Polynomial Calculus with D-Like Operators

  • Published on

  • View

  • Download

Embed Size (px)


  • Polynomial Calculus with D-Like OperatorsAuthor(s): J. W. Burgmeier and R. E. PratherSource: The American Mathematical Monthly, Vol. 82, No. 7 (Aug. - Sep., 1975), pp. 730-737Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2318730 .Accessed: 29/09/2013 09:23

    Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp


    JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact support@jstor.org.


    Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access toThe American Mathematical Monthly.


    This content downloaded from on Sun, 29 Sep 2013 09:23:45 AMAll use subject to JSTOR Terms and Conditions

  • 730 J. W. BURGMEIER AND R. E. PRATHER [Aug.-Sept.

    16. Marston Morse, John von Neumann and Luther P. Eisenhart, Report to the President of the National Academy of Sciences, 28 May 1948.

    17. Laurence F. Schmeckebier, Congressional Apportionment, The Brookings Institution, Washington, D. C. 1941.

    18. George Washington, The Writings of George Washington, Vol. 32 (March 10, 1792-June 30, 1793), John C. Fitzpatrick, editor, United States Government Printing Office, Washington, D. C. 1939, pp. 16-17.

    19. Daniel Webster, The Writings and Speeches of Daniel Webster, vol. VI, National Edition, Little, Brown & Company, Boston Mass., 1903, pp. 102-123.

    20. Walter F. Willcox, The Apportionment of Representatives, The American Economic Review, vol. VI, No. 1, Supplement (March 1916) 3-16.

    21. - , Last Words on the Apportionment Problem, Legislative Reapportionment, vol. 17, No. 2 (1952) issue of Law and Contemporary Problems, pp. 290-302.

    22. Comment, Apportionment of the House of Representatives, Yale Law Journal, 58 (1949) 1360-1386.




    1. Introduction. The newcomer to numerical analysis is usually impressed on finding an unexpected formal analogy [1] between the Taylor and Newton series expansions:

    f(x) - E (Dkf)(a)(x a) k =

    f(X) A (Akf)(a) (X - a)(k

    We say unexpected because the difference operator

    (f)(x) =f(x + l)-f(x) and its associated factorial polynomials

    x"' x(x -l) (x-j+l j! =j!

    seem somewhat removed from their counterparts in the differential calculus. True, the formal identity

    D k A = e D _ I = D + D +, . D +... 2! k

    and the Stirling numbers s (i,j) defined by x'j) = 2=5s (i,j)x' do help to provide a connection. And yet, the question surely arises as to whether these two "expansion systems" are merely isolated curiosities, or instead, singular but typical examples from a family of such systems. Naturally, we wish to infer that it is the latter.

    In order to establish an appropriate setting for the investigation, we first seek to extract the common features of the linear operators D and A (and their associated polynomials xj/j!, xWp/j!, resp.) on the space P' of all polynomials. In both cases, the associated polynomials form a simple basis (there being just one polynomial for each degree) and the operators are linear and strictly unit-degree-decreasing (abbreviated u.d.d. with linearity understood) over P'. And so, we begin with a brief analysis of these u.d.d. or "derivative-like" operators and their expansion capability relative to various simple bases. We are then led to impose a succession of familiar differential properties, leading ultimately to a characterization of the derivative among all u.d.d. operators.

    This content downloaded from on Sun, 29 Sep 2013 09:23:45 AMAll use subject to JSTOR Terms and Conditions


    Though some of our results might have been anticipated, we feel that the investigation does provide new insights as to the essential uniqueness of the classical differential calculus. Assuming this is so, we have tried to make our presentation self-contained and reasonably inclusive.

    2. Expansion systems. In any linear function space, a sequence (Li) of linear functionals and a companion sequence (gi) of functions are said to be bi-orthonormal [2] if

    Lig1 = Sij.

    Perhaps the most common situations of this kind occur in the inner-product spaces where the functionals Li take the form Lif = (f, g). Then the bi-orthonormality accounts for the familiar orthogonal expansions. But the same is seen to be true for the generalized Taylor-Newton expansions:

    THEOREM 1. Let L be a u.d.d. operator and (g1) a simple basis in P'. Then every f E P' has a representation

    f(x) = >j(Lkf)(a)gk(x) k =O

    iff the functionals Lif = (Lif)(a) and the polynomials gj(x) are bi-orthonormal. Proof. Assuming these representations exist, we would have in particular


    g1(x) = >(LLgj)(a)gi(x). i =O

    And because (g1) is a basis it would follow that

    Ligj = (L1g1)(a) = 5ii. Conversely, if we assume the bi-orthonormality and write


    f(X) = E akgk (X) k =O

    for any f E P", then applying the functional Lj yields 00

    (L f)(a) = E ak(Ljgk)(a) = aj k =O

    so that f has the representation claimed. O As a consequence, L and (gj) are said to constitute an expansion system (at a) provided that

    (L gj)(a) = Sij. An expansion system (at 0) will be called a Maclaurin system. A more explicit characterization of these expansion systems is useful in the sequel, namely

    THEOREM 2. L and (gj) constitute an expansion system (at a) iff (i) go= 1 (ii) gj (a)=O (j > 0). (iii) Lgj = gj-, (j> 0).

    Proof. First suppose that all these conditions are met. Then

    [gj_i(a) = < (iP J Conversely, if L and (gj) form an expansion system (at a) we have (i) go = go(a) = (Logo)(a) = Soo 1, (ii) gj (a) = (Logj)(a) = 50=O (J > O),

    This content downloaded from on Sun, 29 Sep 2013 09:23:45 AMAll use subject to JSTOR Terms and Conditions

  • 732 J. W. BURGMEIER AND R. E. PRATHER [Aug.-Sept.

    and finally, considering Theorem 1 (*ii 00 (i1i) (Lgj)(x) = >_(L k (Lgj ))(a )gk (x)



    = ,(L k+gj)(a)gk(x) = gj,(x). O k=O

    COROLLARY. Let (g1) be a simple basis satisfying (i), (ii). Then there is a unique u.d.d. operator L such that L and (gj) constitute an expansion system (at a). Conversely, to each u.d.d. operator L and each real number a, there corresponds a unique simple basis (gj) for which L and (gj) is an expansion system (at a).

    In particular, each u.d.d. operator L may be identified with that (unique) simple basis for which the pair constitute a Maclaurin system. And in considering the resulting infinite matrix representa- tion, it should be clear that any such L will have the same "Jordan canonical form" as the derivative. We obtain somewhat more information in this same vein if we define the Stirling transformations S: P" -> P" to be those linear operators with the following properties:

    (I) S(1) = 1, (II) Po and each pn are S-invariant subspaces, (III) S is invertible. Here, pn is the space of all polynomials of degree at most n, and PO represents the subspace of

    polynomials which vanish at the origin. Then, just as the factorial polynomials may be viewed as a change of basis via the matrix s(i,j) of Stirling numbers, we have the following generalization:

    THEOREM 3. Let M and (hj) be a Maclaurin system and S a Stirling transformation. If

    g = Sh L = SMS1

    then L and (gj) comprise another Maclaurin system. Conversely, if the two Maclaurin systems are given, there exists a Stirling transformation S such that

    g = Sh L = SMS1.

    Proof. First we observe that L is u.d.d. and gn = Shn has degree -' n by (II). But if gn had degree < n we would contradict (III). So (g,) is again a simple basis, and moreover,

    (i) go = Sho = S(1) = 1 using (I), (ii) g1(O)=Sh,(O)=O by (II) (>O). (iii) Lg1 = SMS-lg1 = SMhj = Sh1-l = gj-l,

    so that according to Theorem 2, L and (gj) constitute a new Maclaurin system. Conversely, if L, (g1) and M, (hj) satisfy the conditions of Theorem 2, we have only to define S

    by the change of basis g = Sh. Then the invertibility of S and the invariance of each Pn is assured, and furthermore

    (I) S(1) =Sho = go = 1, (II) f(O) = 0

    => (Sf)(0) = Ej(Mkf)(O)(Shk)(O) k =O

    = (M?f)(O)go(O) = f(0) = 0

    so that S is a Stirling transformation. Finally, for any f E P" we use its Maclaurin representation relative to M, (hj) to obtain

    (S-1LS)(f) = Y,(Mkf)(O)(S-lLS)hk = ,(Mkf)(O)hk-l = M(f). C1 k k

    This content downloaded from on Sun, 29 Sep 2013 09:23:45 AMAll use subject to JSTOR Terms and Conditions


    COROLLARY. The only Maclaurin systems L, (gn) are

    gn -S(xn/n !) L = SDS-' for Stirling transformations S.

    3. Taylor operators. We have seen that each u.d.d. operator L may be identified with a unique simple basis (g1) so that the pair comprise a Maclaurin system. As we move to another point of expansion, say a 7 0, there will again exist a unique simple basis resulting in an expansion system (at a). But these new polynomials, in general, will bear little relationship to the sequence (g,). In order to achieve the desired universality for the expansion basis, we find that the commutativity property of the following definition is needed.

    The u.d.d. operator L is called a Taylor operator if

    LTa = TaL

    for every real number a. Here Ta is the translation

    (Taf)(x) = f(x - a).

    Considering the universality requirement as exemplified by the Taylor-Newton expansion, our above definition is justified by the following result.

    THEOREM 4. Let L be a u.d.d. operator and (gj) the unique simple basis for which L, (gj) constitute a Maclaurin system. Then L, (gf(x - a)) is an expansion system (at each a) if L is a Taylor operator.

    Proof. First suppose that LTa = TaL. Then considering the simple basis (gj(x - a)) one has (i) go(x - a ) = go(x) -1 (ii) g,(x - a)(a) = gj(0) =0 ( > 0), (iii) Lg1(x -a) = LTag(x) = TaLgj(x) = Tagj-i(x) =gi-(x -a)

    so that L, (gj (x - a)) is an expansion system (at a), by virtue of Theorem 2. On the other hand, if the usual properties (i), (ii), (iii) hold relative to L and (gj(x - a)), then for

    any f E P' we may consider its Maclaurin expansion

    f(x) = >j(Lkf)(O)gk(x) k =O

    and use (iii) to obtain 00

    (L Taf)(x) = > (Lkf)(0)(LTagk)(x) k =O


    = J(Lkf)(0)gk-I(x - a) k =O

    = Ta>(Lkf)(0)gk 1(x) = (TaLf)(X). E k =O

    Note that this result establishes a relationship for Taylor operators which is analogous to the connection between Maclaurin and Taylor series for D. Consequently, it seems that we have come a step closer to the derivative by our introduction of the Taylor operators (but we still admit the difference operator A). As for seeing just how close, it remains for us to provide an explicit characterization of the Taylor operators. This could perhaps be accomplished by several means. One could try to examine the nature of their associated Maclaurin polynomials (gj). But instead, we have found a direct characterization of the operators themselves to be more illuminating. For this purpose, we need the following lemma.

    This content downloaded from on Sun, 29 Sep 2013 09:23:45 AMAll use subject to JSTOR Terms and Conditions

  • 734 J. W. BURGMEIER AND R. E. PRATHER [Aug.-Sept.

    LEMMA. Let L be a u.d.d. operator having the matrix (Ais) relative to the basis (xj/j!). Then L is a Taylor operator iff

    An,n+k = AO,k (k - 0).

    Proof. Let all matrices be expressed relative to the basis (x'/j !). Then, because L is u.d.d., the matrix (Ai;) is upper-triangular. The same is true for the matrix (ij) = (in(a)) of the translation Ta. In fact,

    [0 (i =k=ik (k -iT

    Equating like powers of a, we obtain Ai,j_r = A i+r,j, (i < j). And this means that the entries along all super-diagonals are constant, as claimed. Since our arguments are obviously reversible, the result follows. LI

    THEOREM 5. L is a Taylor operator iff it has the form 00

    L = E>AkDk k = I

    for a sequence (Ak) with A1 7 0.

    Proof. First we observe that the infinite sum when applied to any f E P' is actually finite, so that questions of convergence do not arise. Now assume that L is any Taylor operator and use the ordinary Taylor expansion

    f(x) = _(Dif)(0) xj

    for f E P'. Applying the operator L and using the notation of the lemma, we have 00j

    (Lf)(x) = (Djf)(O)L . 1=0 J1

    = (Dif)(0)Aij . i=1 1=0 l.

    This content downloaded from on Sun, 29 Sep 2013 09:23:45 AMAll use subject to JSTOR Terms and Conditions


    = .(Djf)(O)ZAoj co co X -k

    = O l?kJ(D f)(0) X

    = Aok(D f)(x) = (YAOkD f)(x). D1 k =I k=1

    Of course, if we take A, = 1 and all other Ak = 0 we obtain the derivative, whereas the choice Ak = 1/k! gives the difference operator. But more importantly, one sees in general the intimate connection of these Taylor operators with the derivative.

    4. Rolle operators. Surely one of the most important results in the differential calculus is the mean value theorem, or equivalently, Rolle's theorem. We now propose to study the more general u.d.d. operators having this same property. Our original interest in this property was motivated by considerations which are explained in the following concluding section.

    We will say that the u.d.d. operator L is a Rolle operator if we have

    f(a) = f(b) = 0 => (Lf)(6) = 0 (some 6 E (a, b))

    for every f E P'. Of course D is a Rolle operator. And one would hope to discover other such operators and to characterize them as was done for the Taylor operators in the previous section. But the following theorem shows that D is (essentially) the only Rolle operator!

    THEOREM 6. Every Rolle operator is a non-zero multiple of D.

    Proof. Let L be a Rolle operator. First we show that the unique sequence (g1) making L, (g1) a Maclaurin system must have the form g1 = a1x'. Certainly this is true for j = 0, 1. But suppose, on the contrary, that not all gi are of this form, and let n be the least integer for which

    gn(X) = X kp (X)

    with p (O) Z 0 and i ? k < n. Then we set f(x) = gn(X) -xkp(a)

    with a as yet unspecified. We have f(0) = f(a) = 0, whereas

    X) = Xk _,n-k _ak-I (Lf)(x ) `=x (an-i X -_ p (a)).

    Since p (0) , 0, we- can now choose a k 0 so that both factors in (Lf)(x) are of one sign for all x between 0 and a, thus contradicting that L is a Rolle operator. We must therefore conclude that gn(x) = anXn for all n.

    We know that ao = 1. And it is clear that L, (g1) is a Maclaurin system with L Rolle if and only if the same is true for cL, (gj/cj). Thus we may continue as if a, = 1 also. Introducing the quotients Pk = ak-Ilak we have

    LXk =fPkX.

    This makes I3S = 1 and we intend to show that 3m = m for all m ' 1....


View more >