Iterative Processes Related to Riordan Arrays: The Reciprocation and the Inversion of Power Series
aa r X i v : . [ m a t h . C O ] M a y ITERATIVE PROCESSES RELATED TO RIORDAN ARRAYS: THERECIPROCATION AND THE INVERSION OF POWER SERIES
ANA LUZ ´ON
Abstract.
We point out how Banach Fixed Point Theorem, and the Picard successiveapproximation methods induced by it, allows us to treat some mathematical methods inCombinatorics. In particular we get, by this way, a proof and an iterative algorithm forthe Lagrange Inversion Formula. Introduction: beginning with a simple question.
The results in this paper are consequences of special interpretations, as fixed point prob-lems, of the two classical reversion processes in the realm of formal power series: the reciprocation , i. e. the reversion for the Cauchy product, and the inversion , i. e. thereversion for the composition of series.The case of the reciprocation was studied in [4] and [5]. To unify our approach we alsosurvey herein some of the previous results on this topic.The aim of this paper is to show how the (Picard) successive approximation method in-duced by Banach’s Fixed Point Theorem (and some mild generalization) allows us to treatsome mathematical methods in combinatorics getting so some associated algorithms. Inparticular:1) To construct all elements in the Riordan group as a consequence of the iterativeprocess obtained to calculate the reciprocal of any power series admitting it. Presentingalso an algorithm and one pseudo-code description for it.2) To construct, approximatively, the inverse of any power series (admitting it) in sucha way that the Lagrange Inversion Formula can be first predicted and finally proved. Wealso describe the corresponding algoritm.For completeness we are going to recall the metric fixed point theorems we will use, see,for example, [1] for the first one and [10] in page 212 for the second one.
Banach Fixed Point Theorem (BFPT).
Let (
X, d ) be a complete metric space and f : X → X contractive. Then f has a unique fixed point x and f n ( x ) → x for every x ∈ X . Key words and phrases.
Banach Fixed Point Theorem; ultrametric; Riordan group; Lagrange InversionFormula. eneralized Banach Fixed Point Theorem (GBFPT). Let (
X, d ) be a completemetric space. Suppose { f n } n ∈ N : X −→ X is a sequence of contractive maps with thesame contraction constant α and suppose that { f n } −→ f (point to point). Then f is α -contractive and for any point z ∈ X the sequence { f n ◦ · · · ◦ f ( z ) } −−−→ n →∞ x , where x isthe unique fixed point of f .Our framework is the following: we consider K a field of characteristic zero and the ringof power series K [[ x ]] with coefficients in K . If g is any series given by g = P n ≥ g n x n ,we recall that the order of g , ω ( g ), is the smallest nonnegative integer number n such that g n = 0 if any exist. Otherwise, that is if g = 0, we say that its order is ∞ . It is well-knownthat the space ( K [[ x ]] , d ) is a complete ultrametric space where the distance between f and g is given by d ( f, g ) = ω ( f − g ) , f, g ∈ K [[ x ]]. Here we understand that ∞ = 0. Moreoverthe distance between f and g is less than or equal to n +1 , i. e. d ( f, g ) ≤ n +1 , if and only iftheir n-degree Taylor polynomials are equals, T n ( f ) = T n ( g ). Finally the sum and productof series are continuous if we consider the corresponding product topology in K [[ x ]] × K [[ x ]].See for example [7], [4] and [5] for these topics.In this paper N represents the set of natural numbers including 0.This work is motivated by the following question: Question 1.
Can we sum the arithmetic-geometric series P ∞ k =1 kx k − using the BanachFixed Point Theorem?.We can sum the geometric series using BFPT . A visual proof of this fact can be found in[13]. Herein we recall an analytic proof: (the peculiar name of the following function willbe justified later on). We consider h m, : ( K [[ x ]] , d ) → ( K [[ x ]] , d ) t xt + 1Since h m, is contractive, in fact d ( h m, ( t ) , h m, ( t )) ≤ d ( t , t ), we iterate at t = 0 andwe obtain: h m, (0) = 1 h m, (0) = x + 1 h m, (0) = x + x + 1 h m, (0) = x + x + x + 1that is, h n +1 m, (0) = n X k =0 x k s the fixed point of h m, is the solution of xt + 1 = t , then t = − x and from BFPT weinduce h n +1 m, (0) = n X k =0 x k −−−−→ n → ∞ − x which is the unique fixed point of the function, in this case, h m, ( t ) ≡ h ( t ) = xt + 1.Now it is natural to wonder Question 1.We organize the paper in the following way:In Section 2 we use BFPT to a suitable function related to Question 1. We do not answerthe question by this way but we find an interesting arithmetical triangle. Later and using
GBFPT we answer the question. Actually we construct the whole Pascal triangle by thismethod, see [4] and [5].In Section 3 we generalize the method above, to construct the Pascal triangle, finding soa way to construct arithmetical triangles T ( f | g ) for any pair of series f and g with nonnull independent terms. Using the usual product of matrices, we identify the well-knownRiordan group, see [4].In the procedure described above we obtain a new parametrization of the elements inthe Riordan group and so a new notation different from the usual ones. In Section 4 wetry to justify the use of our notation alternatively to the usual notation. Our method ofconstruction and our notation allow us to explain easily a way to add and delete columnssuitably in a Riordan array to get another one. For a concrete kind of triangles, thosedenoted by T (1 | a + bx ), we can calculate the inverse only adding adequately new columnsto those triangles. In fact it is an elementary operations method. We end this section givingexpressions for the so called A and Z sequences of a Riordan array. These expressions aregiven in terms of our notation and related to the inversion in the Riordan group.In Section 5 we give the main new results of this paper. We display an algorithm toconstruct the inverse of a series and we show the relation with the Lagrange InversionFormula. In particular we prove that Banach Fixed Point theorem gives rise to the LagrangeInversion Formula.2. Two answers: a curious triangle and an iterative method
To answer the previous question, in this section we are going to recall lightly some examplesand tools widely studied in [4] and [5].It can be easily shown, see page 2257 in [5], that there are not a first-degree polyno-mial with coefficients in K [[ x ]], f ( t ) = g ( x ) t + h ( x ), and any point, x , such that its n + 1-iteration coincides with the partial sum of the arithmetic-geometric series, that is: ∄ f, x / ∀ n, f n +1 ( x ) = P nk =0 ( k + 1) x k . n view of this, we are going to iterate a polynomial whose fixed point is the sum of thearithmetic-geometric series, that is P ∞ k =1 kx k − = − x ) . Since the equality t = − x ) canbe converted to t = 1 + (2 x − x ) t , we consider the polynomial f ( t ) = 1 + (2 x − x ) t , withcoefficients in K [[ x ]], and we initiate the iteration process at t = 0: f (0) = 1 f (0) = 1 + 2 x − x f (0) = 1 + 2 x + 3 x − + x f (0) = 1 + 2 x + 3 x + 4 x − + − x f (0) = 1 + 2 x + 3 x + 4 x + 5 x − + − + x f (0) = 1 + 2 x + 3 x + 4 x + 5 x + 6 x − + − + − x We know that the sequence of iterations converges to the sum of the arithmetic-geometricseries, because f is contractive in ( K [[ x ]] , d ), but we can observe that in each iteration thepartial sum of such series appears plus a remainder. We want to control the difference withthe partial sum. To do this, we display the coefficients of the remainder as a matrix, thatis: − − −
11 6 − −
26 23 − −
57 72 −
39 10 − −
120 201 −
150 59 −
12 1... ... ... ... ... ... . . .
Observing this matrix we recognize:(1) The rule of construction is similar to that of Pascal triangle: each element is twice theabove element minus the element above to the left side, that is: a n,k = 2 a n − ,k − a n − ,k − .(2) The elements in the first column are Eulerian numbers except for the sign.(3) The sum of the elements in any row are triangular numbers with negative sign.(4) For every element, the sum of all elements in its row to the right and all elements abovein its column is zero. That is, P i − k =1 a kj + P nk = j +1 a ik = 0.(5) The general term is a n,j = n + j − P j − k =1 ( − k (cid:0) n + j − − kn + j − k (cid:1) n + j − k . Etc.For an exhaustive development of this triangle see [5].The above approach, using BFPT , does not give us an exact answer to our question. Tofind an adequate answer we consider the
GBFPT . This is our way to do this:For computability facts we consider the sequence of functions, with polynomial coefficients,given by h , ( t ) = xt , h , ( t ) = xt + x , h , ( t ) = xt + x + x , h , ( t ) = xt + x + x + x , h m, ( t ) = xt + x m − X k =0 x k ach function h m, is -contractive, so { h m, } is an equi-contractive sequence of one-degreepolynomials that converges to h ( t ) = xt + x − x , i. e. : { h m, } −→ h ( t ) = xt + x − x It is easy to see that the crossed iterations induced by
GBFPT are just the correspondingpartial sums of the arithmetic-geometric series: h , (0) = 0 h , ( h , (0)) = xh , ( h , ( h , (0))) = x + 2 x h , ( h , ( h , ( h , (0)))) = x + 2 x + 3 x Now using again
GBFPT we obtain that, since xt + x − x = t ⇒ t = x (1 − x ) , then( h m, ◦ · · · ◦ h , )(0) −→ x (1 − x ) these crossed iterations at zero converge to the unique fixed point of h , that is, the sumof the arithmetic-geometric series. So the answer to our question is yes if we are allowedto use the generalized version of the BFPT .Recall that the Pascal triangle is given by:
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1 ... ... ... ... ... ... ... ...( n ) ( n ) ( n ) ( n ) ( n ) ( n ) ( n ) ··· ( nn )... ... ... ... ... ... ... ... ... − x x (1 − x )2 x − x )3 x − x )4 x − x )5 x − x )6 x − x )7 ··· xn − − x ) n We have just constructed the first two columns of Pascal tringle using
BFPT . In fact weneeded only
BFPT to construct the first one and
GBFPT to get the second one. Themain observation is that we can follow this iterative procedure to construct all columns.For example we can repeat the process to construct the third column. To get this goal, weinterpret the above equicontractive sequence h m, in the following way h m, ( t ) = xt + x m − X k =0 x k = xt + xT m − , where T m − , is the m − h m, ( t ) = xt + x m − X k =0 kx k = xt + xT m − , here T m − , is the Taylor polynomial of the second column (which is the arithmetic-geometric series). So, as one can easily prove, the crossed iterations for this sequencecoincide with the partial sums of the third column: h , (0) = 0 h , ( h , (0)) = 0, h , ( h , ( h , (0))) = x , h , ( h , ( h , ( h , (0)))) = x + 3 x , h , ( h , ( h , ( h , ( h , (0))))) = x + 3 x + 6 x ,Using once more GBFPT , we obtain that these crossed iterations converge to the uniquefixed point of the limit function h ( t ) = xt + x x (1 − x ) . Since h ( t ) = xt + x x (1 − x ) ⇒ t = x (1 − x ) then ( h m, · · · h , )(0) = m X k =0 (cid:18) k (cid:19) x k −→ x (1 − x ) Actually, as we said before, we can construct every column of Pascal triangle using thisprocess:
Proposition 2.
For n ≥ , the n-column in Pascal’s triangle is obtained from the (n-1)-column applying the crossed iterations in GBFPT to the sequence { h m,n } m ∈ N where h m,n ( t ) = xt + xT m − ,n − being T m − ,n − the (m-1)-Taylor polynomial of the (n-1)-column. THE GROUP OF ALL ARITHMETICAL TRIANGLES T ( f | g ) . Now we generalize the previous iterative method for any pair of series f = P n ≥ f n x n and g = P n ≥ g n x n such that f = 0 and g = 0. We construct the following arithmeticaltriangle T ( f | g ), in this notation the Pascal triangle is T (1 | − x ), where the role of series1 is played by the series f and the role of 1 − x by g . f f d , f d , d , f d , d , d , ... ... ... ... . . . f fg xfg x fg · · · n [4] we interpreted the calculation of fg as a fixed point problem. Consider the sequence h m, ( t ) = T m (cid:18) g − gg (cid:19) t + T m (cid:18) fg (cid:19) where T m ( f ) is the m degree Taylor polynomial of f . Observe that the sequence of crossediterations has as its limit the unique fixed point of h ( t ) = g − gg t + fg , that is: fg . It isthe first column of T ( f | g ).To construct the second column and the next ones we consider the equicontractive sequences h m,n ( t ) = T m (cid:18) g − gg (cid:19) t + xT m − (cid:18) x n − fg g n − (cid:19) their corresponding limits are h n ( t ) = (cid:16) g − gg (cid:17) t + x (cid:16) x n − fg g n − (cid:17) whose corresponding uniquefixed points are t n = x n − fg n The series t n is just the n-column of our T ( f | g ). Theorem 3.
Let f = X n ≥ f n x n and g = X n ≥ g n x n with g = 0 , then the Riordan matrix T ( f | g ) = ( d n,k ) is given byif k = 0 , then d , = f g , d n, = − g g d n − , − g g d n − , · · · − g n g d , + f n g if k > , then d n,k = − g g d n − ,k − g g d n − ,k · · · − g n − k g d k,k + d n − ,k − g This theorem gives us the following algorithm to construct for columns any Riordan matrix:
Algorithm 4.
Given f = X n ≥ f n x n and g = X n ≥ g n x n with g = 0: Step 1:
Calculate the first column d n, . d , = f g , d n, = − g g d n − , − g g d n − , · · · − g n g d , + f n g Step k:
Calculate the k -column, d n,k , using k − d n,k = − g g d n − ,k − g g d n − ,k · · · − g n − k g d k,k + d n − ,k − g e can write this algorithm in an informal pseudo-code: READ (f,g,n)SET (d,aux)CALCULATE d[0,0]=f[0]/g[0] % We calculate the first columnFOR i=1 to nFOR k=1 to nCALCULATE aux(k,i)=g[i-k]*d[k,0]ENDCALCULATE d(i,0)=1/g[0]*(f[i]-SUM(aux(:,i)))END % We calculate the remaining columnsFOR j=1 to nFOR i=1 to nFOR k=1 to iCALCULATE aux(k,i)=g[i-k]*d[k,j]ENDCALCULATE d(i,j)=1/g[0]*(d(i-1,j-1)-SUM(aux(:,i))ENDENDPRINT(f,g,d)
So to construct the arithmetical triangle T ( f | g ) it is enough to know the ordered pair ofseries f and g , i. e. the data, and the algorithm of dividing two series. Every column isconstructed by the same rule as that in fg but the coefficients of fg are replaced with thecoefficients of the previous column. Except for the first column, here we need and auxiliarycolumn, the coefficients of f .We can consider the matrix T ( f | g ), like in Linear Algebra, as the associated matrix to a K -linear continuous function, see [4]: T ( f | g ) : ( K [[ x ]] , d ) → ( K [[ x ]] , d ) h T ( f | g )( h ) = fg h (cid:16) xg (cid:17) Using the classical definition of composition of maps and the behavior of the associatedmatrix, we can easily find the formulas for the product and the inverse for these triangles. T ( f | g ) T ( f | g ) = T (cid:18) f f (cid:18) xg (cid:19) (cid:12)(cid:12)(cid:12) g g (cid:18) xg (cid:19)(cid:19) ( T ( f | g )) − = T (cid:18) f ( ω − ) (cid:12)(cid:12)(cid:12) g ( ω − ) (cid:19) , ω = (cid:18) xg (cid:19) , ω ◦ ω − = ω − ◦ ω = x o if we consider the set of the all arithmetical triangles with f = 0 and g = 0 andthe usual product of matrices we obtain a group. Actually this group is the well-knownRiordan group. 4. On the T ( f | g ) notation. We have received some critics about our notation. Someone could think that our notationis, in some sense, cumbersome. Of course it depends strongly on the way you approach oryou run into this group. In this section we are going to give some reasons why our notationcould be very adequate. The basic formula relating our to the classical notation is( d ( x ) , h ( x )) = T (cid:18) xdh (cid:12)(cid:12)(cid:12) xh (cid:19) = (cid:18) f ( x ) g ( x ) , xg (cid:19) = T ( f | g ) = ( d i,j ) i,j ≥ The fundamental equality with our notation is:(1) T ( f | g ) = T ( f | T (1 | g )This equality in the other notation is: (cid:18) f ( x ) g ( x ) , xg ( x ) (cid:19) = ( f ( x ) , x ) (cid:18) g ( x ) , xg ( x ) (cid:19) In (1) we can see that every element of the Riordan group [9] can be expressed by meansof the product of a lower triangular Toepliz matrix whose columns are the coefficients ofseries f , shifted conveniently, the matrix T ( f | T (1 | g )described by Rogers in [8]. These last kind of matrices are really similar to the Jabotinskymatrices, see [3]. We want to point out that the structure of every element of the Riordangroup is essentially in the structure of the matrix T (1 | g ). For example, to know a closedformula for the general term of T (1 | g ) gives us at once a closed formula for the generalterm in T ( f | g ). This matrix T (1 | g ), for us, is intrinsically related to the calculation of1 g , which is its first column.A comparative table of both notations is given: Name (d(t),th(t)) T(f | g) Identity (1 , t ) T (1 | (cid:18) − t , t − t (cid:19) T (1 | − t )Appel subgroup element (d(t),t) T ( d | T (cid:18) h | h (cid:19) Bell subgroup element (d(t),td(t)) T (cid:18) | d (cid:19) A curious and symbolically important, for us, property of our notation is the way to give,by means of the parameters, the natural powers of the Pascal triangle:
Proposition 5.
For every n ∈ N , we have T n (1 | − x ) = T (1 | − nx ) roof. Let us proceed by induction.For n = 2: T (1 | − x ) = T (1 | − x ) T (1 | − x ), ω = x − x . So T (1 | − x ) = T (1 | (1 − x )(1 − x − x ) = T (1 | − x )Suppose that T n − (1 | − x ) = T (1 | − ( n − x ), then T n (1 | − x ) = T (1 | − x ) T n − (1 | − x ) = T (1 | − x ) T (1 | − ( n − x ) == T (1 | (1 − x )(1 − ( n − x − x )) = T (1 | − nx ) (cid:3) Another reason why for us our notation is natural, is related to the way we begun to studythese topics. One of the first things we did was to find our curious triangle described inSection 2. From our notation, the description as a Riordan array is: − − −
11 6 − −
26 23 − −
57 72 −
39 10 − −
120 201 −
150 59 −
12 1... ... ... ... ... ... . . . = T (cid:18) − x ) | x − (cid:19) This notation resembles both the problem we were treating and the algorithm of construc-tion.Another thing we can describe easily with our notation is the fact that, with our construc-tion method by columns, we can add new columns to the left for every element of theRiordan group to obtain again a Riordan array intrinsically related to the initial one, forexample: − − −
11 6 − −
26 23 − −
57 72 −
39 10 − −
120 201 −
150 59 −
12 1... ... ... ... ... ... ... . . . = T (cid:18) x − − x ) | x − (cid:19) Note that we added a new column to the left and look at the way the parameters changedin our notation. n general we can construct a family of new Riordan matrices closely related to it. Forexample by definition of Riordan array we get T ( f g | g ) = f f d , f d , d , f d , d , d , f d , d , d , d , ... ... ... ... ... . . . T (cid:18) fg | g (cid:19) = d , d , d , d , d , d , d , d , d , d , d , d , d , d , d , ... ... ... ... ... . . . where f = X n ≥ f n x n and T ( f | g ) = ( d n,k ) n,k ∈ N .By the same way we observe that T (cid:16) fg m | g (cid:17) for m ∈ N is the matrix obtained from T ( f | g )by deleting the first m -rows and m -columns. Moreover T ( f g m | g ) is the unique Riordanmatrix with the property that by deleting the first m -rows and m -columns from it, weobtain T ( f | g ). In fact it can be easily proved the following: Proposition 6.
Let T ( f | g ) = ( d n,k ) n,k ∈ N be a Riordan matrix, and m ∈ Z then T ( f g m | g ) = ( ˜ d n,k ) n,k ∈ N , with ˜ d n,k = [ x n − k ] f g m − k − . Where [ x j ] S stands for the j -coefficient of the formal power series S . We can always embed any T ( f | g ) in a bi-infinite lower triangular matrix. or our example we have ... − − − − · · · − − − − − · · ·− − − − − − − − − − − − −
11 6 − −
26 23 − This construction is very nice in the case of Riordan matrices of the kind T (1 | a + bx ),(treated as change of variables in [5]). For these matrices we have ... ... ... ... ... ... ... a (cid:0) a − bx (cid:1) ← a · · · a (cid:0) a − bx (cid:1) ← a b a · · · a (cid:0) a − bx (cid:1) ← ab ab a · · · a a − bx ← b b b · · · a ← a · · · a (cid:0) − bxa (cid:1) ← − ba a · · · a (cid:0) − bxa (cid:1) ← b a − ba a · · ·↓ ↓ ↓ ↓ ↓ ↓ ↓ ...( a + bx ) ( a + bx ) ( a + bx ) 1 a + bx a + bx ) a + bx ) · · · Note that, as in the Pascal triangle, we can see the above matrix in the following way a T − (1 | a + bx ) (cid:7) T (1 | a + bx ) ! Where a T − (1 | a + bx ) (cid:7) is a T − (1 | a + bx ) placed in the same way as the inverseof the Pascal triangle is placed in the Hexagon of Pascal in page 194 in [2]. Note that T − (1 | a + bx ) = T (1 | − bxa ). This gives us a method to calculate T − (1 | a + bx ) by meansof elementary operations. ur algorithm of construction do not need the so called A and Z sequences of a Riordanarray, see [6] and [11]. But, in our notation, they appear in the expression of the inverse ofthe Riordan array T ( f | g ) giving us specially aesthetic formula: Proposition 7.
Let f = X n ≥ f n x n and g = X n ≥ g n x n be two formal power series with f = 0 and g = 0 . Suppose that A and Z represent the A -sequence and the Z -sequence,respectively, of T ( f | g ) . Then (i) T − (1 | g ) = T (1 | A )(ii) T − ( f | g ) = T (cid:16) g f ( A − xZ ) (cid:12)(cid:12) A (cid:17) Proof. (i) From Theorem 1.3 of [11], the A -sequence is the unique series, with A (0) = 0,such that 1 g = A (cid:18) xg (cid:19) . So xg = xA (cid:18) xg (cid:19) . If ω = xg then ω = xA ( ω ). On the other handas ω − ◦ ω = x and ω = xg then x = ωg , composing with ω − we get ω − = xg ( ω − ) and ω − x = g ( ω − ). So composing with ω − but now in ω = xA ( ω ) we get x = ω − A ( x ) then1 = ω − x A ( x ) so 1 = g ( ω − ) A ( x ) then A ( x ) = 1 g ( ω − ) . Since T − (1 | g ) = T (1 | g ( ω − ) ) = T (1 | A ).(ii) From Theorem 2.3 in [6] we obtain that the Z is determined by the equality ω − Z = 1 − f g ( ω − ) g f ( ω − ) . From here we get 1 f ( ω − ) = g f ( A − xZ ). So T − ( f | g ) = T (cid:18) f ( ω − ) (cid:12)(cid:12) g ( ω − ) (cid:19) = T (cid:18) g f ( A − xZ ) (cid:12)(cid:12) A (cid:19) (cid:3) Corollary 8. T − ( f | g ) = T (1 | A ) T (cid:18) f | (cid:19) Lagrange inversion formula via Banach fixed point Theorem
In the previous section we showed that to calculate the inverse of T (1 | g ) we need, inparticular, to calculate ω − , where ω = xg and then ω − = xg ( ω − ). So we consider thefunction F : x K [[ x ]] → x K [[ x ]] defined by F ( y ) = xg ( y ). Here x K [[ x ]] represents the serieswith null independent term. This function is -contractive since d ( F ( y ) , F ( y )) = 12 ω ( xg ( y ) − xg ( y )) = 12 ω ( g ( y ) − g ( y ))+1 ≤ d ( y , y ) he domain, x K [[ x ]], of F is the closed ball, in ( K [[ x ]] , d ), whose center is the series S = 0and the ratio is . Consequently our domain is also complete with the relative metric. Sothe unique fixed point of F is ω − = (cid:16) xg (cid:17) − and BFPT can be applied.The
BFPT gives us a theoretical iterative process to calculate ω − . To convert this methodinto an effective approximation process we first note that the relation d ( S , S ) ≤ m +1 means that the m degree Taylor polynomials of both series are equals, that is T m ( S ) = T m ( S ). Then we obtain the following algorithm:Suppose g = P n ≥ g n x n . We begin to iterate at S = 0. F (0) = g x . This means that T ( ω − ) = g x . Using the -contractivity of F we get T ( F ( g x )) = T ( ω − ). Since F ( g x ) = g x + g g x + · · · we obtain T ( ω − ) = g x + g g x . Similar arguments allow usto prove that T ( F ( g x + g g x )) = T ( ω − ). The above construction can be summarizedin the following (the notation is as above): Proposition 9. T m ( ω − ) = T m ( F ( T m − ( F ( · · · ( F ( T ( F (0)))) · · · ))))Following this process we get T ( ω − ) = g xT ( ω − ) = g x + g g x T ( ω − ) = g x + g g x + ( g g + g g ) x T ( ω − ) = g x + g g x + ( g g + g g ) x + ( g g + 3 g g g + g g ) x T ( ω − ) = g x + g g x + ( g g + g g ) x + ( g g + 3 g g g + g g ) x + ( g g + 6 g g g +2 g g + 4 g g g + g g ) x If we recall the Cauchy powers of the series g : g ( x ) = g + g x + g x + g x + g x + · · · g ( x ) = g + g x + (2 g g + g ) x + (2 g g + 2 g g ) x · · · g ( x ) = g + 3 g g x + ( g g + g g ) x + (6 g g g + 3 g g + g ) x + · · · g ( x ) = g + 4 g g x + (4 g g + 6 g g ) x + ( g g + g g + g g ) x + · · · g ( x ) = g +5 g g x +(5 g g +10 g g ) x +(20 g g g +10 g g +5 g g ) x + ( g g + g g + g + g g + g g ) x + · · · comparing adequately the coefficients of ω − and the powers of g we obtain the nextrelationships:[ x ] ω − = [ x ] g [ x ] ω − = [ x ] g [ x ] ω − = [ x ] g [ x ] ω − = [ x ] g [ x ] ω − = [ x ] g hese equalities allow us to predict and motivate the classical Lagrange Inversion Formula,see [12] page 36: [ x n +1 ] ω − = 1 n + 1 [ x n ] g n +1 , with ω = xg From now on we denote T j ≡ T j ( ω − ). To show how this process works note that F ( T n ) = x ( g + g T n + g T n + · · · + g n T nn + · · · ) == T n + ( g [ x n ] T n + g [ x n ] T n + · · · + g n [ x n ] T nn ) x n +1 + S n +2 with S n +2 ∈ x n +2 K [[ x ]]So [ x n +1 ] F ( T n ) = n X k =1 [ x k ] g [ x n ]( ω − ) k Because [ x n ]( ω − ) k = [ x n ]( T n ) k for any k ≤ n . Suppose now that we know n [ x n ]( ω − ) k = k [ x n − k ] g n for k ≤ n then [ x n +1 ] F ( T n ) = [ x n +1 ] ω − = 1 n n X k =1 k [ x k ] g [ x n − k ] g n = 1 n [ x n − ] g ′ g n = 1 n + 1 [ x n ] g n +1 Note that in the above development we need to know n [ x n ]( ω − ) k = k [ x n − k ] g n for k ≤ n .In fact we can give a proof of all above using essentially the fact that ω − is a fixed pointof certain contractive function. Theorem 10. (Lagrange inversion via Banach Fixed Point Theorem) Let K be a fieldof characteristic zero. Suppose that ω is a formal power series in K [[ x ]] with ω (0) = 0 and ω ′ (0) = 0 . Then n [ x n ]( ω − ) k = k [ x n − k ] (cid:16) xω (cid:17) n for n, k ∈ N Proof.
Let g = xω . So [ x ] g = 0. As proved before ω − is the unique fixed point of the -contractive function F : x K [[ x ]] → x K [[ x ]] defined by F ( y ) = xg ( y ). Iterating at y = 0we get [ x ] ω − = [ x ] F (0) = [ x ] g If k >
1, note that [ x ]( ω − ) k = 0 and [ x − k ] g = 0 and then[ x ] ω − = k [ x − k ] g Let proceed by induction on n . Suppose that j [ x j ]( ω − ) k = k [ x j − k ] g j for j ≤ n, k ≥ k . Because if j < k ,then [ x j − k ] g j = 0 = [ x j ]( ω − ) k . Then the equality holds trivially. ince ω − = xg ( ω − ), then for any k ( ω − ) k = x k g k ( ω − ). Consequently[ x n +1 ]( ω − ) k = [ x n +1 ] x k g k ( ω − ) = [ x n +1 − k ] g k ( ω − ) = n +1 − k X j =0 [ x j ] g k [ x n +1 − k ]( ω − ) j by the induction hypothesis[ x n +1 ]( ω − ) k = 1 n + 1 − k n +1 − k X j =0 j [ x j ] g k [ x n +1 − k − j ] g n +1 − k Let us call h = g k [ x n +1 ]( ω − ) k = 1 n + 1 − k n +1 − k X j =0 j [ x j ] h [ x n +1 − k − j ] h n +1 − kk = 1 n + 1 − k n +1 − k X j =1 [ x j − ] h ′ [ x n +1 − k − j ] h n +1 − kk == 1 n + 1 − k n − k X j =0 [ x j ] h ′ [ x n − k − j ] h n +1 − kk = 1 n + 1 − k [ x n − k ]( h ′ h n +1 − kk ) == 1 n + 1 − k [ x n − k ] (cid:18) kn + 1 h n +1 − kk (cid:19) ′ = kn + 1 [ x n +1 − k ] h n +1 k = kn + 1 [ x n +1 − k ] g n +1 (cid:3) The development above gives us the following algorithm to calculate the coefficients of thecompositional inverse of ω = xg : Algorithm 11.
Given g = X j ≥ g j x j , with g = 0. Given F ( y ) = xg ( y ), y ∈ x K [[ x ]]. step 1 : (Initial.) T = g x . step k (2 to n) : Calculate the Taylor polynomial of order k of F ( T k − )We can write this algorithm in an informal pseudo-code: READ (g,F,n)SET TCALCULATE for i from 2 to n doT[i]:=convert(series(F(T[i-1],x=,i)),polynom);endPRINT T
Acknowledgment:
The author thanks Manuel A. Mor´on for his helpful comments.I also thank the referee for his/her suggestions and comments which strongly improvedearlier version of this paper. The author was partially supported by the grant MICINN-FIS2008-04921-C02-02. eferences [1] J. Dugundji and A. Granas. Fixed Point Theory.
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