We obtain some recurrence relationships among the partition vectors of the
partial exponential Bell polynomials. On using such results, the $n$-th Adomian
polynomial for any nonlinear operator can be expressed explicitly in terms of
the partial exponential Bell polynomials. Some new identities for the partial
exponential Bell polynomials are obtained by solving certain ordinary
differential equations using Adomian decomposition method

Kataria, K. K.

Vellaisamy, P.

Comptes Rendus Mathematique

English

arXiv

We obtain some recurrence relationships among the partition vectors of the partial exponential Bell polynomials. On using such results, the n-th Adomian polynomial for any nonlinear operator can be expressed explicitly in terms of the partial exponential Bell polynomials. Some new identities for the partial exponential Bell polynomials are obtained by solving certain ordinary differential equations using the Adomian decomposition method. (C) 2017 Academie des sciences. Published by Elsevier Masson SAS. All rights reserved

KATARIA, KK

VELLAISAMY, P

Dspace at IIT Bombay

Correlation between Adomian and partial exponential Bell polynomials

Kuldeep Kumar Kataria

Palaniappan Vellaisamy

Comptes Rendus Mathématique

C. R. Acad. Sci. Paris, Ser. I 355 (2017) 929–936Contents lists available at ScienceDirectC. R. Acad. Sci. Paris, Ser. Iwww.sciencedirect.comCombinatorics/Ordinary differential equationsCorrelation between Adomian and partial exponential Bell polynomialsCorrélation des polynômes d’Adomian et des polynômes de Bell exponentiels partielsKuldeep Kumar Kataria, Palaniappan VellaisamyDepartment of Mathematics, Indian Institute of Technology Bombay, Powai, Mumbai 400076, Indiaa r t i c l e i n f o a b s t r a c tArticle history:Received 29 December 2016Accepted 29 August 2017Available online 12 September 2017Presented by the Editorial BoardWe obtain some recurrence relationships among the partition vectors of the partial exponential Bell polynomials. On using such results, the n-th Adomian polynomial for any nonlinear operator can be expressed explicitly in terms of the partial exponential Bell polynomials. Some new identities for the partial exponential Bell polynomials are obtained by solving certain ordinary differential equations using the Adomian decomposition method.© 2017 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.r é s u m éNous montrons des relations de récurrence entre les vecteurs partition des polynômes de Bell exponentiels partiels. Utilisant ces relations, le n-ième polynôme d’Adomian, pour n’importe quel opérateur non linéaire, s’exprime explicitement en termes des polynômes de Bell exponentiels partiels. On en déduit des identités nouvelles pour ces derniers, via la solution de certaines équations différentielles ordinaires, en utilisant la méthode de décomposition d’Adomian.© 2017 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.1. IntroductionThe Bell polynomials studied by Bell [4,5] are special polynomials in combinatorial analysis, with numerous applications in different areas of mathematics. The incomplete or partial exponential Bell polynomials Bn,k (see [7], [10, p. 96]) in n − k + 1 variables are triangular arrays of polynomials defined byE-mail address: kulkat@math.iitb.ac.in (K.K. Kataria).http://dx.doi.org/10.1016/j.crma.2017.08.0021631-073X/© 2017 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.930 K.K. Kataria, P. Vellaisamy / C. R. Acad. Sci. Paris, Ser. I 355 (2017) 929–936Bn,k(u1, u2, . . . , un−k+1) = n!∑knn−k+1∏j=11k j!(u jj!)k j, (1.1)where the partition set is given by kn ={(k1,k2, . . . ,kn−k+1) : ∑n−k+1j=1 k j = k,∑n−k+1j=1 jk j = n,k j ∈N0}. Here Nm ={x : x ≥ m, x ∈N∪ {0}} and N denotes the set of positive integers. Also, the sumBn(u1, u2, . . . , un) =n∑k=1Bn,k(u1, u2, . . . , un−k+1), (1.2)is called the n-th complete exponential Bell polynomial. For more details, results and some known identities on Bell poly-nomials, we refer the reader to [6, pp. 133–137], [7] and [10, pp. 95–98].Next we briefly explain the Adomian decomposition method (ADM) [2,3], which will be used later to obtain some new identities for Bell polynomials. In ADM, the solution to the functional equationu = f + L(u) + N(u), (1.3)where L and N are linear and nonlinear operators respectively and f is a known function, is expressed in the form of an infinite seriesu =∞∑n=0un. (1.4)The nonlinear term N(u) decomposes asN(u) =∞∑n=0An(u0, u1, . . . , un), (1.5)where An denotes the n-th Adomian polynomial in u0, u1, . . . , un . Also, the series (1.4) and (1.5) are assumed to be abso-lutely convergent. So, (1.3) can be rewritten as∞∑n=0un = f +∞∑n=0L(un) +∞∑n=0An.Thus the uns are obtained by the following recursive relationu0 = f and un = L(un−1) + An−1.The crucial step involved in ADM is the calculation of Adomian polynomials. Adomian [2, pp. 19–21] gave a method for determining these polynomials, by parameterizing u as uλ = ∑∞n=0 unλn and assuming N(uλ) to be analytic in λ, which decomposes as N(uλ) = ∑∞n=0 An(u0, u1, . . . , un)λn . Hence, Adomian polynomials are given byAn(u0, u1, . . . , un) = 1n!∂n N(uλ)∂λn∣∣∣∣λ=0, ∀ n ∈N0. (1.6)An improved version of the above result (see Zhu et al. [13]) is given byAn(u0, u1, . . . , un) = 1n!∂n N(∑nk=0 ukλk)∂λn∣∣∣∣∣λ=0, ∀ n ∈N0. (1.7)Rach [12] suggested the following formula for these polynomials: A0(u0) = N(u0) andAn(u0, u1, . . . , un) =n∑k=1C(k,n)N(k)(u0), ∀ n ∈ N, (1.8)whereC(k,n) =∑knn∏j=1uk jjk j! , (1.9)and the summation is taken over the partition set kn ={(k1,k2, . . . ,kn) : ∑nj=1 k j = k,∑nj=1 jk j = n,k j ∈N0}. Also, N(k)(.)denotes the k-th derivative of the nonlinear term. One can easily show the equivalence of (1.6) and (1.8) using the Faà K.K. Kataria, P. Vellaisamy / C. R. Acad. Sci. Paris, Ser. I 355 (2017) 929–936 931di Bruno’s formula. Recently, Kataria and Vellaisamy [11] obtained simple parameterization methods for generating these Adomian polynomials both explicitly and recursively.In this paper, we first obtain some results related to the partition vectors of the partial exponential Bell polynomials. Then we show that the C(k, n)’s, which are homogeneous polynomials of order k, can be represented in terms of the well-known Bell polynomials. Hence, a closed form expression of the n-th order Adomian polynomial for any nonlinear operator is obtained as a finite sum of the partial exponential Bell polynomials. The significance of this result is that any algorithm or identity for the C(k, n)’s will give the corresponding results for the Bell polynomials and vice versa. Indeed we use the results of Duan [8,9] to obtain some recursive algorithms for the partial exponential Bell polynomials. Also, we use the Adomian decomposition method to solve certain ordinary differential equations to obtain some new identities for the partial exponential Bell polynomials.2. The partition vectors of the Bell polynomialsWe use the following results by Duan [8] for the partition set kn of Adomian polynomials to show some similar results for the partition set kn of Bell polynomials.Lemma 2.1. For 1 ≤ k ≤ n, kn ⊂ Nn0 and 11 = {(1)}, 1n = {(0, 0, . . . , 0, 1)}.Lemma 2.2. For every vector (k1,k2, . . . ,kn) ∈ kn, 2 ≤ k ≤ n, the last (k − 1) entries are zero, i.e. kn−k+2 = kn−k+3 = . . . = kn = 0.Theorem 2.1. For n ∈N2 , if 2 ≤ k ≤[ n2], then kn = 1 ∪ 2 , and if [ n2]< k ≤ n, then kn = 1 , where1 = {(k1 + 1,k2, . . . ,kn−1,0) : (k1,k2, . . . ,kn−1) ∈ k−1n−1}and2 = {(0,k1,k2, . . . ,kn−k,0,0, . . . ,0︸ ︷︷ ︸k−1 times) : (k1,k2, . . . ,kn−k) ∈ kn−k}.Next we obtain some results for the partition set kn of Bell polynomials. The following lemmas are easy to prove.Lemma 2.3. For 1 ≤ k ≤ n, kn ⊂ Nn−k+10 with 1n = 1n for all n ∈N.The following result is evident from Lemma 2.2.Lemma 2.4. Let enj denote the n-tuple vector with unity at the j-th place and zero elsewhere. Thenkn =⎧⎨⎩n−k+1∑j=1k jenj :(k1,k2, . . . ,kn−k+1) ∈ kn⎫⎬⎭ .Similar recurrence relationships among the partition vectors of the partial exponential Bell polynomials hold.Theorem 2.2. For n ∈N2 , if 2 ≤ k ≤[ n2], then kn = 1 ∪ 2 , and if [ n2]< k ≤ n, then kn = 1 , where1 = {(k1 + 1,k2, . . . ,kn−k+1) : (k1,k2, . . . ,kn−k+1) ∈ k−1n−1}and2 = {(0,k1,k2, . . . ,kn−2k+1,0,0, . . . ,0︸ ︷︷ ︸k−1 times) : (k1,k2, . . . ,kn−2k+1) ∈ kn−k}.Proof. The proof is evident on using Lemma 2.2 and Theorem 2.1. A relationship between Adomian polynomials and Bell polynomials was first obtained by Abbaoui et al. [1]. Here we establish a relationship between Adomian and partial exponential Bell polynomials, which is different from the one proved in [1].932 K.K. Kataria, P. Vellaisamy / C. R. Acad. Sci. Paris, Ser. I 355 (2017) 929–936Theorem 2.3. Let An, n ≥ 1, be the n-th Adomian polynomial for the nonlinear term N(u). ThenAn(u0, u1, . . . , un) = 1n!n∑k=1Bn,k(1!u1,2!u2, . . . , (n − k + 1)!un−k+1)N(k)(u0). (2.1)Proof. On setting kn−k+ j = 0 for j = 2, 3, . . . , k and using Lemma 2.4, we haveBn,k(1!u1,2!u2, . . . , (n − k + 1)!un−k+1) = n!∑knn−k+1∏j=1uk jjk j! = n!∑knn∏j=1uk jjk j! = n!C(k,n).Hence,C(k,n) = 1n! Bn,k(1!u1,2!u2, . . . , (n − k + 1)!un−k+1). (2.2)The proof completes on using (1.8). From (2.2) it is clear that the homogeneous polynomials of order k, i.e. the C(k, n)’s are the partial exponential Bell polynomials. On using (2.2), the following results follow from the recursive algorithm of C(k, n) given by Corollary 1 of Duan [8].Corollary 2.1. For all n ∈N, we have Bn,1(1!u1, 2!u2, . . . , n!un) = n!un and Bn,n(u1) = un1 . When [ n2]< k ≤ n and n ∈N2 , we haveBn,k(1!u1,2!u2, . . . , (n − k + 1)!un−k+1) = n Bn−1,k−1(1!u1,2!u2, . . . , (n − k + 1)!un−k+1)∣∣k1→k1+1 .For 2 ≤ k ≤ [ n2 ] and n ∈N4 , we haveBn,k(1!u1,2!u2, . . . , (n − k + 1)!un−k+1) = n Bn−1,k−1(1!u1,2!u2, . . . , (n − k + 1)!un−k+1)∣∣k1→k1+1+ (n)k Bn−k,k(1!u2,2!u3, . . . , (n − 2k + 1)!un−2k+2),where (n)k = n(n − 1) . . . (n − k + 1) denotes the falling factorials.An alternate recursive algorithm for the Bell polynomials follows from Corollary 3 and 4 of Duan [9].Corollary 2.2. Let n be any positive integer. Then for 2 ≤ k ≤ n, we haveBn,k(1!u1,2!u2, . . . , (n − k + 1)!un−k+1)=n−k∑j=0( j + 1)(n − 1) ju j+1 Bn− j−1,k−1(1!u1,2!u2, . . . , (n − j − k + 1)!un− j−k+1).Alternatively, for 2 ≤ k ≤ n − 1, we haveBn,k(1!u1,2!u2, . . . , (n − k + 1)!un−k+1) = u1 Bn−1,k−1(1!u1,2!u2, . . . , (n − k + 1)!un−k+1)+n−k∑j=1( j + 1)u j+1 ∂∂u jBn−1,k(1!u1,2!u2, . . . , (n − k)!un−k).Next we show that the n-th Adomian polynomial for exponential non-linearity can be expressed in terms of the n-th complete exponential Bell polynomial. By using (2.1), the Adomian polynomials for N(u) = eu are A0(u0) = eu0 andA1(u0, u1) = B1,1(1!u1)eu0 = u1eu0 ,A2(u0, u1, u2) = eu02![B2,1(1!u1,2!u2) + B2,2(1!u1)] =(u2 + u212)eu0 ,A3(u0, u1, u2, u3) = eu03![B3,1(1!u1,2!u2,3!u3) + B3,2(1!u1,2!u2) + B3,3(1!u1)] =(u3 + u1u2 + u316)eu0 ,...An(u0, u1, . . . , un) = eu0n!n∑k=1Bn,k(1!u1,2!u2, . . . , (n − k + 1)!un−k+1) = eu0n! Bn(1!u1,2!u2, . . . ,n!un).K.K. Kataria, P. Vellaisamy / C. R. Acad. Sci. Paris, Ser. I 355 (2017) 929–936 933Now we give two recursive algorithms for the complete exponential Bell polynomials.Theorem 2.4. Let n be any positive integer. ThenBn(1!u1, . . . ,n!un) =n−1∑k=0(k + 1)(n − 1)kuk+1 Bn−k−1(1!u1, . . . , (n − k − 1)!un−k−1),andBn(1!u1, . . . ,n!un) = u1 Bn−1(1!u1, . . . , (n − 1)!un−1) +n−1∑k=1(k + 1)uk+1 ∂∂ukBn−1(1!u1, . . . , (n − 1)!un−1).Proof. From Corollary 1 and 2 of Duan [9], we have the following recursive algorithms for Adomian polynomials:An(u0, u1, . . . , un) = 1nn−1∑k=0(k + 1)uk+1 ∂∂u0An−k−1(u0, u1, . . . , un−k−1),An(u0, u1, . . . , un) = 1nn−1∑k=0(k + 1)uk+1 ∂∂ukAn−1(u0, u1, . . . , un−1), n ≥ 1.The result follows by choosing the nonlinear term N(u) = eu in Theorem 2.3 and using (2.1) in the above expressions for Adomian polynomials. 3. Some identities of the Bell polynomialsWe now state and prove some new identities for the partial exponential Bell polynomials. This is achieved by solving certain ordinary differential equations using the Adomian decomposition method.Theorem 3.1. Let α, β be any real numbers and n be any positive integer. Thenn∑k=1(−β)k Bn,k(0!α,−1!α2β, . . . , (−1)n−k (n − k)!αn−k+1βn−k) = n!(−αβ)n. (3.1)Proof. Consider the following ordinary differential equationdudx= αe−βu, u(0) = 1, |αβx| < eβ . (3.2)Equivalently,u(x) = u(0) + αx∫0e−βu(t) dt. (3.3)In (3.3), the nonlinear term is N(u) = e−βu . Substituting u = ∑∞n=0 un and N(u) = ∑∞n=0 An in the above equation and applying ADM, we get∞∑n=0un(x) = u(0) + α∞∑n=0x∫0An (u0(t), u1(t), . . . , un(t)) dt.Therefore, for all n ≥ 0,un+1(x) = αx∫0An (u0(t), u1(t), . . . , un(t)) dt. (3.4)On comparing u0 = u(0) = 1, hence u1, u2, u3, u4 are recursively obtained as934 K.K. Kataria, P. Vellaisamy / C. R. Acad. Sci. Paris, Ser. I 355 (2017) 929–936A0 (u0(x)) = e−βu0 = 1eβ, u1(x) = αx∫0A0 (u0(t)) dt = αeβx,A1 (u0(x), u1(x)) = −βu1e−βu0 = − αβe2βx, u2(x) = αx∫0A1 (u0(t), u1(t)) dt = − α2β2e2βx2,A2 (u0(x), u1(x), u2(x)) = 12!(β2u21 − 2βu2)e−βu0 = α2β2e3βx2,u3(x) = αx∫0A2 (u0(t), u1(t), u2(t)) dt = α3β23e3βx3,A3 (u0(x), u1(x), u2(x), u3(x)) = 13!(−β3u31 + 6β2u1u2 − 6βu3)e−βu0 = −α3β3e4βx3,u4(x) = αx∫0A3 (u0(t), u1(t), u2(t), u3(t)) dt = −α4β34e4βx4,and so on, where the Adomian polynomials A1, A2, . . . are calculated using (1.7).By using the method of separation of variables, it is easy to see that u(x) = 1 + β−1 ln(1 + αβe−β x) is the solution to (3.2). Since |αβe−β x| < 1, from the Taylor series expansion of ln(1 + αβe−β x), we haveu(x) = 1 +∞∑n=1(−1)n+1 αnβn−1neβnxn = u(0) +∞∑n=1un(x). (3.5)By the uniqueness of the solution to the differential equation (3.2), the series solution obtained by ADM is consistent with (3.5). Therefore, the n-th component of u(x) isun(x) = (−1)n+1 αnβn−1neβnxn, n ≥ 1. (3.6)Now using Rach’s formula (1.8) in (3.4), we obtainun+1(x) = αx∫0n∑k=1N(k)(u0)∑knn∏j=1uk jj (t)k j! dt =αeβ(n+1)x∫0tn dtn∑k=1(−β)k∑knn∏j=11k j!((−1) j+1α jβ j−1j)k j= αxn+1(n + 1)eβ(n+1)n∑k=1(−β)k∑knn−k+1∏j=11k j!((−1) j+1( j − 1)!α jβ j−1j!)k j,where the last two steps follow from (3.6) and Lemma 2.4, respectively. Finally, by using (3.6) and rearranging the terms, we getn∑k=1(−β)k∑knn−k+1∏j=1n!k j!((−1) j+1( j − 1)!α jβ j−1j!)k j= n!(−αβ)n,which is the required identity. Substituting {α, β} ∈ {{−1, 1}, {1, 1}, {−1, −1}, {1, −1}} in (3.1), we obtain the following results, respectively.Corollary 3.1. For any positive integer n, the following identities hold:n∑k=1(−1)k Bn,k(−0!,−1!, . . . ,−(n − k)!) = n!,n∑(−1)k Bn,k(0!,−1!, . . . , (−1)n−k(n − k)!) = n!(−1)n,k=1K.K. Kataria, P. Vellaisamy / C. R. Acad. Sci. Paris, Ser. I 355 (2017) 929–936 935n∑k=1Bn,k(−0!,1!, . . . , (−1)n−k+1(n − k)!) = n!(−1)n, (3.7)n∑k=1Bn,k(0!,1!, . . . , (n − k)!) = n!. (3.8)Remark 3.1. The identities given by (3.7) and (3.8) above can be expressed in terms of the n-th complete exponential Bell polynomials as follows:Bn(−0!,1!, . . . , (−1)n(n − 1)!) = n!(−1)n and Bn(0!,1!, . . . , (n − 1)!) = n!.Note that (3.8) is a known result, i.e. Bn,k(0!, 1!, . . . , (n − k)!) = s(n, k), where the s(n, k)’s are Stirling numbers of the first kind.Next we obtain an identity of the partial exponential Bell polynomials in terms of the falling factorials.Theorem 3.2. Let α be any non-zero real number and n be any positive integer. Thenn∑k=1(1 − 1α)kBn,k((α)1, (α)2, . . . , (α)n−k+1) = (α − 1)n,where (α)k = α(α − 1)(α − 2) . . . (α − k + 1) denotes the falling factorial.Proof. Consider the following ordinary differential equationdudx= αu1−1/α, u(0) = 1, |x| < 1. (3.9)Equivalently,u(x) = u(0) + αx∫0u1−1/α(t)dt. (3.10)In (3.10), there is no linear term, but the nonlinear term is N(u) = u1−1/α . Substituting u = ∑∞n=0 un and N(u) = ∑∞n=0 Anin the above equation and applying ADM, we get∞∑n=0un(x) = u(0) + α∞∑n=0x∫0An (u0(t), u1(t), . . . , un(t)) dt.Therefore, for all n ≥ 0,un+1(x) = αx∫0An (u0(t), u1(t), . . . , un(t)) dt. (3.11)On comparing u0 = u(0) = 1 and hence u1, u2, u3, u4 are recursively obtained asA0 (u0(x)) = u1−1/α0 = 1, u1(x) = (α)1x,A1 (u0(x), u1(x)) =(1 − 1α)u−1/α0 u1 =(α)2αx, u2(x) = (α)22! x2,A2 (u0(x), u1(x), u2(x)) = 12!(1 − 1α)(− 1αu−1/α−10 u21 + 2u−1/α0 u2)= (α)32!α x2, u3(x) = (α)33! x3,A3 (u0(x), u1(x), u2(x), u3(x)) = 13!(1 − 1α)((1α+ 1α2)u−1/α−20 u31 −4αu−1/α−10 u1u2 + 6u−1/α0 u3− 2αu−1/α−10 u1u2)= (α)43!α x3,u4(x) = (α)44! x4,and so on.936 K.K. Kataria, P. Vellaisamy / C. R. Acad. Sci. Paris, Ser. I 355 (2017) 929–936Also, using the method of separation of variables, it is easy to see that u(x) = (1 + x)α is the solution to (3.9). Since |x| < 1, expand (1 + x)α by using generalized binomial theoremu(x) = 1 +∞∑n=1(α)nn! xn = u(0) +∞∑n=1un(x). (3.12)By the uniqueness of the solution to the differential equation (3.9), the series solution obtained by ADM is consistent with (3.12). Therefore, the n-th component of u(x) isun(x) = (α)nn! xn, n ≥ 0. (3.13)Now, from (3.11) and (3.13), we have(α)n+1(n + 1)! xn+1 = αx∫0An (u0(t), u1(t), . . . , un(t)) dt = αx∫0n∑k=1N(k)(u0)∑knn∏j=1uk jj (t)k j! dt (using (1.8) and (1.9))= αx∫0tn dtn∑k=1(1 − 1α)k∑knn∏j=11k j!((α) jj!)k j(using (3.13))= αn + 1 xn+1n∑k=1(1 − 1α)k∑knn−k+1∏j=11k j!((α) jj!)k j,where the last step follows on using Lemma 2.4. On rearranging the terms, we getn∑k=1(1 − 1α)k∑knn!n−k+1∏j=11k j!((α) jj!)k j= (α − 1)n,which is the required identity. AcknowledgementsThe authors wish to thank Dr. Randolph Rach for providing insightful comments on initial version of the manuscript, and especially for pointing reference [1]. The research of the first author was supported by UGC fellowship No. F.2-2/98(SA-I), Govt. of India.References[1] K. Abbaoui, Y. Cherruault, V. Seng, Practical formulae for the calculus of multivariable Adomian polynomials, Math. Comput. Model. 22 (1) (1995) 89–93.[2] G. Adomian, Nonlinear Stochastic Operator Equations, Academic Press, Orlando, FL, USA, 1986.[3] G. Adomian, Solving Frontier Problems of Physics: The Decomposition Method, Kluwer Academic, Dordrecht, The Netherlands, 1994.[4] E.T. Bell, Partition polynomials, Ann. of Math. 29 (1–4) (1927) 38–46.[5] E.T. Bell, Exponential polynomials, Ann. of Math. 35 (2) (1934) 258–277.[6] L. Comtet, Advanced Combinatorics: The Art of Finite and Infinite Expansions, D. Reidel Publishing Co., Dordrecht, The Netherlands, 1974.[7] D. Cvijović, New identities for the partial Bell polynomials, Appl. Math. Lett. 24 (9) (2011) 1544–1547.[8] J.-S. Duan, Recurrence triangle for Adomian polynomials, Appl. Math. Comput. 216 (4) (2010) 1235–1241.[9] J.-S. Duan, Convenient analytic recurrence algorithms for the Adomian polynomials, Appl. Math. Comput. 217 (13) (2011) 6337–6348.[10] M. Hazewinkel, Encyclopedia of Mathematics, Supplement I, Kluwer Academic, Dordrecht, The Netherlands, 1997.[11] K.K. Kataria, P. Vellaisamy, Simple parametrization methods for generating Adomian polynomials, Appl. Anal. Discrete Math. 10 (1) (2016) 168–185.[12] R. Rach, A convenient computational form for the Adomian polynomials, J. Math. Anal. Appl. 102 (2) (1984) 415–419.[13] Y. Zhu, Q. Chang, S. Wu, A new algorithm for calculating Adomian polynomials, Appl. Math. Comput. 169 (1) (2005) 402–416.

Kataria, Kuldeep Kumar

Vellaisamy, Palaniappan

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Correlation between Adomian and Partial Exponential Bell Polynomials

Correlation between Adomian and Partial Exponential Bell Polynomials

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