri+1 (x 0 ± e) and Ri_1 (x0 ± e) have different signs. Therefore the eventual change of sign of Ri(x), (i~1) does not effect the value of v*(x). Thus the only points where v*(x) can alter are zeros of * R (x) and then V (x) actually alters (if this zero is 0 simple). * Now V (-· ro) == n (all * c. are positive) and V (+ oo) = 0, l therefore R0 (x) has n real different zeros ·1c:e1rthermore, we see that between each zero of R0 (x) ther~ must lie a zero of R1 (x); otherwise R1 (x) should have the same sign in (xk - ~ , ¾:+1 + ~), where e is a sufficiently small positive number .• and the change of v* (x) in xk is then cancelled by the change in xk+1 · So we have shown that R0 (x) and R1 (x) have only real zeros which separate one another; this proves the lemma stated in the beginning of this section. J. Proof of the theorem In this section we shall prove the theorem, mentioned in the introduction, concerning the zeros of the (n,n) Padd polynomial belonging to the exponential function. The theorem is~ The zeros of Pn(z) all lie in the left complex half-plane · Re(z) < 0. In order to prove this theorem we use the following function-theoretic theorem~ Let D be a simply connected domain with boundary C (double--9-point free) and f(z) holomorphi~ in D, continuous in D + C and./, O on C. If the change of argument of f(z) is a, when f{z) is continued along C in the positive sense, then the number of zeros of f(z) in Dis equal to A 2 7C' • This theorem will now be applied to the Pade polynomial Pn(z), with the contour C, which consists of a part of the imaginary axis and a semi-circle with centre in z~o and radius Ras shown in fig.1. A Z=O B fig.1 theoretic theorem. The direction of the contour C is positive. We choose R so la~ge that C encloses all the zeros • ~ , 1 • .. of Pn(z) which lie in the left half plane. From the lemma proved in section 2 we know already that Pn{z) is~ O on the imaginary axis, so that we may apply the above stated function~ Moreover we assume R so large that only the main term (2n) ... (n+1) of Pn(z) is important. We now continue Pn(z) along C starting in A with Z= Rei~/2 • The argument of Pn(z) is ;ri~ in A and after continuing Pn(z) along the semi,-;-circle unto '9 the argument is ~'JC , so tne increaJ:'l is n,; . J -10-The path BA along the imaginary axis will now be studied. We decompose Pn(ix) in its real and imaginary parts. (5) Pn(ix) = Fn(x) + i Gn(x), where F (x) and G (x) are polynomials with real coefficients. n n It is clear that the function ~(x) defined by ~(x) is the tangent of the argument of Pn(ix). When n is even Fn has degree n and G0 degree n-1, when n is odd then Fn has degree n-1 and Gn degree n. Thus we see that when n is even ~ (x) + 0 for x-+ - oo,, and ~ (x) t 0 for x➔ + a,-. When n is odd ~ (x) ~ - oo for x-+- - en and ~ (x) 1' oo for X-+ CD • We now make use of the lemma shown in section 2, which states that F and G have only real zeros which separate . n n shown one another, therefore ~(x) takes the form in the figures 2 and 3 for n is even and n is odd respectively. fig.2 n is even n-1 zeros n vertical asymptotes / fig.3 n is odd n zeros n-1 vertical asymptotes. -11-From these figures it follows immediately that the argument of P (z) is increased with n;e when P (z) 1s n n continued along th~ imaginary axis from B to A. Hence the total increase of the argument is 2n7t' , thus according to the stated function-theoretic theorem, the number of zeros of Pn(z) which lie in Dis n, or all the zeros of Pn(z) lie in the left half of the complex plane, which proves the theorem. 4. Application of the theorem We now apply the theorem to the functions Fn(s) defined in formula {2) P (s) - P (-s) Fn(s) = n s P (~) • n Applying the inverse Laplace transformation to the functions Fn(s) we obtain: (14) +i <D+<F p ( ) p ( ) fn ( t ) = L -1 [ Fn ( s ) J 1 J st n s - n -s = 21ti _1 {U)!cr s Pn(s) Where q should be chosen such that all the poles of the intEgrand in (14) lie at the left hand side of the path of integration. ds, A consequence of the theorem is that~ can be taken negative ( S=O is not a pole!) We consider now the following integral: J,..,{ ( )}2 1 fcof (t)+iJoo+e~t P n (s) - P n (-s) I=o fn t dt= 2~1 ___ P_n_(_s ... )--'" o n -ioo + CJ" 5 Interchanging the order of integration, we +. (D-t(f _ ....!_ 1J Pn(s) - Pn(-s) I- 2~i . s P (s) 1 = 21ti -l CD+Cf" n P n ( s) - P n ( -s) ds s 2 Pn(s) get ds = ds dt. -12-where Lis a contour consisting of L1 and L2 as shown in fig. 4; fig.4 References [1 J O. Perron: [?] M.Marden: [3] H.S.Wall: [!+] H.S.Wall: The path of integration L encloses only one simple pole~= O with residue 1, hence 00 ~ {f0 (t)} 2dt = 1. which proves formula (4). Die Lehre von den Kettenbrttchen. 2" Auflage B.G.Teubner Berlin 1929 The geometry of zeros of a polynomial in a complex domain. Math. Society New York (1949) America Analytical theory of Continued Fracti-ons. van Nostrand New York (1948) Polynomials whose zeros have negative real parts. Amer. Math. Monthly~ pp. 308-322. 

On certain Padé-polynomials in connection with the block function

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