WARING S PROBLEM FOR LINEAR POLYNOMIALS AND LAURENT POLYNOMIALS

ROCKY MOUNTAIN JOURNAL OF MATHEMATICS Volume 35, Number 5, 005 WARING S PROBLEM FOR LINEAR POLYNOMIALS AND LAURENT POLYNOMIALS DONG-IL KIM ABSTRACT Waring s problem is about representing any function in a class of functions as a sum of kth powers of nonconstant functions in the same class We allow complex coefficients p in these kind of problems Consider 1 f i(z k = z p and f i(z k =1 Suppose that k Let p 1 and p be the smallest numbers of functions that give the above identities WK Hayman obtained lower bounds of p 1 and p for polynomials, entire functions, rational functions and meromorphic functions First, we consider Waring s problem for linear polynomials and get p 1 = k and p = k + 1 Then, we study Waring s problem for Laurent polynomials and obtain lower bounds of p 1 and p 1 Introduction 11 Waring s problem Waring s problem deals with representing any function in a class of functions as a sum of kth powers of nonconstant functions in the same class We allow complex coefficients in these problems Let k and n be natural numbers Consider the equation of the form (111 n f i (z k = Q(z, where f 1, f,,f n and Q are nonconstant polynomials with complex coefficients Suppose that (11 f 1 (z k + f (z k + + f n (z k = z Then we obtain (113 f 1 (Q(z k + f (Q(z k + + f n (Q(z k = Q(z Received by the editors on February 19, 003, and in revised form on May 1, 003 Copyright c 005 Rocky Mountain Mathematics Consortium 1533

1534 D-I KIM by the substitution of Q(z forz Hence any nonconstant polynomial Q(z can be represented by the sum of nkth powers of nonconstant polynomials Therefore studying the form (11 is important By applying finite differences, Newman and Slater [11] showed that the identity function z can be always represented as a sum of kkth powers of nonconstant polynomials The following theorem about the (k 1th differences of x k can be found in the book by Hardy and Wright [6] Theorem 111 We have (114 k 1 ( k 1 ( 1 k 1 r (x + r k = k! x + d, r r=0 where d is an integer independent of x Infactd =(1/(k 1(k! By substituting (z d/k! for x in the equation (114, we can represent z as a sum of kkth powers of nonconstant linear polynomials Newman and Slater credit this finite difference argument to S Hurwitz who has conjectured that the number k of kth powers of nonconstant polynomials is the minimum needed for the representation of the identity function z Also, Heilbronn has conjectured that k is minimal even if entire functions are allowed [8] We can consider other classes for Waring s problem We denote the sets of polynomials, entire functions, rational functions, and meromorphic functions by P, E, R and M respectively as in [9] By a meromorphic function we mean a meromorphic function in the whole complex plane The finite difference argument implies that Waring s problems for P, E, R and M are solvable Theorem 11 [9] Suppose that k and that n Let f 1, f,,f n be nonconstant functions in class C satisfying (115 n f i (z k = z,

WARING S PROBLEM 1535 where C is one of the classes P, E, R and M Suppose that p C (k is the smallest number n satisfying (115 Then (116 (117 (118 (119 p P (k > 1 k + + 1 4, k 3, p E (k 1 k + + 1 4, k, p R (k > k +1, k, p M (k k +1, k The inequality (116 was found first by Newman and Slater [11], and (118 was found first by Green [4] Hayman used Cartan s theorem [1] to prove his theorem [9] 1 Fermat type of Waring s problem We consider the case that Q(z in the equation (111 is a nonzero constant Without loss of generality, Q 1 Then (11 f 1 (z k + f (z k + + f n (z k =1 Equations of the form (11 are called Fermat type equations The finite difference argument implies that any nonzero constant can be represented by a sum of (k +1 kth powers of nonconstant polynomials [11] Theorem 11 [9] Suppose that k and that n Let f 1, f,,f n be nonconstant functions in class C satisfying (1 n f i (z k =1, where C is one of the classes P, E, R and M Suppose that P C (k is

1536 D-I KIM the smallest number n satisfying (1 Then (13 (14 (15 (16 P P (k > 1 k + + 1 4, P E (k 1 k + + 1 4, P R (k > k +1, P M (k k +1 Newman and Slater noted that, together with the construction of Molluzzo, the inequality (13 gives the correct order of magnitude for P P (k [11] Theorem 1 [10] We have [ P P (k (4k +1 1/] Suppose that n andthatk For given n and a class C of functions, consider the following question: For which integers k, do there exist nonconstant functions f 1, f,,f n in the class C satisfying (11? We quote what Gundersen summarized in his survey paper [5] Consider the equation (17 f 1 k + f k + f 3 k =1 The following theorem is a collection of results of Toda [1], Fujimoto [3], Green [4], Newman and Slater [11] and Hayman[9] Theorem 13 For the four classes P, E, R, M, we have the following results for equation (17: (a If k 9, then there do not exist three nonconstant meromorphic functions f 1, f, f 3 that satisfy (17 (b If k 8, then there do not exist three nonconstant rational functions f 1, f, f 3 that satisfy (17

WARING S PROBLEM 1537 (c If k 7, then there do not exist three nonconstant entire functions f 1, f, f 3 that satisfy (17 (d If k 6, then there do not exist three nonconstant polynomials f 1, f, f 3 that satisfy (17 If the number n in the equation (11 increases, then the number of open questions increases [5] Waring s problem for linear polynomials 1 The representation of a function by linear polynomials Theorem 11 [11] Suppose that f 1 k +f k + +f n k = Q where Q is a polynomial, not identically zero, and f 1, f,,f n are nonconstant polynomials Then we have deg Q D(k n n(n 1 + n+ where D is the largest of the degrees of the f i s Further if all the f i s are linear we have (11 deg Q k n(n 1 If Q(z = z and D = 1, then, from the inequalities (11 and (116, we get (1 n 1+ 8k 7 > 1 k + + 1 4, k 3 For example, if k =5thenwehaven 4 We improve (1 as follows Theorem 1 Suppose that k and that n Let f 1, f,f n be nonconstant linear polynomials satisfying n (13 f i (z k = z

1538 D-I KIM Suppose that p is the smallest number n satisfying (13 Then p = k Proof The finite difference argument implies that the identity function z can be always represented by a sum of kkth powers of nonconstant linear polynomials Thus, p k Ifk =andp =1,thenweget f 1 = z, which is impossible Hence p andsop = Consider the case k 3 We suppose that p<kand will obtain a contradiction Write f i (z =a i + b i z for all i According to the minimality of p, all the f k i are linearly independent Thus we can have a i =0 for at most one i Thenf i (z k =(a i + b i z k =a k i (1 + (b i /a i z k if a i 0 Suppose that a k i = α i and that b i /a i = β i for each i Suppose that a p =0anda i 0for1 i p 1 Then p f i (z k = p (a i + b i z k p 1 = b k p z k + p 1 = b k p z k + = b p k z k + j=0 α i (1 + β i z k k ( α i k j j=0 β j i z j k ( ( p 1 k z j j α i β i j Since the righthand side is equal to z, we get, in particular, the system of equations (14 p 1 β i j α i =0 for j k 1 Since p<k,weusep 1 equations Now consider α i for 1 i p 1 as unknowns Then the coefficients form a square matrix M 1 whose

WARING S PROBLEM 1539 determinant is given by β 1 β β p 1 3 3 3 β 1 β β p 1 M 1 = p p p β 1 β β p 1 1 1 1 β 1 β β p 1 = β 1 β β p 1 β 1 β β p 1 p p p β 1 β β p 1 Since the last determinant for M 1 is the van der Monde determinant [], we get M 1 = β 1 β β p 1 i<j(β j β i Since all the f k i are linearly independent, we have β i β j for i j and we get M 1 0 Hence the system (14 of homogeneous linear equations has only the trivial solution and so α i =0foralliwith 1 i p 1 This is a contradiction Suppose that a i 0foralli Then p k ( ( p k f i (z k = z j j α i β i j j=0 Since the righthand side is equal to z, weget (15 p β j i α i =0 for j k By using p equations, we have a coefficient matrix M whose determinant is given by β 1 β β p 3 3 3 β 1 β β p M = p+1 p+1 p+1 β 1 β β p

1540 D-I KIM Since M 0, the system (15 has only the trivial solution and so α i =0foralli This is a contradiction The representation of a constant by linear polynomials If Q(z =1andD = 1 in Theorem 11, then from the inequalities (11 and (13 we get (1 n 1+ 8k +1 > 1 + k + 1 4, k For example, if k =3,thenwehaven 3 We improve (1 as follows Theorem 1 Suppose that k and that n Let f 1, f,,f n be nonconstant linear polynomials satisfying ( n f i (z k =1 Suppose that p is the smallest number n satisfying ( p = k +1 Then Example We define ω = e πi/(k+1 Then k+1 ν=1 ( 1+ω ν k z =1 (k +1 1/k Thus p k +1 The proof of Theorem 1 is similar to that of Theorem 1, and we omit it

WARING S PROBLEM 1541 3 Waring s problem for Laurent polynomials 31 Definitions and Cartan s theory Definition 311 A Laurent polynomial is a function of the form t (311 f(z = a n z n, n=s where s and t are integers with s t and a n C with a s 0 a t Hence a Laurent polynomial need not be a polynomial, in spite of the name There is some interest in the representation For example, if (31 f 1 k + f k + + f n k =1 holds for nonconstant Laurent polynomials f 1, f,,f n, we can replace z by e z to get (31 in terms of transcendental entire functions We use notations from the Nevanlinna theory that can be found in Hayman s book [7] We define log + x = max{0, log x} for x 0 We write n(t, f for the number of poles of f(z in z t counting multiplicities, and π m(r, f= 1 log + f(re iθ dθ, π 0 r n(t, f n(0,f N(r, f= dt + n(0,flogr, 0 t T (r, f=m(r, f+n(r, f Theorem 31 [1] Suppose that p Let F 1, F,,F p be p linearly independent entire functions Suppose that there is no point z such that F i (z =0for all i, where1 i p Set (313 F p+1 = p F i (z,

154 D-I KIM (314 T (r = 1 π π 0 sup log F i (re iθ dθ sup log F i (0 1 i p 1 i p Let n i (r be the number of zeros of F i (z in z r, whereazeroof order d is counted exactly min{d, p 1} times For 1 i p +1,set (315 N i (r = r 0 n i (t n i (0 t dt + n i (0 log r Then (316 T (r holds with p+1 N i (r+s(r (317 S(r = 1 π π 0 max log Δ (α1,α,,α p (re iθ dθ + O(1, where the maximum is taken for all choices of {α 1,α,,α p } of distinct numbers α 1,α,,α p from {1,,,p+1}, andwherethe Wronskian W (F α1,f α,,f αp =F α1 F α F αp Δ (α1,α,,α p with (318 Δ (α1,α,,α p (z 1 1 1 (F α 1 /(F α1 (F α /(F α (F α p /(F αp = (F α 1 /(F α1 (F α /(F α (F α p /(F αp (F α (p 1 1 /(F α1 (F α (p 1 /(F α (F α (p 1 p /(F αp

WARING S PROBLEM 1543 3 The representation of a function by Laurent polynomials Theorem 31 Suppose that k 3 and that n Let f 1,f,,f n be nonconstant Laurent polynomials satisfying (31 n f i (z k = z Suppose that at least one of the f i is not a polynomial and that p is the smallest number n satisfying (31 Then p p>kfor k 3, ie, p>1/+ k +1/4 for k 3 Proof We proceed with the proof similarly as in the proof of Theorem 11 Let p be the smallest number n satisfying (31 Consider (3 f 1 k + f k + + f p k = z For each i with 1 i p, wecanrewrite f i (z = t i n=s i a n z n = g i (z z si, where g i (z is a polynomial We can suppose that there exists an i such that s i 1 Otherwise, all f i (z are polynomials Let m =min 1 i p s i Thenm 1 From (3, we get (33 h 1 k + h k + + h p k = z 1 mk, where h i (z =f i (z z m Note that each h i (z isapolynomial We write h i k = F i,where1 i p, andf p+1 = z 1 mk Then (33 becomes (34 F 1 + F + + F p = F p+1 The functions F 1, F,,F p have no common zero Otherwise, there exists z 0 C such that 0 = F 1 (z 0 =F (z 0 = = F p (z 0 If z 0 0,

1544 D-I KIM then (34 does not hold This is a contradiction Assume that z 0 =0 There exists i such that s i = m Now, by choosing i so that s i = m, we get h i (z =f i (zz s i = ( a ti z t i + a ti 1z t i 1 + + a si z s i z s i = a ti z t i s i + a ti 1z t i s i 1 + + a si Since a si 0,wehaveF i (z 0 =h i (z 0 k = h i (0 k = a k si 0,whichis a contradiction The functions F 1, F,,F p form a linearly independent set Otherwise, one of the functions is a linear combination of the others Then z can be represented as a sum of p 1 kth powers This contradicts the minimality of p There exists i with t i 1 Otherwise, we get t i 0 for all i Then we obtain f k 1 + f k + + f k p = O(1 as z But the righthand side is z, which tends to by (3 This is a contradiction We write α =degh 1 =max 1 i p deg h i without loss of generality There exists h j with j 1suchthatdegh j = α Otherwise, deg h 1 > max 1<j p deg h j We write h 1 (z = a α z α + Then h 1 (z k = a k α z αk + Thena k α z αk cannot be canceled by h k +h k k 3 + +h p and z 1 mk since 1 mk αk If F i (z 0 = 0 with the order of zero d, then we count this zero min{d, p 1} times instead of the order of z 0 for n i (r Each zero of F i (z =(f i (z z m k has order k at least Thus d is a multiple of k If k p 1, then we do not need to check this case because z always can be represented by k polynomials Suppose that k>p 1 Since p 1 <k d, wehavemin{d, p 1} = p 1andthepointz 0 is counted p 1timesforN i (r Then we get N i (r p 1 = 1 p 1 ( r ( r 0 1 k 0 = 1 ( k N r, 1 F i n i (t n i (0 t dt + n i (0 log r n (t, (1/F i n (0, (1/F i t dt + n (0, 1Fi log r

WARING S PROBLEM 1545 since z 0 is counted once for (N i (r/(p 1 and d/k times for (1/kN (r, (1/F i So we have N i (r p 1 k N (r, 1Fi By Jensen s formula, we have N (r, 1Fi = T (r, 1Fi m (r, 1Fi = T (r, F i +O(1 m (r, 1Fi Since F i is entire, we get Hence N (r, 1Fi T (r, F i =N(r, F i +m(r, F i = 1 π = 1 π = 1 π 1 π π 0 π 0 π 0 π 0 log + F i (re iθ dθ ( log + F i (re iθ log + 1 F i (re iθ dθ + O(1 log F i (re iθ dθ + O(1 sup 1 i p = T (r+o(1 Therefore, for each i with 1 i p, wehave log F i (re iθ dθ sup log F i (0 + O(1 1 i p (35 N i (r p 1 T (r+o(1 k Consider r n p+1 (t n p+1 (0 N p+1 (r = dt + n p+1 (0 log r t 0 Since F p+1 (z =z 1 mk and 1 mk 1+k>p,wegetn p+1 (0 = p 1 and (36 N p+1 (r =(p 1 log r

1546 D-I KIM Now, recall π S(r = 1 max log Δ (α1,α π,,α p (re iθ dθ + O(1, 0 where the maximum is taken for all choices of {α 1,α,,α p } of distinct numbers α 1, α,, α p from {1,,,p+1} Consider Δ (α1,α,,α p (z 1 1 1 (F α 1 /(F α1 (F α /(F α (F α p /(F αp = (F α 1 /(F α1 (F α /(F α (F α p /(F αp (F α (p 1 1 /(F α1 (F α (p 1 /(F α (F α (p 1 p /(F αp First, consider the case {α 1,α,,α p } = {1,,,p} Wewrited i = deg F i,where1 i p Then we can rewrite F i = A di i j=1 (z z j, where A i is a constant Then F i /F i = d i j=1 (z z j 1 = O(z 1 For 1 l p 1, F (l i = F (l i F i F (l 1 i F (l 1 i F (l i F i F i = O(z l The function Δ (α1,α,,α p is a sum of products of the form ±F (j i /F i, where 1 i p and 1 j p 1 Therefore we get Δ (α1,α,,α p (z = O(z 1 O(z O(z (p 1 =O(z (p(p 1/ Now consider the other cases Choose p different numbers {α 1,α,, α p } from {1,,,p+1} such that α p = p + 1 Then, since F p+1 = z 1 mk,weget Δ (α1,α,,α p (z 1 1 1 1 (F α 1 /(F α1 (F α /(F α (F α p 1 /(F αp 1 O(z 1 = (F α 1 /(F α1 (F α /(F α (F α p 1 /(F αp 1 O(z (F α (p 1 1 /(F α1 (F α (p 1 /(F α (F α (p 1 p 1 /(F αp 1 O(z (p 1 = ( O(z 1 O(z O(z (p 1 =O z (p(p 1/

WARING S PROBLEM 1547 Thus, considering all choices of {α 1,α,, α p },weobtain (37 max Δ(α1,α,,α p = O (z p(p 1/ The equation (317 together with (37 gives (38 S(r p(p 1 log r + O(1 From Cartan s theorem, (35, (36, and (38, we get (39 T (r p(p 1 k T (r+(p 1 log r p(p 1 log r + O(1 Thus (310 ( 1 p(p 1 ( T (r (p 1 1 p log r + O(1 k We define d j =degh j,where1 j p Since there exists j with t j 1 and m 1, we have α =max 1 j p deg h j =max 1 j p (t j m Then we have h j (re iθ c j r deg h j = c j r d j as r,wherec j is a positive constant Hence log F j = k log h j k log c j r d j kd j log r + O(1 as r Thus sup 1 j p log F j = sup 1 j p k log h j k = kα log r + O(1 as r Hence we get (311 T (r kα log r + O(1 ( max d j log r + O(1 1 j p as r Now, the inequality (310 together with (311 gives ( (k p(p 1 α log r (p 1 1 p log r + O(1

1548 D-I KIM as r Ifp =,weobtaink p(p 1 = If p>, the righthand side is negative for large r, so that k < p(p 1 This completes the proof of Theorem 31 Since the class L of Laurent polynomials is not closed under composition, we cannot deduce the representability of all functions in L from that of z On the other hand if W C (k denotes the minimum number n, such that any element of the class C can be represented as the sum of nkth powers of functions in C, then W L (k W P (k k In fact if f(z =P (z/z nk is a Laurent polynomial, where P (z isan ordinary polynomial, then we have P (z = p (P ν (z k ν=1 where p = W P (k andthep ν are polynomials, Then f(z = p ( Pν (z ν=1 is the required representation of f(z in terms of Laurent polynomials Thus W L (k p = W P (k z n k 33 The representation of a constant by Laurent polynomials Theorem 331 Suppose that k 3 and that n Let f 1,f,,f n be nonconstant Laurent polynomials satisfying (331 n f i (z k =1 Suppose that at least one of the f i is not a polynomial and that p is the smallest number n satisfying (331 Then p p>kfor k 3, ie, p>1/+ k +1/4 for k 3

WARING S PROBLEM 1549 The proof of Theorem 331 is similar to that of Theorem 31, so we omit it 34 The proof of Hayman s theorem We conclude this paper by discussing the proof of Hayman s theorem [9, p],whichis Theorem 11 in this paper, for polynomials Suppose that k and that n Let f 1,f,,f n be nonconstant polynomials satisfying n f i (z k = z Suppose that p is the smallest number n satisfying the above equation Since F 1 = f k 1,F = f k,,f p = f k p and F p+1 = z, weget (341 N i (r p 1 T (r+o(1 k for 1 i p, and (34 N p+1 (r =logr Hayman considered only 1 1 1 (F 1 /F 1 (F /F (F p/f p Δ(z = (F 1 /F 1 (F /F (F p /F p (F (p 1 1 /F 1 (F (p 1 /F (F p (p 1 /F p and set (343 S(r = 1 π log Δ(re iθ dθ + O(1 π 0 Then ( Δ(z =O z (p(p 1/ Now, we estimate Δ(z more precisely We define d i = degf i, where 1 i p, and F i = a di z d i + a di 1z d i 1 + Then F i = a d i d i z d i 1 + a di 1(d i 1z d i + Hence, F i = a d i d i z di 1 + a di 1(d i 1z di + F i a di z d i + adi 1z d = d i i 1 + z + O(z

1550 D-I KIM Similarly, F i = a d i d i (d i 1z di + a di 1(d i 1(d i z di 3 + F i a di z d i + adi 1z d i 1 + = d i(d i 1 z + O(z 3 Similarly, we can write F (l i = d ( i(d i 1 (d i (l 1 1 F i z l + O z l+1, where 1 l p 1 Consider the determinant δ where each (i, j-entry is the leading term of the (i, j-entry in Δ Then δ(z = 1 1 d 1 /z d /z (d 1 (d 1 1/z (d (d 1/z (d 1 (d 1 1 (d 1 p +/z p 1 (d (d 1 (d p +/z p 1 1 d p /z (d p (d p 1/z (d p (d p 1 (d p p +/z p 1 By applying elementary row operations to δ we get δ = ( z (p(p 1/ δ, where 1 1 1 d 1 d d p δ = d 1 d d p p 1 p 1 p 1 d 1 d d p Since the last determinant δ is the van der Monde determinant [], we get δ = i<j(d j d i

WARING S PROBLEM 1551 Since there exist d i and d j such that d i = d j =max 1 n p d n,where i j, weobtainδ =0andweget ( (344 Δ(z =O z ( p(p 1/ 1 The equation (343 together with (344 gives ( p(p 1 (345 S(r 1 log r + O(1 If it were sufficient to consider only this Δ as Hayman [9, p4]claimed, it would follow from Cartan s theorem as stated by Hayman [9, p3], (341, (34, and (345 that ( p(p 1 p(p 1 (346 T (r T (r+logr + 1 log r + O(1 k Thus ( p(p 1 p(p 1 (347 1 T (r log r + O(1 k For the case k = and p =, we know that (z +(1/4 (z (1/4 = z Let F 1 (z =(z +(1/4, F (z = (z (1/4, and F 3 (z =z Then we obtain Δ(z = 1 1 F 1/F 1 F /F = (z (1/4 (z +(1/4 ( 1 1 = (z (1/4 (z +(1/4 = O z as z Thus, by the inequality (347, we get 0 log r + O(1 as r This is a contradiction Thus we need to consider the maximum of Δ(z for all choices of {α 1,α } from {1,, 3} to use Cartan s theorem Since Δ (1,3 (z = 1 1 F 1/F 1 F 3/F 3 = 1 (z +(1/4 + 1/4 z (z +(1/4 ( ( ( 1 1 1 = O + O z z = O z

155 D-I KIM as z and Δ (,3 (z = 1 1 F /F F 3/F 3 = 1 (z (1/4 + 1/4 z (z (1/4 ( ( ( 1 1 1 = O + O z z = O z as z,themaximumof Δ(z for all choices of {α 1,α } from {1,, 3} is O(1/z and not O(z The above consideration shows the sharpness of Cartan s theorem: it is really necessary to define w in the proof of Cartan s theorem [1, p 13] and S(r in (317 as the maximum involving several determinants, not only using one determinant Hayman [9, p 4] incorrectly quoted Cartan s theorem in terms of one determinant only Luckily, he estimated this one determinant in a cruder fashion than he could have, and his cruder upper bound is, in fact, an upper bound for the required maximum Therefore the final conclusions of Hayman remain valid Acknowledgments The author wishes to thank Professor Aimo Hinkkanen for encouragement, suggestions and discussions The author would also like to thank the referee for several valuable suggestions, including the example in connection with Theorem 1 REFERENCES 1 H Cartan, Sur les zéros des combinaisons linéaires de p fonctions holomorphes données, Math Cluj 7 (1933, 5 31 CW Curtis, Linear algebra, Springer-Verlag, New York, 1984, pp 161 16 3 H Fujimoto, On meromorphic maps into the complex projective space,jmath Soc Japan 6 (1974, 7 88 4 M Green, Some Picard theorems for holomorphic maps to algebraic varieties, Amer J Math 97 (1975, 43 75 5 G Gundersen, Complex functional equations, in Complex differential and functional equations, Proc of the Summer School (Mekrijärvi 000 (Ilpo Laine, ed, University of Joensuu, Dept of Math Report Ser No 5, 003, pp 1 50 6 GH Hardy and EM Wright, An introduction to the theory of numbers, Clarendon Press, Oxford, 1960, pp 35 36 7 WK Hayman, Meromorphic functions, Clarendon Press, Oxford, 1964 8, Research problems in function theory, Athlone Press, London, 1967, p 16

WARING S PROBLEM 1553 9, Waring s problem für analytische Funktionen, Bayer Akad Wiss Math Natur Kl Sitzungsber 1984 (1985, 1 13 10 J Molluzzo, Monotonicity of quadrature formulas and polynomial representation, Doctoral Thesis, Yeshiva University, 197 11 D Newman and M Slater, Waring s problem for the ring of polynomials, J Number Theory 11 (1979, 477 487 p 1 N Toda, On the functional equation i=0 a if n i i =1,Tôhoku Math J 3 (1971, 89 99 Department of Mathematics, University of Illinois, 1409 West Green Street, Urbana, IL 61801 E-mail address: dikim@hallymackr and dikim@mathuiucedu