# Taylor’s Expansion for Composite Functions

Authors: Le Thi Phuong Ngoc; Nguyen Anh Triet
Journal: The Scientific World Journal
 (2013)

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## Abstract

We build a Taylor’s expansion for composite functions. Some applications are introduced, where the proposed technique allows the authors to obtain an asymptotic expansion of high order in many small parameters of solutions.


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## Body

## 1. Introduction

Letf∈CN+1([0,1]×ℝ+×ℝ3×ℝ+3;ℝ) and uα∈W={v∈L∞(0,T;H2(0, 1)):vt∈L∞(0,T;H1(0,1))},α∈ℤ+n,|α|≤N.Let two functionsh(x,t,ε) andF(x,t,ε), which depend on (x,t) and ε=(ε1,…,εn), be defined as follows:
(1)h(x,t,ε)=∑‍|α|≤Nuα(x,t)εα,
where
(2)εα=ε1α1⋯εnαn,ε=(ε1,…,εn)∈ℝn,α=(α1,…,αn)∈ℤ+n,|α|=α1+⋯+αn,F(x,t,ε)=f(x,t,h,hx,ht,∥h∥2,∥hx∥2,∥ht∥2),
where notation ∥·∥ stands for the norm in L2(0,1).The functionF(x,t,ε) has the form of a composite function as follows:
(3)F(x,t,ε)=f(x,t,g1(ε),g2(ε),g3(ε),g4(ε),g5(ε),g6(ε)),
where
(4)g1(ε)=h(x,t,ε)=∑‍|α|≤Nuα(x,t)εα,g2(ε)=hx(x,t,ε)=∑‍|α|≤Nuαx(x,t)εα,g3(ε)=ht(x,t,ε)=∑‍|α|≤Nuαt(x,t)εα,g4(ε)=∥h(t,ε)∥2=∥∑‍|α|≤Nuα(t)εα∥2=∑‍|δ|≤2N∑‍|α|≤N∑‍|β|≤N∑‍α+β=δ〈uα(t),uβ(t)〉εδ=∑‍|α|≤2Ngα[4](t)εα,g5(ε)=∥hx(t,ε)∥2=∥∑‍|α|≤Nuαx(t)εα∥2=∑‍|δ|≤2N∑‍|α|≤N∑‍|β|≤N∑‍α+β=δ〈uαx(t),uβx(t)〉εδ=∑‍|α|≤2Ngα[5](t)εα,g6(ε)=∥ht(t,ε)∥2=∥∑‍|α|≤Nuαt(t)εα∥2=∑‍|δ|≤2N∑‍|α|≤N∑‍|β|≤N∑‍α+β=δ〈uαt(t),uβt(t)〉εδ=∑‍|α|≤2Ngα[6](t)εα,
where notation 〈·,·〉 stands for the scalar product in L2(0,1).Under the above assumptions, in order to obtain an asymptotic expansion of high order in many small parameters of solutions in recent papers [1–4], we need to solve the following problem.Problem 1.
Establish the functionsFα=Fα(x,t),|α|≤N (independent of ε) such that for ∥ε∥=ε12+⋯+εn2 being small enough,
(5)F(x,t,ε)=∑‍|α|≤NFα(x,t)εα+O(∥ε∥N+1);
it means that
(6)∥F-∑‍|α|≤NFαεα∥L∞(0,T,L2(0,1))≤CN∥ε∥N+1,
where the constant CN is independent from ε.
By Taylor’s expansion forF(x,t,ε), it implies that
(7)Fα(x,t)=1α!DαF(x,t,0),|α|≤N.However, forF(x,t,ε) which is given in (3), it is very difficult to calculate DαF(x,t,0), and so, the Problem 1 cannot proceed. Whichmethod can be used for solving the Problem 1? To answer, let us note that the function F(x,t,ε)has the form of a composite function;this viewsuggests thatwe need to construct the Taylor-Maclaurin expansion of the composite function. So, first in Section 2 we solve the following problem.Problem 2.
LetΩ be an open subset of ℝnand 0∈Ω. Letg=(g1,…,gp)∈CN+1(Ω;ℝp)andf∈CN+1(ℝp;ℝ).Seek the representation formula for fog, such that for ∥x∥ being small enough,
(8)(fog)(x)=f(g1(x),…,gp(x))=∑‍|α|≤Ndαxα+O(∥x∥N+1),
where dα,|α|≤N are calculated from the values of the given functions f;g1,…,gpand of their derivatives at a suitable point.Next in Section3, as an application of the method used, we study the Problem 1. This technique is also a great help for the authors to obtain an asymptotic expansion as they want in recent papers [1–4].


## 2. Solving the Problem2

We use the following notations. For a multi-indexα=(α1,…,αn)∈ℤ+n and x=(x1,…,xn)∈ℝn, we put
(9)|α|=α1+⋯+αn,α!=α1!⋯αn!,∥x∥=x12+⋯+xn2,xα=x1α1⋯xnαn,α,β∈ℤ+n,α≤β⟺αi≤βi∀i=1,…,n.The following lemma is useful to solve the Problems1 and 2.Lemma 3.
Letm,N∈ℕ and aα∈ℝ,α∈ℤ+n,1≤|α|≤N. Then
(10)(∑‍1≤|α|≤Naαxα)m=∑‍m≤|α|≤mNTN(m)[a]αxα,
where the coefficients TN(m)[a]α,m≤|α|≤mN depending on a=(aα),α∈ℤ+n,1≤|α|≤N are defined by the recurrent formulas
(11)TN(m)[a]α={uα,1≤|α|≤N,m=1,∑‍β∈Aα(m)(N)aα-βTN(m-1)[u]β,m≤|α|≤mN,m≥2,Aα(m)(N)={β∈ℤ+p:β≤α,1≤|α-β|≤N,m-1≤|β|≤(m-1)N}.The proof of Lemma3 can be found in [2].Now, using Taylor’s expansion of the functionfaround the point g=(g1,…,gp)∈ℝp, we obtain that
(12)f(g1+H1,…,gp+Hp)=f(g1,…,gp)+∑‍1≤|α|≤N1α!Dαf(g1,…,gp)Hα+RN[f,H],H=(H1,…,Hp)∈ℝp,∥H∥issmallenough,RN[f,H]=(N+1)×∑‍|α|=N+11α!Hα∫01‍(1-θ)NDα×f(g+θH)dθ.Similarly, we use Maclaurin’s expansion ofgi as follows:
(13)Hi=gi(x)-gi(0)=∑‍1≤|β|≤N1β!Dβgi(0)xβ+RN[gi,x]=Pi+Ri,x∈ℝn,kkkkkkkkk∥x∥issmallenough,Pi=∑‍1≤|β|≤N1β!Dβgi(0)xβ,Ri=RN[gi,x]=(N+1)∑‍|β|=N+11β!xβ∫01‍(1-θ)NDβgi(θx)dθ.Substituting (13) into (12), we get
(14)f(g1(x),…,gp(x))=f(g1(0),…,gp(0))+∑‍1≤|α|≤N1α!Dαf(g1(0),…,gp(0))Hα+RN[f,H]=f(g(0))+∑‍1≤|α|≤N1α!Dαf(g(0))Pα+∑‍1≤|α|≤N1α!Dαf(g(0))(Hα-Pα)+RN[f,H]=f(g(0))+∑‍1≤|α|≤N1α!Dαf(g(0))Pα+∥x∥N+1RN(1)[f,g,H,R],
where
(15)∥x∥N+1RN(1)[f,g,H,R]=∑‍1≤|α|≤N1α!Dαf(g(0))(Hα-Pα)+RN[f,H],RN[f,H]=(N+1)∑‍|α|=N+11α!Hα∫01‍(1-θ)NDα×f(g(0)+θH)dθ.Applying Lemma3, with a=σ(i)=(σβ(i)),σβ(i)=(1/β!)Dβgi(0),1≤|β|≤N, it implies that
(16)Piαi=(∑‍1≤|β|≤N1β!Dβgi(0)xβ)αi≡(∑‍1≤|β|≤Nσβ(i)xβ)αi=∑αi≤|γi|≤αiN‍TN(α1)[σ(i)]γixγi.Hence,(17)Pα=P1α1⋯Ppαp=(∑‍1≤|β|≤N1β!Dβg1(0)xβ)α1⋯(∑‍1≤|β|≤N1β!Dβgp(0)xβ)αp=∑α1≤|γ1|≤α1N‍TN(α1)[σ(1)]γ1xγ1⋯∑αp≤|γp|≤αpN‍TN(αp)[σ(p)]γpxγp=∑α1≤|γ1|≤α1N,⋮αp≤|γp|≤αpN‍TN(α1)[σ(1)]γ1⋯TN(αp)[σ(p)]γpxγ1+⋯+γp=∑α1≤|γ1|≤α1N,⋮αp≤|γp|≤αpN‍TN(α1)[σ(1)]γ1⋯TN(αp)[σ(p)]γpxδ=∑α1≤|γ1|≤α1N,⋮αp≤|γp|≤αpN‍∑γ1+⋯+γp=δ‍TN(α1)[σ(1)]γ1⋯TN(αp)[σ(p)]γpxδ=∑|α|≤|δ|≤|α|N‍∑α1≤|γ1|≤α1N,⋮αp≤|γp|≤αpN∑γ1+⋯+γp=δ‍TN(α1)[σ(1)]γ1⋯TN(αp)[σ(p)]γpxδ‍≡∑|α|≤|δ|≤|α|N‍Φδ[α,N,σ(1),…,σ(p)]xδ=∑|α|≤|δ|≤N‍Φδ[α,N,σ(1),…,σ(p)]xδ+∑N+1≤|δ|≤|α|N‍Φδ[α,N,σ(1),…,σ(p)]xδ,Φδ[α,N,σ(1),…,σ(p)]=∑α1≤|γ1|≤α1N,⋮αp≤|γp|≤αpN‍∑γ1+⋯+γp=δ‍TN(α1)[σ(1)]γ1⋯TN(αp)[σ(p)]γp.Consequently,(18)f(g(x))=f(g(0))+∑‍1≤|α|≤N1α!Dαf(g(0))Pα+∥x∥N+1RN(1)[f,g,H,R]=f(g(0))+∑‍1≤|α|≤N1α!Dαf(g(0))×∑|α|≤|δ|≤|α|N‍Φδ[α,N,σ(1),…,σ(p)]xδ+∥x∥N+1RN(1)[f,g,H,R]=f(g(0))+∑‍1≤|α|≤N1α!Dαf(g(0))×∑|α|≤|δ|≤N‍Φδ[α,N,σ(1),…,σ(p)]xδ+∑‍1≤|α|≤N1α!Dαf(g(0))×∑N+1≤|δ|≤|α|N‍Φδ[α,N,σ(1),…,σ(p)]xδ+∥x∥N+1RN(1)[f,g,H,R]=f(g(0))+∑‍1≤|α|≤N1α!Dαf(g(0))×∑|α|≤|δ|≤N‍Φδ[α,N,σ(1),…,σ(p)]xδ+∥x∥N+1RN(2)[f,g,H,R,x]=f(g(0))+∑‍1≤|δ|≤N∑1≤|α|≤|δ|‍1α!Dαf(g(0))×Φδ[α,N,σ(1),…,σ(p)]xδ+∥x∥N+1RN(2)[f,g,H,R,x]=f(g(0))+∑‍1≤|δ|≤Ndδxδ+∥x∥N+1RN(2)[f,g,H,R,x].Clearly, the Problem2 is solved with
(19)dδ=∑1≤|α|≤|δ|‍1α!Dαf(g(0))Φδ[α,N,σ(1),…,σ(p)].


## 3. Solving the Problem1

As an application of the method used in Section2 for p=6,Ω=(0,1),ε→=(ε1,…,εn)∈ℝn, and f=f(x,t,g1,…,g6)∈CN+1(ℝ3×ℝ+3;ℝ), for eachfixed (x,t)∈[0,1]×ℝ+;gi∈CN+1(Ω;ℝ), i=1,2,3, for eachfixed (x,t)∈[0,1]×ℝ+:gi∈CN+1(Ω;ℝ+),i=4,5,6, for eachfixed t∈ℝ+, we now investigate the Problem 1.For eachfixed (x,t)∈[0,1]×ℝ+, using Taylor’s expansion of the function f around the point g=(g1,…,g6)∈ℝ3×ℝ+3 up to order N,we obtain that
(20)f(x,t,g+H)=f(x,t,g1+H1,…,g6+H6)=f(x,t,g1,…,g6)+∑‍1≤|α|≤N1α!Dαf(x,t,g1,…,g6)Hα+∥H∥N+1RN(1)[f,H],(21)∥H∥N+1RN(1)[f,H]=(N+1)∑‍|α|=N+11α!Hα∫01‍(1-θ)NDα×f(g+θH)dθ,
where H=(H1,…,H6)∈ℝ6,∥H∥ is small enough, Diαif=∂f/∂gi,i=1,2,…,6.Next, the precise structure of the representation formulas forg1,…,g6will be needed below to continue.For each fixed(x,t)∈[0,1]×ℝ+: we have the following.(i) The representation formula forgi(ε)=gi(x,t,ε),i=1,2,3.We rewriteg1(ε) as follows
(22)g1(ε)=∑‍|α|≤Nuα(x,t)εα≡g1(0)+∑‍1≤|β|≤Ngβ[1](x,t)εβ≡g1(0)+H1,
in which
(23)g1(0)=g1(x,t)=u0(x,t),|β|=0,gβ[1](x,t)=uβ(x,t),1≤|β|≤N.It is similar tog2(ε),g3(ε); we write
(24)g2(ε)=∑‍|α|≤Nuαx(x,t)εα≡g2(0)+∑‍1≤|β|≤Ngβ[2](x,t)εβ≡g2(0)+H2,g3(ε)=∑‍|α|≤Nuαt(x,t)εα≡g3(0)+∑‍1≤|β|≤Ngβ[3](x,t)εβ≡g3(0)+H3,
where
(25)g2(0)=g2(x,t)=u0x(x,t),g3(0)=g3(x,t)=u0t(x,t),|β|=0,gβ[2](x,t)=uβx(x,t),gβ[3](x,t)=uβt(x,t),1≤|β|≤N.For eachfixed t∈ℝ+: we have the following.(ii) The representation formula forgi(ε)=gi(t,ε),i=4,5,6.Write(26)g4(ε)=∥∑‍|α|≤Nuα(t)εα∥2=∑‍|α|≤N∑‍|β|≤N〈uα(t),uβ(t)〉εα+β=∥u0(t)∥2+∑‍1≤|δ|≤NGδ[u(t)]εδ+∥ε∥N+1RN(1)[u,ε]≡g4(0)+∑‍1≤|β|≤Ngβ[4](t)εβ+R4≡g4(0)+H4≡g4(0)+P4+R4,(27)g4(0)=∥u0(t)∥2,gδ[4](t)=Gδ[u(t)]=∑‍|α|≤N∑‍|β|≤N∑‍α+β=δ〈uα(t),uβ(t)〉,1≤|δ|≤N,Gδ[u(t)]=∑‍|α|≤N∑‍|β|≤N∑‍α+β=δ〈uα(t),uβ(t)〉,1≤|δ|≤2N,R4=∥ε∥N+1RN(1)[u,ε]=∑‍N+1≤|δ|≤2NGδ[u(t)]εδ.Similarly,(28)g5(ε)=∥∑‍|α|≤Nuαx(t)εα∥2=∑‍|α|≤N∑‍|β|≤N〈uαx(t),uβx(t)〉εα+β=∥u0x(t)∥2+∑‍1≤|δ|≤NGδ[ux(t)]εδ+∥ε∥N+1RN(1)[ux,ε]≡g5(0)+∑‍1≤|β|≤Ngβ[5](t)εβ+R5≡g5(0)+H5≡g5(0)+P5+R5,
in which
(29)g5(0)=∥u0x(t)∥2,gδ[5](t)=Gδ[ux(t)]=∑‍|α|≤N∑‍|β|≤N∑‍α+β=δ〈uαx(t),uβx(t)〉,1≤|δ|≤N,R5=∥ε∥N+1RN(1)[ux,ε]=∑‍N+1≤|δ|≤2NGδ[ux(t)]εδ,(30)g6(ε)=∥∑‍|α|≤Nuαt(t)εα∥2=∑‍|α|≤N∑‍|β|≤N〈uαt(t),uβt(t)〉εα+β=∥u0t(t)∥2+∑‍1≤|δ|≤NGδ[ut(t)]εδ+∥ε∥N+1RN(1)[ut,ε]≡g6(0)+∑‍1≤|β|≤Ngβ[6](t)εβ+R6≡g6(0)+H6≡g6(0)+P6+R6,
where
(31)g6(0)=∥u0t(t)∥2,gδ[6](t)=Gδ[ut(t)]=∑‍|α|≤N∑‍|β|≤N∑‍α+β=δ〈uαt(t),uβt(t)〉,1≤|δ|≤N,R6=∥ε∥N+1RN(1)[ut,ε]=∑‍N+1≤|δ|≤2NGδ[ut(t)]εδ.Then(22), (24), (26), (28), and (30) imply that
(32)gi(ε)-gi(0)=Hi=Pi+Ri=∑‍1≤|β|≤Ngβ[i]εβ+Ri,Ri=0,i=1,2,3,Rj=O(∥ε∥N+1),j=4,5,6.We also need the following lemma.Lemma 4.
For allα=(α1,…,αn)∈ℤ+n,|α|≥1, then
(33)Hα=H1α1H2α2H3α3H4α4H5α5H6α6=∑|α|≤|δ|≤N‍Φδ[α,N,σ(1),…,σ(6)]εδ+O(∥ε∥N+1),
where
(34)Φδ[α,N,σ(1),…,σ(6)]=∑α1≤|γ1|≤α1N,⋮α6≤|γ6|≤α6N‍∑γ1+⋯+γ6=δ‍TN(α1)[σ(1)]γ1⋯TN(α6)[σ(6)]γ6,σ(i)=(σβ(i)),σβ(i)=gβ[i],1≤|β|≤N,i=1,2,…,6.Proof of Lemma4.
We have(35)Hα=H1α1H2α2H3α3H4α4H5α5H6α6=P1α1P2α2P3α3(P4+R4)α4(P5+R5)α5(P6+R6)α6=P1α1P2α2P3α3P4α4P5α5P6α6+P1α1P2α2P3α3×[(P4+R4)α4(P5+R5)α5(P6+R6)α6-P4α4P5α5P6α6].
Note thatRj=O(∥ε∥N+1),j=4,5,6, so
(36)(P4+R4)α4(P5+R5)α5(P6+R6)α6-P4α4P5α5P6α6=(P4+R4)α4(P5+R5)α5(P6+R6)α6-P4α4(P5+R5)α5(P6+R6)α6+P4α4(P5+R5)α5(P6+R6)α6-P4α4P5α5(P6+R6)α6+P4α4P5α5(P6+R6)α6-P4α4P5α5P6α6=[(P4+R4)α4-P4α4](P5+R5)α5(P6+R6)α6+P4α4[(P5+R5)α5-P5α5](P6+R6)α6+P4α4P5α5[(P6+R6)α6-P6α6]=R4[∑j=0α4-1‍(P4+R4)jP4α4-1-j]×(P5+R5)α5(P6+R6)α6+R5[∑j=0α5-1‍(P5+R5)jP5α5-1-j]P4α4(P6+R6)α6+R6[∑j=0α6-1‍(P6+R6)jP6α6-1-j]P4α4P5α5=∥ε∥N+1RN(1)[H,α,ε]=O(∥ε∥N+1).
Combining (35) and (36) yields
(37)Hα=H1α1H2α2H3α3H4α4H5α5H6α6=P1α1P2α2P3α3(P4+R4)α4(P5+R5)α5(P6+R6)α6=P1α1P2α2P3α3P4α4P5α5P6α6+∥ε∥N+1RN(1)[H,α,ε]=(∑‍1≤|β|≤Ngβ(1)εβ)α1⋯(∑‍1≤|β|≤Ngβ(6)εβ)α6+∥ε∥N+1RN(1)[H,α,ε]=∑|α|≤|δ|≤N‍Φδ[α,N,σ(1),…,σ(6)]εδ+∑N+1≤|δ|≤|α|N‍Φδ[α,N,σ(1),…,σ(6)]εδ+∥ε∥N+1RN(1)[H,α,ε]≡∑|α|≤|δ|≤N‍Φδ[α,N,σ(1),…,σ(6)]εδ+∥ε∥N+1RN(2)[H,α,ε],
where
(38)∥ε∥N+1RN(2)[H,α,ε]=∑N+1≤|δ|≤|α|N‍Φδ[α,N,σ(1),…,σ(6)]εδ+∥ε∥N+1RN(1)[H,α,ε].
Lemma4 is proved.Substituting (37) into (20), we obtain
(39)F(x,t,εα)=f(x,t,h,hx,ht,∥h∥2,∥hx∥2,∥ht∥2)=f(x,t,g+H)=f(x,t,g(0)+H)=f(x,t,g1(0)+H1,…,g6(0)+H6)=f(x,t,g(0))+∑‍1≤|α|≤N1α!Dαf(x,t,g(0))Hα+∥H∥N+1RN(1)[f,H]=f(x,t,g(0))+∑‍1≤|α|≤N1α!Dαf(x,t,g(0))×[∑|α|≤|δ|≤N‍Φδ[α,N,σ(1),…,σ(6)]εδ+∥ε∥N+1RN(2)[H,α,ε]∑|α|≤|δ|≤N‍]+∥H∥N+1RN(1)[f,H]=f(x,t,g(0))+∑‍1≤|α|≤N1α!Dαf(x,t,g(0))×[∑|α|≤|δ|≤N‍Φδ[α,N,σ(1),…,σ(6)]εδ]+∥ε∥N+1∑‍1≤|α|≤N1α!Dαf(x,t,g(0))RN(2)[H,α,ε]+∥H∥N+1RN(1)[f,H]=f(x,t,g(0))+∑‍1≤|α|≤N1α!Dαf(x,t,g(0))×∑|α|≤|δ|≤N‍Φδ[α,N,σ(1),…,σ(6)]εδ+∥ε∥N+1RN(3)[f,g,H,ε]=f(x,t,g(0))+∑1≤|δ|≤N‍∑‍|α|≤|δ|1α!Dαf(x,t,g(0))Φδ×[α,N,σ(1),…,σ(6)]εδ+∥ε∥N+1RN(3)[f,g,H,ε→]=f(x,t,g(0))+∑1≤|δ|≤N‍Fδ(x,t)εδ+∥ε∥N+1RN(3)[f,g,H,ε],
where
(40)∥ε∥N+1RN(3)[f,g,H,ε]=∥ε∥N+1∑‍1≤|α|≤N1α!Dαf(x,t,g(0))RN(2)[H,α,ε]+∥H∥N+1RN(1)[f,H],Fδ(x,t)=∑‍|α|≤|δ|1α!Dαf(x,t,g(0))Φδ[α,N,σ(1),…,σ(6)],δ∈ℤ+6,1≤|δ|≤N.Sinceuα∈W={v∈L∞(0,T,H2(0,1)):vt∈L∞(0,T,H1(0,1))},α∈ℤ+n,|α|≤N, hence uα,uαx,uαt∈L∞(0,T,H1(0,1))⊂L∞(QT). Then from (39) and (40)1, we get
(41)∥F(ε)-∑‍|δ|≤NFδεδ∥L∞(0,T,L2(0,1))=∥ε∥N+1∥RN(3)[f,g,H,ε]∥L∞(0,T,L2(0,1))≤CN∥ε∥N+1,
for ∥ε∥ is small enough; consequently the Problem 1 is solved.Remark 5.
(i) This result improves the one in [1], for n=1.
(ii) In case ofF(x,t,ε)=μ(x,t,h,∥hx∥2), where μ∈CN+1([0,1]×ℝ+×ℝ×ℝ+;ℝ),n=p, this result is also obtained in [3].



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*Source: 536280-2013-09-25.xml*