# Rational Solutions of Riccati Differential Equation with Coefficients Rational

Authors: Nadhem Echi
Journal: ISRN Applied Mathematics
 (2011)

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## Abstract

This paper presents a simple and efficient method for determining the rational solution of Riccati differential equation with coefficients rational. In case the differential Galois group of the differential equation(El):y′′=ry, r∈ℂ(x) is reducible, we look for the rational solutions of Riccati differential
equation θ′+θ2=r, by reducing the number of checks to be made and by accelerating the search for the partial fraction decomposition of the solution reserved for the poles of θ
which are false poles of r. This partial fraction decomposition of solution can be used to code r. The examples demonstrate the effectiveness of the method.


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## Body

## 1. Introduction

The quadratic Riccati differential equation:(1.1)(ER):σ′=p2σ2+p1σ+p0,
where p0,p1, and p2 are in a differential field 𝕂,p2≠0. The quadratic Riccati differential equation is first converted to a reduced Riccati differential equation:(1.2)(Er):θ′+θ2=r,
where θ=-p2σ-(1/2)a, with a=(p2′/p2)+p1 and r=(1/4)a2-(1/2)a′-p2p0.Furthermore, we puty′/y=θ, reduced Riccati differential equation (1.2) is converted to a second-order linear ordinary differential equation(1.3)(El):y′′=ry.
If we have a particular solution nonzero of (El) then general solution is y=cu where c′=λ/u2, λ constant (see [1–5]).In the paper, we base ourselves mainly on the work of Kovacic [2] where the differential Galois group of the differential equation (El) is reducible and we take 𝕂=ℂ(X).The case where every solution of(El) is Liouvillian corresponds to the case where reduced Riccati differential equation (Er) has algebraic solution over 𝕂. The case where differential Galois group is reducible corresponds to the case where the Riccati differential equation (Er) has the rational solution (u′/u)u solution of (El). The solution u of (El) is rational fraction if and only if u′/u is rational fraction with simples poles, integers residues and negative degree.The fieldℂ(X)[u] is differential extension of ℂ(X) by exponential of an integral and if v′=1/u2 then (u,v) two solutions of (El) are linearly independent over field of constants ℂ. The ordinary extension ℂ(X)[u,v] is differential extension of ℂ(X)[u], by an integral. ℂ(X)[u,v] is Picard-Vessiot extension of ℂ(X)[u] for the differential equation (El) (see [2, 3, 6–8]). The existence of rational solution u′/u of Riccati differential equation (Er) and  research primitive of 1/u2give all solutions of (Er).This paper presents a simple and efficient method for determining the solution of Riccati differential equation with coefficients rational. In case the differential Galois group of the differential equation(El):y′′=ry, r∈ℂ(x) is reducible, we look for the rational solutions of Riccati differential equation θ′+θ2=r, by reducing the number of checks to be made and by accelerating the search for the partial fraction decomposition of the solution reserved for the poles of θ which are false poles of r. This partial fraction decomposition of solution can be used to code r. The examples demonstrate the effectiveness of the method.


## 2. Form of Rational Solution of Equation:(Er)

Letr∈ℂ(x),r≠0 be rational fraction and let θ∈ℂ(x) be the rational solution of Riccati differential equation (Er):θ′+θ2=r.


### 2.1. Study in the Polec of Multiplicity ν of θ

We put(2.1)θ=τ(x-c)ν,whereτ(c)≠0.
We have(2.2)r=θ′+θ2=(x-c)-2ν[(τ-ν2(x-c)ν-1)2-ν24(x-c)2ν-2+τ′(x-c)ν].
Thus (2.3)(τ-ν2(x-c)ν-1)2=(x-c)2νr+(x-c)ν[ν24(x-c)ν-2-τ′].Case 1 (ν≥2).
The function(x-c)2νr defines and is equal to τ(c)2 at c.
Thusc is pole of multiplicity 2ν of r where
(2.4)lim⁡x→c(x-c)2νr=(lim⁡x→c(x-c)νθ)2.Case 2 (ν=1).
We have(2.5)(τ-12)2=(x-c)2r+14-(x-c)τ′.
The function(x-c)2r defines and is equal to τ(c)(τ(c)-1) at c.Situation 1(limx→c(x-c)2r≠0,-1/4)
c is double pole of r and the residue τ(c) of θ at c have tow possibility values as follows:
(2.6)(τ(c)-12)2=lim⁡x→c(x-c)2r+14.
Thus, c is double pole of r and the residue of θ at simple pole c equals
(2.7)τ(c)=αc+12,
where
(2.8)αc2=lim⁡x→c(x-c)2r+14.Situation 2(limx→c(x-c)2r=-1/4)
c is double pole of r and the residue of θ at c is 1/2.Situation 3(limx→c(x-c)2r=0)
c is a simple pole of r or is not pole of r and the residue of θ at simple pole c equals 1.Proposition 2.1.
Letθ∈ℂ(x) such as θ′+θ2=r.
(1)
The fraction:r=N/D with N and D being polynomials relatively prime
(2.9)D=D1D22D32D42,
where D1,D2,D3, and D4 are polynomials relatively prime pairwise. D1,D2, and D3 are with simple roots, and D4 is without simple root
(2.10)∀c∈Root(D2)lim⁡x→c⁡(x-c)2r=-14,∀c∈Root(D3)lim⁡x→c⁡(x-c)2r≠-14.(2)
(a)
Letν≥2. c pole of multiplicity ν of θ⇔c∈Root(D4).(b)
c simple pole of θ with residue ≠1,1/2⇔c∈Root(D3), the residue of θ where c equals αc+(1/2) where
(2.11)αc2=lim⁡x→c(x-c)2r+14.(c)
c simple pole of θ with residue =1/2⇔c∈Roots(D2).(d)
c simple pole of θ with residue =1⇔c∈Roots(D1) or c pole of θ and false pole of r.Corollary 2.2.
One assumes thatr=N/D with N and D polynomials being relatively prime, D=D1D22D32D42 where D1,D2,D3, and D4 are polynomials relatively prime pairwise, D1,D2, and D3 are with simple roots, and D4 is without simple root
(2.12)∀c∈Root(D2)lim⁡x→c⁡(x-c)2r=-14,∀c∈Root(D3)lim⁡x→c⁡(x-c)2r≠-14.
a rational fraction θ. Verify θ′+θ2=r is the shape
(2.13)θ=E(θ)+∑c∈Roots(D4)θc+∑c∈Roots(D3)αc+1/2x-c+12D2′D2+D1′D1+D0′D0
with D0 being monic polynomial with simple roots and the roots are false poles of r.


### 2.2. Study in the Infinity

Lett∈ℂ(x),t≠0 rational fraction such as (dt/dx)+t2=0.Case 1.
We assume thatdo(θ)<0.
We havedo(θ′)<0, and do(θ2)<0 thus do(r)<0. We put
(2.14)θ=tσ(t)withσbeingrationalfractiondefinedat0.
We have
(2.15)σ(0)=lim⁡x→∞xθ=sumtheresiduesofθ,r=θ′+θ2=(σ2-σ)t2-σ′t3,lim⁡x→∞x2r=σ(0)2-σ(0).
Thus dor≤-2 and (σ(0)-(1/2))2=limx→∞x2r+(1/4).
If:limx→∞x2r=-1/4 then the sum of residues of θ: σ(0)=1/2.
If:limx→∞x2r≠-1/4 then the sum of residues of θ: σ(0)=α∞+(1/2) where
(2.16)α∞2=lim⁡x→∞x2r+14.Case 2.
We assume thatdo(θ)=0.
E(θ) constant ≠0. We put
(2.17)θ=E(θ)+tσ(t),withσbeingrationalfractiondefinedat0.
We have
(2.18)σ(0)=lim⁡x→∞x(θ-E(θ))=thesumofresiduesofθ,r=θ′+θ2=(E(θ))2+2E(θ)tσ+(σ2-σ)t2-σ′t3.
So E(r) constant equals E(θ)2:
(2.19)2E(θ)[sumtheresiduesofθ]=sumtheresiduesofr.Case 3.
We assume thatdo(θ)>0.
We put(2.20)ν=do(θ)≥1,θ=t-νσ(t)
with σ being rational fraction defined at 0. The scalar σ(0) is the dominant coefficient of E(θ). We have
(2.21)r=θ′+θ2=t-2ν[σ2+νtν+1σ-tν+2σ′].
Thus,
(2.22)t2νr=σ(0)2+o(t).
So dor=2ν=2doθ and σ2(o) the dominant coefficient of E(r).Proposition 2.3.
Letθ∈ℂ(x) such as θ′+θ2=r.
(1)
do(θ)<0⇔do(r)<0. Thus do(r)≤-2 and
(2.23)thesumofresiduesofθ={12iflim⁡x→∞x2r=-14,α∞+12iflim⁡x→∞x2r≠-14.(2)
do(θ)=0⇔do(r)=0. In the caseE(θ) is square root of E(r):
(2.24)2E(θ)(sumofresiduesofθ)=sumofresiduesofr=lim⁡x→∞x(r-E(r)).(3)
We havedo(θ)>0⇔do(r)>0. In the case
(a)
do(r)=2do(θ);(b)
the dominant coefficient ofE(θ) is square root of E(r).


### 2.3. Determination ofE(θ); do(r)=2ν>0

We assume thatr is a rational fraction of degree 2ν>0 and θ∈ℂ(x) such as θ′+θ2=r.Leta be the dominant coefficient of E(r). Thus r~ax2ν if x tends to ∞: (2.25)t2νE(r)a=1+a1t+⋯+a2νt2ν.
The Taylor's expansion of order ν+1 at 0 (2.26)(t2νE(r)a)1/2=1+s1t+⋯+sν+1tν+1+o(tν+1).
We have (2.27)t2νr=t2νE(r)+o(t2ν)=t2νE(r)+o(tν+1)=a[(t2νE(r)a)1/2]2+o(tν+1).
We have θ=t-νσ(t) with σ rational fraction defined at 0, σ(0)2=a and (2.28)(σ+ν2tν+1)2=t2νr+ν24t2ν+2+tν+2σ′=a[1+s1t+⋯+sν+1tν+1]2+o(tν+1)σ+ν2tν+1=σ(0)[1+s1t+⋯+sν+1tν+1]+o(tν+1).
Thus(2.29)θ=t-νσ=σ(0)[t-ν+s1t-(ν-1)+⋯+sν+1t]-ν2t+o(t).
Imply(2.30)E(θ)=σ(0)[t-ν+s1t-(ν-1)+⋯+sν],σ(0)sν+1-ν2=sumofresiduesofθ.Proposition 2.4.
Letr be a rational fraction of degree 2ν>0, θ∈ℂ(x) such as θ′+θ2=r and a the dominant coefficient of E(r). If
(2.31)(t2νE(r)a)1/2=1+s1t+⋯+sν+1tν+1+o(tν+1).
Then
(2.32)E(θ)=α[t-ν+s1t-(ν-1)+⋯+sν],αsν+1-ν2=sumofresiduesofθ,
where
(2.33)α2=a.


## 2.1. Study in the Polec of Multiplicity ν of θ

We put(2.1)θ=τ(x-c)ν,whereτ(c)≠0.
We have(2.2)r=θ′+θ2=(x-c)-2ν[(τ-ν2(x-c)ν-1)2-ν24(x-c)2ν-2+τ′(x-c)ν].
Thus (2.3)(τ-ν2(x-c)ν-1)2=(x-c)2νr+(x-c)ν[ν24(x-c)ν-2-τ′].Case 1 (ν≥2).
The function(x-c)2νr defines and is equal to τ(c)2 at c.
Thusc is pole of multiplicity 2ν of r where
(2.4)lim⁡x→c(x-c)2νr=(lim⁡x→c(x-c)νθ)2.Case 2 (ν=1).
We have(2.5)(τ-12)2=(x-c)2r+14-(x-c)τ′.
The function(x-c)2r defines and is equal to τ(c)(τ(c)-1) at c.Situation 1(limx→c(x-c)2r≠0,-1/4)
c is double pole of r and the residue τ(c) of θ at c have tow possibility values as follows:
(2.6)(τ(c)-12)2=lim⁡x→c(x-c)2r+14.
Thus, c is double pole of r and the residue of θ at simple pole c equals
(2.7)τ(c)=αc+12,
where
(2.8)αc2=lim⁡x→c(x-c)2r+14.Situation 2(limx→c(x-c)2r=-1/4)
c is double pole of r and the residue of θ at c is 1/2.Situation 3(limx→c(x-c)2r=0)
c is a simple pole of r or is not pole of r and the residue of θ at simple pole c equals 1.Proposition 2.1.
Letθ∈ℂ(x) such as θ′+θ2=r.
(1)
The fraction:r=N/D with N and D being polynomials relatively prime
(2.9)D=D1D22D32D42,
where D1,D2,D3, and D4 are polynomials relatively prime pairwise. D1,D2, and D3 are with simple roots, and D4 is without simple root
(2.10)∀c∈Root(D2)lim⁡x→c⁡(x-c)2r=-14,∀c∈Root(D3)lim⁡x→c⁡(x-c)2r≠-14.(2)
(a)
Letν≥2. c pole of multiplicity ν of θ⇔c∈Root(D4).(b)
c simple pole of θ with residue ≠1,1/2⇔c∈Root(D3), the residue of θ where c equals αc+(1/2) where
(2.11)αc2=lim⁡x→c(x-c)2r+14.(c)
c simple pole of θ with residue =1/2⇔c∈Roots(D2).(d)
c simple pole of θ with residue =1⇔c∈Roots(D1) or c pole of θ and false pole of r.Corollary 2.2.
One assumes thatr=N/D with N and D polynomials being relatively prime, D=D1D22D32D42 where D1,D2,D3, and D4 are polynomials relatively prime pairwise, D1,D2, and D3 are with simple roots, and D4 is without simple root
(2.12)∀c∈Root(D2)lim⁡x→c⁡(x-c)2r=-14,∀c∈Root(D3)lim⁡x→c⁡(x-c)2r≠-14.
a rational fraction θ. Verify θ′+θ2=r is the shape
(2.13)θ=E(θ)+∑c∈Roots(D4)θc+∑c∈Roots(D3)αc+1/2x-c+12D2′D2+D1′D1+D0′D0
with D0 being monic polynomial with simple roots and the roots are false poles of r.


## 2.2. Study in the Infinity

Lett∈ℂ(x),t≠0 rational fraction such as (dt/dx)+t2=0.Case 1.
We assume thatdo(θ)<0.
We havedo(θ′)<0, and do(θ2)<0 thus do(r)<0. We put
(2.14)θ=tσ(t)withσbeingrationalfractiondefinedat0.
We have
(2.15)σ(0)=lim⁡x→∞xθ=sumtheresiduesofθ,r=θ′+θ2=(σ2-σ)t2-σ′t3,lim⁡x→∞x2r=σ(0)2-σ(0).
Thus dor≤-2 and (σ(0)-(1/2))2=limx→∞x2r+(1/4).
If:limx→∞x2r=-1/4 then the sum of residues of θ: σ(0)=1/2.
If:limx→∞x2r≠-1/4 then the sum of residues of θ: σ(0)=α∞+(1/2) where
(2.16)α∞2=lim⁡x→∞x2r+14.Case 2.
We assume thatdo(θ)=0.
E(θ) constant ≠0. We put
(2.17)θ=E(θ)+tσ(t),withσbeingrationalfractiondefinedat0.
We have
(2.18)σ(0)=lim⁡x→∞x(θ-E(θ))=thesumofresiduesofθ,r=θ′+θ2=(E(θ))2+2E(θ)tσ+(σ2-σ)t2-σ′t3.
So E(r) constant equals E(θ)2:
(2.19)2E(θ)[sumtheresiduesofθ]=sumtheresiduesofr.Case 3.
We assume thatdo(θ)>0.
We put(2.20)ν=do(θ)≥1,θ=t-νσ(t)
with σ being rational fraction defined at 0. The scalar σ(0) is the dominant coefficient of E(θ). We have
(2.21)r=θ′+θ2=t-2ν[σ2+νtν+1σ-tν+2σ′].
Thus,
(2.22)t2νr=σ(0)2+o(t).
So dor=2ν=2doθ and σ2(o) the dominant coefficient of E(r).Proposition 2.3.
Letθ∈ℂ(x) such as θ′+θ2=r.
(1)
do(θ)<0⇔do(r)<0. Thus do(r)≤-2 and
(2.23)thesumofresiduesofθ={12iflim⁡x→∞x2r=-14,α∞+12iflim⁡x→∞x2r≠-14.(2)
do(θ)=0⇔do(r)=0. In the caseE(θ) is square root of E(r):
(2.24)2E(θ)(sumofresiduesofθ)=sumofresiduesofr=lim⁡x→∞x(r-E(r)).(3)
We havedo(θ)>0⇔do(r)>0. In the case
(a)
do(r)=2do(θ);(b)
the dominant coefficient ofE(θ) is square root of E(r).


## 2.3. Determination ofE(θ); do(r)=2ν>0

We assume thatr is a rational fraction of degree 2ν>0 and θ∈ℂ(x) such as θ′+θ2=r.Leta be the dominant coefficient of E(r). Thus r~ax2ν if x tends to ∞: (2.25)t2νE(r)a=1+a1t+⋯+a2νt2ν.
The Taylor's expansion of order ν+1 at 0 (2.26)(t2νE(r)a)1/2=1+s1t+⋯+sν+1tν+1+o(tν+1).
We have (2.27)t2νr=t2νE(r)+o(t2ν)=t2νE(r)+o(tν+1)=a[(t2νE(r)a)1/2]2+o(tν+1).
We have θ=t-νσ(t) with σ rational fraction defined at 0, σ(0)2=a and (2.28)(σ+ν2tν+1)2=t2νr+ν24t2ν+2+tν+2σ′=a[1+s1t+⋯+sν+1tν+1]2+o(tν+1)σ+ν2tν+1=σ(0)[1+s1t+⋯+sν+1tν+1]+o(tν+1).
Thus(2.29)θ=t-νσ=σ(0)[t-ν+s1t-(ν-1)+⋯+sν+1t]-ν2t+o(t).
Imply(2.30)E(θ)=σ(0)[t-ν+s1t-(ν-1)+⋯+sν],σ(0)sν+1-ν2=sumofresiduesofθ.Proposition 2.4.
Letr be a rational fraction of degree 2ν>0, θ∈ℂ(x) such as θ′+θ2=r and a the dominant coefficient of E(r). If
(2.31)(t2νE(r)a)1/2=1+s1t+⋯+sν+1tν+1+o(tν+1).
Then
(2.32)E(θ)=α[t-ν+s1t-(ν-1)+⋯+sν],αsν+1-ν2=sumofresiduesofθ,
where
(2.33)α2=a.


## 3. Determination of Partial Fraction Decomposition

Letr=N/D be rational fraction with N and D polynomials being relatively prime, D=D1D22D32D42 where D1,D2,D3, and D4 polynomials being relatively prime pair-wise, D1,D2, and D3 with simple roots and D4 without simple root(3.1)∀c∈Root(D2)lim⁡x→c⁡(x-c)2r=-14,∀c∈Root(D3)lim⁡x→c⁡(x-c)2r≠-14.
Let θ∈ℂ(x) be rational fraction. Verify θ′+θ2=r.


### 3.1. CasedoD3=0 and doD4=0

We haver=N/D1D22.This case corresponds to the fact that one polec of r is simple or double with(3.2)lim⁡x→c(x-c)2r=-14.Proposition 3.1.
One assumesdor<0 and doD3=doD4=0. One has:
(3.3)lim⁡x→∞x2r+14=(q2)2
with q being positive integer of parity against that of doD2(3.4)θ=12D2′D2+D1′D1+D0′D0
with D0 polynomial of degree
(3.5)doD0=12(q+1-doD2)-doD1.Proof.
We have(3.6)θ=12D2′D2+D1′D1+D0′D0.
The sum of residues of θ equal (1/2)doD2+doD1+doD0=α∞+(1/2) with α∞2=limx→∞x2r+(1/4). In particular α∞=q/2 with q∈ℕ.
Remark 3.2.
Iflimx→∞x2r=-1/4 then doD2=1, doD1=doD0=0, θ=(1/2)(1/(x-c)), and r=-1/4(x-c)2.Proposition 3.3.
One assumesdor=0 and doD3=doD4=0.
(1)
E(θ) square root of E(r) such as p=(1/E(θ))limx→∞x[r-E(r)] is positive integer of same parity as doD2.(2)
One has(3.7)θ=E(θ)+12D2′D2+D1′D1+D0′D0
with D0 being polynomial of degree
(3.8)doD0=12p-doD1-12doD2.Proof.
We have(3.9)θ=E(θ)+12D2′D2+D1′D1+D0′D0.
We have
(3.10)2E(θ)sumofresidues(θ)=sumofresidues(r)=lim⁡x→∞x[r-E(r)],2E(θ)[12doD2+doD1+doD0]=lim⁡x→∞x[r-E(r)].
Thus
(3.11)doD2+2doD1+2doD0=1E(θ)lim⁡x→∞x[r-E(r)].
If r is constant then D1=D2=D0=1 and θ are constant.
Ifr is nonconstant then r is not polynomial thus
(3.12)p=1E(θ)lim⁡x→∞x[r-E(r)]
is a positive integer of same parity as doD2.Proposition 3.4.
We assumedor=2ν>0 and doD3=doD4=0.
Leta be the dominant coefficient of E(r) and one considers the Taylor's expansion at infinity:
(3.13)(t2νE(r)a)1/2=1+s1t+⋯+sν+1tν+1+o(tν+1).(1)
(3.14)4asν+12=p2,
where p is a positive integer, p≥doD2+2doD1+ν, and same parity of doD2+2doD1+ν.(2)
One has(3.15)θ=E(θ)+12D2′D2+D1′D1+D0′D0
with D0 being polynomial of degree
(3.16)doD0=p-ν-doD22-doD1
and α the dominant coefficient of E(θ) where
(3.17)p=2αsν+1.Proof.
We have(3.18)E(θ)=α[t-ν+s1t-(ν-1)+⋯+sν],αsν+1-ν2=sumofresiduesofθ,θ=E(θ)+12D2′D2+D1′D1+D0′D0.
Thus
(3.19)12doD2+doD1+doD0=αsν+1-ν2αsν+1=p/2 with p positive integer. Thus p≥doD2+2doD1+ν, p and doD2+2doD1+ν have the same parity.


### 3.2. Case:D3=X-c and doD4=0

This case corresponds to the fact that a pole ofr is simple or double with only double pole c such as:(3.20)lim⁡x→c(x-c)2r≠-14,(3.21)r=ND1D22(x-c)2.Proposition 3.5.
Consider (3.21) and let θ∈ℂ(x) be a rational fraction. Verify θ′+θ2=r. One assumes dor<0. Accordingly, in view of (2.11) and (2.16) one has
(3.22)θ=αc+(1/2)x-c+12D2′D2+D1′D1+D0′D0
with D0 being polynomial of degree
(3.23)doD0=λ-12doD2-doD1,
where
(3.24)λ=α∞-αc
one half positive integer of same parity as doD2.Proof.
We have(3.25)lim⁡x→∞xθ=αc+12+12doD2+doD1+doD0,(αc+12doD2+doD1+doD0)2=lim⁡x→∞x2r+14=α∞2.
Thus
(3.26)12doD2+doD1+doD0=α∞-αc.Proposition 3.6.
Consider (3.21) and let θ∈ℂ(x) be a rational fraction. Verify: θ′+θ2=r. One assumes dor=0. One has:
(3.27)θ=E(θ)+αc+(1/2)x-c+12D2′D2+D1′D1+D0′D0
with E2(θ)=E(r) and D0 being polynomial of degree
(3.28)doD0=12λ-12doD2-doD1-12,
where
(3.29)λ=1E(θ)lim⁡x→∞x[r-E(r)]-2αcλ is positive integer of parity against that of doD2.Proof.
We have(3.30)2E(θ)[αc+12+12doD2+doD1+doD0]=lim⁡x→∞x[r-E(r)].
Thus
(3.31)1+doD2+2doD1+2doD0=1E(θ)lim⁡x→∞x[r-E(r)]-2αc.Proposition 3.7.
Consider (3.21) and let θ∈ℂ(x) be a rational fraction. Verify θ′+θ2=r. One assumes dor=2ν>0. Let a be the dominant coefficient of E(r), let α be the dominant coefficient of E(θ), t=1/x and Taylors expansion:
(3.32)(t2νE(r)a)1/2=1+s1t+⋯+sν+1tν+1+o(tν+1).
One has
(3.33)θ=E(θ)+αc+1/2x-c+12D2′D2+D1′D1+D0′D0,
with D0 being polynomial of degree:
(3.34)doD0=αsν+1-αc-1+ν+doD22-doD1,α2=a.Proof.
We have(3.35)E(θ)=α[t-ν+s1t-(ν-1)+⋯+sν],αsν+1-ν2=sumofresiduesofθ.
Thus
(3.36)θ=E(θ)+αc+(1/2)x-c+12D2′D2+D1′D1+D0′D0αsν+1-ν2=αc+12+12doD2+doD1+doD0.
Thus
(3.37)doD0=αsν+1-αc-1+ν+doD22-doD1.


### 3.3. Case: (doD3≠0 and doD4≠0) or (doD3≥2  and   doD4=0)

D1D22 and D3D4 are polynomials being relatively prime. Thus there are only two polynomials N1 and N2 such as (3.38)D1D22D32D42(r-E(r))=N1(D1D22)+N2(D3D4),doN1<do(D3D4).
Thus(3.39)r=E(r)+N1D32D42+N2D1D22D3D4.
We have(3.40)do(N1D32D42)=do(N1D3D41D3D4)<do(1D3D4)≤-2.
Because D4 does not have simple roots verify doD4=0 or doD4≥2. Thus do(D3D4)≥2. Thus(3.41)lim⁡x→∞x2N1D32D42=0,do(r-E(r))<0,do(N2D1D22D3D4)<0.
We consider the rational fraction(3.42)F=E(r)+N1D32D42-δ,
where(3.43)δ=14[1k(D3′D3)2+(1k+1)(D3′D3)′],k=doD3≠0.Proposition 3.8.
One assumes{doD3≠0doD4≠0 or {doD3≥2doD4=0. (1)
Ifdor≥0 then: doF=dor and E(F)=E(r).(2)
Ifdor<0 then: doF=-2 where limx→∞x2F=1/4.(3)
For allc root of D3 one has
(3.44)lim⁡x→c(x-c)2F=lim⁡x→c(x-c)2r+14.(4)
For allc root of D4 of multiplicity ν one has
(3.45)(x-c)2νr=(x-c)2νF+o((x-c)ν-1).Proof.
(1)r-F=N2/D1D22D3D4+δ is from negative degree.
(2)limx→∞x2F=limx→∞(x2N1/D32D42)-limx→∞x2δ=-limx→∞x2δ=1/4.
(3) Letc be root of D3(3.46)lim⁡x→∞(x-c)2(r-F)=lim⁡x→cN2D1D22D4(x-c)2D3+lim⁡x→c(x-c)2δ=lim⁡x→c(x-c)2δ=-14.
(4) Letc be root of D4(3.47)(x-c)2ν(r-F)=N2D1D22D3(x-c)2νD4+(x-c)2νδ=o((x-c)ν-1).Lemma 3.9.
LetZ be nonzero rational fraction. Σ is a finished set such as
(3.48)Σ∩[Roots(Z)∪poles(Z)]=∅.(1)
There existsOopen connected set on which one has a square root holomorphic of Z, containing for every c∈Σ a half-right closed by origin c.(2)
If, besides,Z is from even degree then one can choose O of complementary compact.Proof.
We fixc0 is not an element of Σ. Let Σ′ be a finished set containing roots, poles of Z and c0. They consider all rights linked to the different pairs of points of Σ∪Σ′. They choose a point w not being on these rights. For all c∈Σ∪Σ′, straight line which joins right w and c does not contain any other point of Σ∪Σ′.
Case 1 (doZ even).
ReplacingZ by rational fraction Z/(x-c0)doZ.
AssumedoZ=0. We put K=⋃c∈Σ′[w,c]. K is compact-connected.
We putO=ℂ∖K. O is open at infinity such as for all c∈Σ reaching right w in c private of w is contained in O.
Ifγ is a shoelace of Othen K is included in one connected component of ℂ∖γ. Thus:(3.49)12iπ∫γZ′Z(x)dx=±∑c∈Σ′residue(Z′Z,c)=±doZ=0.
Thus there exist the primitive of Z′/Z and the determination of logarithm of Z and the square root of Z in O.
Case 2 (doZ odd).
They use the previous case in replacingZ by rational fraction Z/(x-c0).Notation 3.3.
We choose a square root of the polynomial of even degreeN1+D32D42[E(r)-δ] on a connected open at infinity, roots of D3 and root of D4. We put
(3.50)(F)1/2=1D3D4(N1+D32D42[E(r)-δ])1/2.Proposition 3.10.
One assumes{doD3≠0doD4≠0 or {doD3≥2doD4=0.
One has(3.51)θ=E(θ)+∑c∈Root(D3)∪Root(D4)εc(partialfractionof(F)1/2atc)+12(D2D3D4)′D2D3D4+D1′D1+D0′D0,
where εc=±1.Proof.
Forc root of multiplicity ν≥2 of D4(3.52)((x-c)νθ-ν2(x-c)ν-1)2=((x-c)ν(F)1/2)2+o((x-c)ν-1).
Thus
(3.53)(x-c)νθ-ν2(x-c)ν-1=ɛc(x-c)ν(F)1/2+o((x-c)ν-1),θ-ν2(x-c)=ɛc(F)1/2+o(1x-c).
Partial fraction of θ, associated in root c of (D4), minus ν/2(x-c), is in sign meadows that of (F)1/2.
Forc root of D3(3.54)[(residueofθatc)-12]2=lim⁡x→c((x-c)(F)1/2)2.
Thus
(3.55)[(residueofθatc)-12]=εc(residueof(F)1/2atc).
Partial fraction of θ, associated in root c of (D3), minus 1/2(x-c), is in sign meadows that of (F)1/2. Thus
(3.56)θ=E(θ)+∑c∈Root(D3)∪Root(D4)εc(partialfractionof(F)1/2atc)+12(D2D3D4)′D2D3D4+D1′D1+D0′D0.Proposition 3.11.
One assumesdor<0 and {doD3≠0doD4≠0 or {doD3≥2doD4=0.
One has(3.57)θ=∑c∈Root(D3)∪Root(D4)εc(partialfractionof(F)1/2atc)+12(D2D3D4)′D2D3D4+D1′D1+D0′D0,
where D0 being a polynomial of degree
(3.58)[∑c∈Roots(D3)∪Roots(D4)ɛc(residueof(F)1/2atc)+12do(D2D3D4)+doD1+doD0-12]2=lim⁡x→∞x2r+14.Proof.
We have(3.59)lim⁡x→∞⁡xθ=∑c∈Roots(D3)∪Roots(D4)εc(residueof(F)1/2atc)+12do(D2D3D4)+doD1+doD0,lim⁡x→∞x2r+14=(lim⁡x→∞xθ-12)2.
Thus
(3.60)[∑c∈Roots(D3)∪Zéros(D4)ɛc(residueof(F)1/2atc)+12do(D2D3D4)+doD1+doD0-12]2=lim⁡x→∞x2r+14.Proposition 3.12.
One assumesdor=0. One has
(3.61)θ=E(θ)+∑c∈Root(D3)∪Root(D4)ɛc(partialfractionof(F)1/2atc)+12(D2D3D4)′D2D3D4+D1′D1+D0′D0,
where E2(θ)=E(r) and D0 being a polynomial of degree:
(3.62)doD0=12E(θ)lim⁡x→∞x[r-E(r)]-[∑c∈R(D3)∪R(D4)ɛc(residueof(F)1/2atc)+12do(D2D3D4)+doD1].Proof.
We have(3.63)E2(θ)=E(r),2E(θ)∑cresidue ofθatc=∑cresidueofratc.
Thus
(3.64)2E(θ)[∑c∈R(D3)∪R(D4)ɛc(residueof(F)1/2atc)+12do(D2D3D4)+doD1+doD0]=lim⁡x→∞x(r-E(r)).Proposition 3.13.
One assumesdor=2ν>0. One has
(3.65)θ=E(θ)+∑c∈Root(D3)∪Root(D4)ɛc(partialfractionof(F)1/2atc)+12(D2D3D4)′D2D3D4+D1′D1+D0′D0,
where D0 being a polynomial of degree:
(3.66)doD0=αsν+1-ν2-[∑c∈R(D3)∪R(D4)ɛc(residueof(F)1/2atc)+12do(D2D3D4)+doD1].Proof.
Leta be dominant coefficient of E(r), and we have αsν+1-(ν/2)= sum of residues of θ,
(3.67)θ=E(θ)+∑c∈Zéros(D3)∪Zéros(D4)εc(partialfractionof(F)1/2atc)+12(D2D3D4)′D2D3D4+D1′D1+D0′D0.
Thus
(3.68)αsν+1-ν2=[∑c∈R(D3)∪R(D4)ɛc(residueof(F)1/2atc)+12do(D2D3D4)+doD1+doD0].
Thus
(3.69)doD0=αsν+1-ν2-[∑c∈R(D3)∪R(D4)ɛc(residueof(F)1/2atc)+12do(D2D3D4)+doD1].


### 3.4. CasedoD3=0 and doD4≠0

D1D22 and D4 are polynomials being relatively prime. Thus there are only two polynomials N1 and N2 such as (3.70)D1D22D32D42(r-E(r))=N1(D1D22)+N2D4,doN1<doD4.
We have(3.71)r=E(r)+N1D42+N2D1D22D4,lim⁡x→∞x2N1D42=0.
We consider the rational fraction:(3.72)F=E(r)+N1D42-14doD4(D4′D4)′.Proposition 3.14.
One assumes{doD3=0doD4≠0.(1)
Ifdor≥0 then: doF=dor and E(F)=E(r).(2)
Ifdor<0 then: doF=-2 where limx→∞x2F=1/4.(3)
For allc root of D4 of multiplicity ν one has
(3.73)(x-c)2νr=(x-c)2νF+o((x-c)ν-1).Proof.
(1)r-F=(N2/D1D22D4)+(1/4doD4)(D4′/D4)′ is from negative degree.
(2)(3.74)lim⁡x→∞x2F=lim⁡x→∞N1D42-lim⁡x→∞x214doD4(D4′D4)′=-lim⁡x→∞x214doD4(D4′D4)′=14.
(3) Letc be root of D4:
(3.75)(x-c)2ν(r-F)=N2D1D22(x-c)2νD4+(x-c)2ν14doD4(D4′D4)′=o((x-c)ν-1).Notation 3.4.
We choose a square root of the polynomial of even degree:N1+D42[E(r)-(1/4doD4)(D4′/D4)′] on a connected open at infinity, root of D4. We put
(3.76)(F)1/2=1D4(N1+D42[E(r)-14doD4(D4′D4)′])1/2.Proposition 3.15.
One assumesdor<0. One has
(3.77)θ=∑c∈Roots(D4)ɛc(partialfractionof(F)1/2atc)+12(D2D4)′D2D4+D1′D1+D0′D0,
where ɛc=±1 and D0 being a polynomial of degree:
(3.78)[∑c∈Root(D4)ɛc(residueof(F)1/2atc)+12do(D2D4)+doD1+doD0-12]2=lim⁡x→∞x2r+14.Proof.
(3.79)lim⁡x→∞xθ=∑c∈Roots(D4)ɛc(residueof(F)1/2atc)+12do(D2D4)+doD1+doD0,lim⁡x→∞⁡x2r+14=(lim⁡x→∞⁡xθ-12)2.
Thus
(3.80)[∑c∈Roots(D4)ɛc(residueof(F)1/2atc)+12do(D2D4)+doD1+doD0-12]2=lim⁡x→∞x2r+14.Proposition 3.16.
One assumesdor=0. One has
(3.81)θ=E(θ)+∑c∈Roots(D4)ɛc(partialfractionof(F)1/2atc)+12(D2D4)′D2D4+D1′D1+D0′D0,
where ɛc=±1 and D0 being a polynomial of degree
(3.82)doD0=12E(θ)lim⁡x→∞x[r-E(r)]-[∑c∈Roots(D4)ɛc(residueof(F)1/2atc)+12do(D2D4)+doD1].Proof.
We have(3.83)E2(θ)=E(r),2E(θ)∑cresidueofθatc=∑cresidueofratc.
Thus
(3.84)2E(θ)[∑c∈Roots(D4)ɛc(residueof(F)1/2atc)+12do(D2D4)+doD1+doD0]=lim⁡x→∞x(r-E(r)).Proposition 3.17.
One assumesdor=2ν>0. One has
(3.85)θ=E(θ)+∑c∈Roots(D4)ɛc(partialfractionof(F)1/2atc)+12(D2D4)′D2D4+D1′D1+D0′D0,
where ɛc=±1 and D0 being a polynomial of degree
(3.86)doD0=αsν+1-ν2-[∑c∈Roots(D4)ɛc(residueof(F)1/2atc)+12do(D2D4)+doD1].Proof.
Leta dominant coefficient of E(r). E(θ)=α[t-ν+s1t-(ν-1)+⋯+sν]. αsν+1-(ν/2)=sumresidueofθ, where α2=a. Thus
(3.87)θ=E(θ)+∑c∈Roots(D4)ɛc(partialfractionof(F)1/2atc)+12(D2D4)′D2D4+D1′D1+D0′D0.
Thus
(3.88)αsν+1-ν2=[∑c∈Roots(D4)ɛc(residueof(F)1/2atc)+12do(D2D4)+doD1+doD0].
Thus
(3.89)doD0=αsν+1-ν2-[∑c∈Roots(D4)ɛc(residueof(F)1/2atc)+12do(D2D4)+doD1].


## 3.1. CasedoD3=0 and doD4=0

We haver=N/D1D22.This case corresponds to the fact that one polec of r is simple or double with(3.2)lim⁡x→c(x-c)2r=-14.Proposition 3.1.
One assumesdor<0 and doD3=doD4=0. One has:
(3.3)lim⁡x→∞x2r+14=(q2)2
with q being positive integer of parity against that of doD2(3.4)θ=12D2′D2+D1′D1+D0′D0
with D0 polynomial of degree
(3.5)doD0=12(q+1-doD2)-doD1.Proof.
We have(3.6)θ=12D2′D2+D1′D1+D0′D0.
The sum of residues of θ equal (1/2)doD2+doD1+doD0=α∞+(1/2) with α∞2=limx→∞x2r+(1/4). In particular α∞=q/2 with q∈ℕ.
Remark 3.2.
Iflimx→∞x2r=-1/4 then doD2=1, doD1=doD0=0, θ=(1/2)(1/(x-c)), and r=-1/4(x-c)2.Proposition 3.3.
One assumesdor=0 and doD3=doD4=0.
(1)
E(θ) square root of E(r) such as p=(1/E(θ))limx→∞x[r-E(r)] is positive integer of same parity as doD2.(2)
One has(3.7)θ=E(θ)+12D2′D2+D1′D1+D0′D0
with D0 being polynomial of degree
(3.8)doD0=12p-doD1-12doD2.Proof.
We have(3.9)θ=E(θ)+12D2′D2+D1′D1+D0′D0.
We have
(3.10)2E(θ)sumofresidues(θ)=sumofresidues(r)=lim⁡x→∞x[r-E(r)],2E(θ)[12doD2+doD1+doD0]=lim⁡x→∞x[r-E(r)].
Thus
(3.11)doD2+2doD1+2doD0=1E(θ)lim⁡x→∞x[r-E(r)].
If r is constant then D1=D2=D0=1 and θ are constant.
Ifr is nonconstant then r is not polynomial thus
(3.12)p=1E(θ)lim⁡x→∞x[r-E(r)]
is a positive integer of same parity as doD2.Proposition 3.4.
We assumedor=2ν>0 and doD3=doD4=0.
Leta be the dominant coefficient of E(r) and one considers the Taylor's expansion at infinity:
(3.13)(t2νE(r)a)1/2=1+s1t+⋯+sν+1tν+1+o(tν+1).(1)
(3.14)4asν+12=p2,
where p is a positive integer, p≥doD2+2doD1+ν, and same parity of doD2+2doD1+ν.(2)
One has(3.15)θ=E(θ)+12D2′D2+D1′D1+D0′D0
with D0 being polynomial of degree
(3.16)doD0=p-ν-doD22-doD1
and α the dominant coefficient of E(θ) where
(3.17)p=2αsν+1.Proof.
We have(3.18)E(θ)=α[t-ν+s1t-(ν-1)+⋯+sν],αsν+1-ν2=sumofresiduesofθ,θ=E(θ)+12D2′D2+D1′D1+D0′D0.
Thus
(3.19)12doD2+doD1+doD0=αsν+1-ν2αsν+1=p/2 with p positive integer. Thus p≥doD2+2doD1+ν, p and doD2+2doD1+ν have the same parity.


## 3.2. Case:D3=X-c and doD4=0

This case corresponds to the fact that a pole ofr is simple or double with only double pole c such as:(3.20)lim⁡x→c(x-c)2r≠-14,(3.21)r=ND1D22(x-c)2.Proposition 3.5.
Consider (3.21) and let θ∈ℂ(x) be a rational fraction. Verify θ′+θ2=r. One assumes dor<0. Accordingly, in view of (2.11) and (2.16) one has
(3.22)θ=αc+(1/2)x-c+12D2′D2+D1′D1+D0′D0
with D0 being polynomial of degree
(3.23)doD0=λ-12doD2-doD1,
where
(3.24)λ=α∞-αc
one half positive integer of same parity as doD2.Proof.
We have(3.25)lim⁡x→∞xθ=αc+12+12doD2+doD1+doD0,(αc+12doD2+doD1+doD0)2=lim⁡x→∞x2r+14=α∞2.
Thus
(3.26)12doD2+doD1+doD0=α∞-αc.Proposition 3.6.
Consider (3.21) and let θ∈ℂ(x) be a rational fraction. Verify: θ′+θ2=r. One assumes dor=0. One has:
(3.27)θ=E(θ)+αc+(1/2)x-c+12D2′D2+D1′D1+D0′D0
with E2(θ)=E(r) and D0 being polynomial of degree
(3.28)doD0=12λ-12doD2-doD1-12,
where
(3.29)λ=1E(θ)lim⁡x→∞x[r-E(r)]-2αcλ is positive integer of parity against that of doD2.Proof.
We have(3.30)2E(θ)[αc+12+12doD2+doD1+doD0]=lim⁡x→∞x[r-E(r)].
Thus
(3.31)1+doD2+2doD1+2doD0=1E(θ)lim⁡x→∞x[r-E(r)]-2αc.Proposition 3.7.
Consider (3.21) and let θ∈ℂ(x) be a rational fraction. Verify θ′+θ2=r. One assumes dor=2ν>0. Let a be the dominant coefficient of E(r), let α be the dominant coefficient of E(θ), t=1/x and Taylors expansion:
(3.32)(t2νE(r)a)1/2=1+s1t+⋯+sν+1tν+1+o(tν+1).
One has
(3.33)θ=E(θ)+αc+1/2x-c+12D2′D2+D1′D1+D0′D0,
with D0 being polynomial of degree:
(3.34)doD0=αsν+1-αc-1+ν+doD22-doD1,α2=a.Proof.
We have(3.35)E(θ)=α[t-ν+s1t-(ν-1)+⋯+sν],αsν+1-ν2=sumofresiduesofθ.
Thus
(3.36)θ=E(θ)+αc+(1/2)x-c+12D2′D2+D1′D1+D0′D0αsν+1-ν2=αc+12+12doD2+doD1+doD0.
Thus
(3.37)doD0=αsν+1-αc-1+ν+doD22-doD1.


## 3.3. Case: (doD3≠0 and doD4≠0) or (doD3≥2  and   doD4=0)

D1D22 and D3D4 are polynomials being relatively prime. Thus there are only two polynomials N1 and N2 such as (3.38)D1D22D32D42(r-E(r))=N1(D1D22)+N2(D3D4),doN1<do(D3D4).
Thus(3.39)r=E(r)+N1D32D42+N2D1D22D3D4.
We have(3.40)do(N1D32D42)=do(N1D3D41D3D4)<do(1D3D4)≤-2.
Because D4 does not have simple roots verify doD4=0 or doD4≥2. Thus do(D3D4)≥2. Thus(3.41)lim⁡x→∞x2N1D32D42=0,do(r-E(r))<0,do(N2D1D22D3D4)<0.
We consider the rational fraction(3.42)F=E(r)+N1D32D42-δ,
where(3.43)δ=14[1k(D3′D3)2+(1k+1)(D3′D3)′],k=doD3≠0.Proposition 3.8.
One assumes{doD3≠0doD4≠0 or {doD3≥2doD4=0. (1)
Ifdor≥0 then: doF=dor and E(F)=E(r).(2)
Ifdor<0 then: doF=-2 where limx→∞x2F=1/4.(3)
For allc root of D3 one has
(3.44)lim⁡x→c(x-c)2F=lim⁡x→c(x-c)2r+14.(4)
For allc root of D4 of multiplicity ν one has
(3.45)(x-c)2νr=(x-c)2νF+o((x-c)ν-1).Proof.
(1)r-F=N2/D1D22D3D4+δ is from negative degree.
(2)limx→∞x2F=limx→∞(x2N1/D32D42)-limx→∞x2δ=-limx→∞x2δ=1/4.
(3) Letc be root of D3(3.46)lim⁡x→∞(x-c)2(r-F)=lim⁡x→cN2D1D22D4(x-c)2D3+lim⁡x→c(x-c)2δ=lim⁡x→c(x-c)2δ=-14.
(4) Letc be root of D4(3.47)(x-c)2ν(r-F)=N2D1D22D3(x-c)2νD4+(x-c)2νδ=o((x-c)ν-1).Lemma 3.9.
LetZ be nonzero rational fraction. Σ is a finished set such as
(3.48)Σ∩[Roots(Z)∪poles(Z)]=∅.(1)
There existsOopen connected set on which one has a square root holomorphic of Z, containing for every c∈Σ a half-right closed by origin c.(2)
If, besides,Z is from even degree then one can choose O of complementary compact.Proof.
We fixc0 is not an element of Σ. Let Σ′ be a finished set containing roots, poles of Z and c0. They consider all rights linked to the different pairs of points of Σ∪Σ′. They choose a point w not being on these rights. For all c∈Σ∪Σ′, straight line which joins right w and c does not contain any other point of Σ∪Σ′.
Case 1 (doZ even).
ReplacingZ by rational fraction Z/(x-c0)doZ.
AssumedoZ=0. We put K=⋃c∈Σ′[w,c]. K is compact-connected.
We putO=ℂ∖K. O is open at infinity such as for all c∈Σ reaching right w in c private of w is contained in O.
Ifγ is a shoelace of Othen K is included in one connected component of ℂ∖γ. Thus:(3.49)12iπ∫γZ′Z(x)dx=±∑c∈Σ′residue(Z′Z,c)=±doZ=0.
Thus there exist the primitive of Z′/Z and the determination of logarithm of Z and the square root of Z in O.
Case 2 (doZ odd).
They use the previous case in replacingZ by rational fraction Z/(x-c0).Notation 3.3.
We choose a square root of the polynomial of even degreeN1+D32D42[E(r)-δ] on a connected open at infinity, roots of D3 and root of D4. We put
(3.50)(F)1/2=1D3D4(N1+D32D42[E(r)-δ])1/2.Proposition 3.10.
One assumes{doD3≠0doD4≠0 or {doD3≥2doD4=0.
One has(3.51)θ=E(θ)+∑c∈Root(D3)∪Root(D4)εc(partialfractionof(F)1/2atc)+12(D2D3D4)′D2D3D4+D1′D1+D0′D0,
where εc=±1.Proof.
Forc root of multiplicity ν≥2 of D4(3.52)((x-c)νθ-ν2(x-c)ν-1)2=((x-c)ν(F)1/2)2+o((x-c)ν-1).
Thus
(3.53)(x-c)νθ-ν2(x-c)ν-1=ɛc(x-c)ν(F)1/2+o((x-c)ν-1),θ-ν2(x-c)=ɛc(F)1/2+o(1x-c).
Partial fraction of θ, associated in root c of (D4), minus ν/2(x-c), is in sign meadows that of (F)1/2.
Forc root of D3(3.54)[(residueofθatc)-12]2=lim⁡x→c((x-c)(F)1/2)2.
Thus
(3.55)[(residueofθatc)-12]=εc(residueof(F)1/2atc).
Partial fraction of θ, associated in root c of (D3), minus 1/2(x-c), is in sign meadows that of (F)1/2. Thus
(3.56)θ=E(θ)+∑c∈Root(D3)∪Root(D4)εc(partialfractionof(F)1/2atc)+12(D2D3D4)′D2D3D4+D1′D1+D0′D0.Proposition 3.11.
One assumesdor<0 and {doD3≠0doD4≠0 or {doD3≥2doD4=0.
One has(3.57)θ=∑c∈Root(D3)∪Root(D4)εc(partialfractionof(F)1/2atc)+12(D2D3D4)′D2D3D4+D1′D1+D0′D0,
where D0 being a polynomial of degree
(3.58)[∑c∈Roots(D3)∪Roots(D4)ɛc(residueof(F)1/2atc)+12do(D2D3D4)+doD1+doD0-12]2=lim⁡x→∞x2r+14.Proof.
We have(3.59)lim⁡x→∞⁡xθ=∑c∈Roots(D3)∪Roots(D4)εc(residueof(F)1/2atc)+12do(D2D3D4)+doD1+doD0,lim⁡x→∞x2r+14=(lim⁡x→∞xθ-12)2.
Thus
(3.60)[∑c∈Roots(D3)∪Zéros(D4)ɛc(residueof(F)1/2atc)+12do(D2D3D4)+doD1+doD0-12]2=lim⁡x→∞x2r+14.Proposition 3.12.
One assumesdor=0. One has
(3.61)θ=E(θ)+∑c∈Root(D3)∪Root(D4)ɛc(partialfractionof(F)1/2atc)+12(D2D3D4)′D2D3D4+D1′D1+D0′D0,
where E2(θ)=E(r) and D0 being a polynomial of degree:
(3.62)doD0=12E(θ)lim⁡x→∞x[r-E(r)]-[∑c∈R(D3)∪R(D4)ɛc(residueof(F)1/2atc)+12do(D2D3D4)+doD1].Proof.
We have(3.63)E2(θ)=E(r),2E(θ)∑cresidue ofθatc=∑cresidueofratc.
Thus
(3.64)2E(θ)[∑c∈R(D3)∪R(D4)ɛc(residueof(F)1/2atc)+12do(D2D3D4)+doD1+doD0]=lim⁡x→∞x(r-E(r)).Proposition 3.13.
One assumesdor=2ν>0. One has
(3.65)θ=E(θ)+∑c∈Root(D3)∪Root(D4)ɛc(partialfractionof(F)1/2atc)+12(D2D3D4)′D2D3D4+D1′D1+D0′D0,
where D0 being a polynomial of degree:
(3.66)doD0=αsν+1-ν2-[∑c∈R(D3)∪R(D4)ɛc(residueof(F)1/2atc)+12do(D2D3D4)+doD1].Proof.
Leta be dominant coefficient of E(r), and we have αsν+1-(ν/2)= sum of residues of θ,
(3.67)θ=E(θ)+∑c∈Zéros(D3)∪Zéros(D4)εc(partialfractionof(F)1/2atc)+12(D2D3D4)′D2D3D4+D1′D1+D0′D0.
Thus
(3.68)αsν+1-ν2=[∑c∈R(D3)∪R(D4)ɛc(residueof(F)1/2atc)+12do(D2D3D4)+doD1+doD0].
Thus
(3.69)doD0=αsν+1-ν2-[∑c∈R(D3)∪R(D4)ɛc(residueof(F)1/2atc)+12do(D2D3D4)+doD1].


## 3.4. CasedoD3=0 and doD4≠0

D1D22 and D4 are polynomials being relatively prime. Thus there are only two polynomials N1 and N2 such as (3.70)D1D22D32D42(r-E(r))=N1(D1D22)+N2D4,doN1<doD4.
We have(3.71)r=E(r)+N1D42+N2D1D22D4,lim⁡x→∞x2N1D42=0.
We consider the rational fraction:(3.72)F=E(r)+N1D42-14doD4(D4′D4)′.Proposition 3.14.
One assumes{doD3=0doD4≠0.(1)
Ifdor≥0 then: doF=dor and E(F)=E(r).(2)
Ifdor<0 then: doF=-2 where limx→∞x2F=1/4.(3)
For allc root of D4 of multiplicity ν one has
(3.73)(x-c)2νr=(x-c)2νF+o((x-c)ν-1).Proof.
(1)r-F=(N2/D1D22D4)+(1/4doD4)(D4′/D4)′ is from negative degree.
(2)(3.74)lim⁡x→∞x2F=lim⁡x→∞N1D42-lim⁡x→∞x214doD4(D4′D4)′=-lim⁡x→∞x214doD4(D4′D4)′=14.
(3) Letc be root of D4:
(3.75)(x-c)2ν(r-F)=N2D1D22(x-c)2νD4+(x-c)2ν14doD4(D4′D4)′=o((x-c)ν-1).Notation 3.4.
We choose a square root of the polynomial of even degree:N1+D42[E(r)-(1/4doD4)(D4′/D4)′] on a connected open at infinity, root of D4. We put
(3.76)(F)1/2=1D4(N1+D42[E(r)-14doD4(D4′D4)′])1/2.Proposition 3.15.
One assumesdor<0. One has
(3.77)θ=∑c∈Roots(D4)ɛc(partialfractionof(F)1/2atc)+12(D2D4)′D2D4+D1′D1+D0′D0,
where ɛc=±1 and D0 being a polynomial of degree:
(3.78)[∑c∈Root(D4)ɛc(residueof(F)1/2atc)+12do(D2D4)+doD1+doD0-12]2=lim⁡x→∞x2r+14.Proof.
(3.79)lim⁡x→∞xθ=∑c∈Roots(D4)ɛc(residueof(F)1/2atc)+12do(D2D4)+doD1+doD0,lim⁡x→∞⁡x2r+14=(lim⁡x→∞⁡xθ-12)2.
Thus
(3.80)[∑c∈Roots(D4)ɛc(residueof(F)1/2atc)+12do(D2D4)+doD1+doD0-12]2=lim⁡x→∞x2r+14.Proposition 3.16.
One assumesdor=0. One has
(3.81)θ=E(θ)+∑c∈Roots(D4)ɛc(partialfractionof(F)1/2atc)+12(D2D4)′D2D4+D1′D1+D0′D0,
where ɛc=±1 and D0 being a polynomial of degree
(3.82)doD0=12E(θ)lim⁡x→∞x[r-E(r)]-[∑c∈Roots(D4)ɛc(residueof(F)1/2atc)+12do(D2D4)+doD1].Proof.
We have(3.83)E2(θ)=E(r),2E(θ)∑cresidueofθatc=∑cresidueofratc.
Thus
(3.84)2E(θ)[∑c∈Roots(D4)ɛc(residueof(F)1/2atc)+12do(D2D4)+doD1+doD0]=lim⁡x→∞x(r-E(r)).Proposition 3.17.
One assumesdor=2ν>0. One has
(3.85)θ=E(θ)+∑c∈Roots(D4)ɛc(partialfractionof(F)1/2atc)+12(D2D4)′D2D4+D1′D1+D0′D0,
where ɛc=±1 and D0 being a polynomial of degree
(3.86)doD0=αsν+1-ν2-[∑c∈Roots(D4)ɛc(residueof(F)1/2atc)+12do(D2D4)+doD1].Proof.
Leta dominant coefficient of E(r). E(θ)=α[t-ν+s1t-(ν-1)+⋯+sν]. αsν+1-(ν/2)=sumresidueofθ, where α2=a. Thus
(3.87)θ=E(θ)+∑c∈Roots(D4)ɛc(partialfractionof(F)1/2atc)+12(D2D4)′D2D4+D1′D1+D0′D0.
Thus
(3.88)αsν+1-ν2=[∑c∈Roots(D4)ɛc(residueof(F)1/2atc)+12do(D2D4)+doD1+doD0].
Thus
(3.89)doD0=αsν+1-ν2-[∑c∈Roots(D4)ɛc(residueof(F)1/2atc)+12do(D2D4)+doD1].


## 4. Recurrent Method at Infinity

### 4.1. Presentation ofD0 as Determinant

Proposition 4.1.
Letc1,…,cm be complex constants. We put
(4.1)P(x)=(x-c1)⋯(x-cm)=xm-p1xm-1+p2xm-2-⋯+(-1)mpm.
For j∈ℕ, we put σj=c1j+⋯+cmj and
(4.2)Δ=|1c1⋅⋅c1m-11c2⋅⋅c2m-1⋅⋅⋅⋅1cm⋅⋅cmm-1|.(1)
One has(4.3)Δ2=|σ0σ1⋅⋅σm-1σ1σ2⋅⋅σm⋅⋅⋅⋅σm-1σm⋅⋅σ2m-2|.(2)
One has(4.4)Δ2P(x)=|σ0⋅⋅σm-11⋅⋅x⋅⋅⋅⋅⋅⋅σm⋅⋅σ2m-1xm|.Proof.
Δ,|σ0σ1··σm-1σ1σ2··σm····σm-1σm··σ2m-2| are polynomials at c1,…,cm with real coefficients. Thus, Δ2P(x) and |σ0··σm-11··x······σm··σ2m-1xm| are polynomials at c1,…,cm,x with real coefficients. Thus, to have both identities we can assume c1,…,cm,x are reals. Furthermore, we can restrict to the open of Zariski c1,…,cm,x distinct real nonzero.
We have for allj=1,…,m,cjm=p1cjm-1-p2cjm-2-⋯-(-1)mpm.
We put(4.5)Δ(x)=|1c1⋅⋅c1m-1c1m1c2⋅⋅c2m-1c2m⋅⋅⋅⋅⋅1cm⋅⋅cmm-1cmm1x⋅⋅xm-1xm|,Δ(x)=|1c1⋅⋅c1m-101c2⋅⋅c2m-10⋅⋅⋅⋅⋅⋅1cm⋅⋅cmm-101x⋅⋅xm-1P(x)|=ΔP(x).
We put(4.6)vj=(c1j-1..cmj-1),forj=1,…,m.
The scalar product
(4.7)〈vj,vk〉=c1j+k-2+⋯+cmj+k-2=σj+k-2.
The  Gram matrix of v1,…,vm is
(4.8)G=(σ0σ1⋅⋅σm-1σ1σ2⋅⋅σm⋅⋅⋅⋅σm-1σm⋅⋅σ2m-2)⋅
Thus
(4.9)Δ2=|σ0σ1..σm-1σ1σ2..σm....σm-1σm..σ2m-2|.
We put
(4.10)vj(x)=(c1j-1⋅⋅cmj-1xj-1),j=1,…,m+1.
We notice
(4.11)〈vj(x),vk(x)〉=〈vj,vk〉+xj+k-2=σj+k-2+xj+k-2.
The jth column of Gram of matrix which defines Δ(x) is
(4.12)(σj-1σj⋅⋅σj+m-1)+xj-1(1x⋅⋅xm).
Thus
(4.13)Δ2(x)=|σ0σ1⋅⋅σmσ1σ2⋅⋅σm+1⋅⋅⋅⋅σmσm+1⋅⋅σ2m|+|1σ1⋅⋅σmxσ2⋅⋅σm+1⋅⋅⋅⋅xmσm+1⋅⋅σ2m|+⋯+|σ0⋅σm-1xmσ1⋅σmxm+1⋅⋅⋅⋅σm⋅σ2m-1x2m|=∑i,jxi+j-2cofactori,j(M),
where
(4.14)M=(σ0σ1⋅⋅σmσ1σ2⋅⋅σm+1⋅⋅⋅⋅σmσm+1⋅⋅σ2m).
We obtain
(4.15)Δ2(x)=Δ2P2(x)=(1x⋯xm)Com(M)(1x⋅⋅xm)
The cofactor(m+1,m+1) of M is Δ2. The adjoint of M is nonzero. We prove that adjoint of M of rank 1.  det(M)=0. Because: σm=p1σm-1-p2σm-2-⋯-(-1)mpmσ0 and for all k≥m;σk=p1σk-1-p2σk-2-⋯-(-1)mpmσk-m. Thus, the (m+1)th column of M is is a linear combination of other column.
We consider the matrix(4.16)E1,1=(10⋅00⋅⋅⋅0⋅⋅0),(4.17)M+E1,1=(σ0+1σ1⋅⋅σmσ1σ2⋅⋅σm+1⋅⋅⋅⋅σmσm+1⋅⋅σ2m),det⁡(M+E1,1)=det⁡(M)+|1σ1⋅⋅σm0σ2⋅⋅σm+1⋅⋅⋅⋅0σm+1⋅⋅σ2m|=|σ2σ3⋅⋅σm+1σ3σ4⋅⋅σm+2⋅⋅⋅⋅σm+1σm+2⋅⋅σ2m|⋅
It is the determinant of Gram of
(4.18)(c1⋅⋅cm),…,(c1m⋅⋅cmm).
Thus
(4.19)det⁡(M+E1,1)=|c1⋅⋅⋅c1m⋅⋅⋅⋅⋅⋅⋅⋅cm⋅⋅⋅cmm|2⋅
But
(4.20)|c1⋅⋅⋅c1m⋅⋅⋅⋅⋅⋅⋅⋅cm⋅⋅⋅cmm|=(-1)m-1pm|c1⋅⋅⋅c1m-11⋅⋅⋅⋅1⋅⋅⋅⋅cm⋅⋅⋅cmm-11|=pmΔ.
Thus det(M+E1,1)=pm2Δ2. Or pm=c1⋯cm is scalar nonzero, M+E1,1 is invertible matrix.
WithM symmetric matrix, Com(M) symmetric matrix, we have
(4.21)MCom(M)=Com(M)M=det⁡(M)I=0,Ibeingmatriceidentity.
Thus
(4.22)(M+E1,1)Com(M)=E1,1Com(M).
Or M+E1,1 is invertible and Com(M)≠0, Com(M) of rank 1.
Thus(4.23)cofactori,j(M)=cofactori,m+1(M)cofactorm+1,m+1(M)cofactorm+1,j(M).
For i=1,…,m+1,j=1,…,m+1, we put: Ci,j=cofactori,j(M),
(4.24)Ci,jΔ2=Ci,m+1Δ2Cm+1,jΔ2.
Thus:
(4.25)P(x)2=∑1≤i,j≤m+1Ci,jΔ2xi+j-2=∑1≤i,j≤m+1Ci,m+1Δ2Cj,m+1Δ2xi+j-2=(∑1≤i≤m+1Ci,m+1Δ2xi-1)2.
Or P is monic polynomial and Cm+1,m+1/Δ2=1(4.26)P(x)=∑iCi,m+1Δ2xi-1=1Δ2|σ0⋅⋅⋅σm-11σ1⋅⋅⋅σmx⋅⋅⋅⋅⋅⋅σm⋅⋅⋅σ2m-1xm|.
We obtain
(4.27)Δ2P(x)=|σ0⋅⋅⋅σm-11σ1⋅⋅⋅σmx⋅⋅⋅⋅⋅⋅σm⋅⋅⋅σ2m-1xm|⋅


### 4.2. Taylor Expansion ofP′/P

Proposition 4.2.
Letc1,…,cm be pairwise distinct constants.
(1)
We put, forj=0,…,m: σj=c1j+⋯+cmj and P(x)=(x-c1)⋯(x-cm)(4.28)(P′P)′+(P′P)2=2∑i≠j1ci-cj1x-ci.(2)
We putt=1/x. Taylor expansion on the neighborhood of infinity:
(a)
(4.29)(P′P)′+(P′P)2=∑l=2∞[(∑μ=0l-2σμσl-2-μ)-(l-1)σl-2]tl,(b)
(4.30)xP′P=∑l=0∞σltl.Proof.
We haveP′/P=∑i=1m1/(x-ci).
(1)
(4.31)(P′P)′+(P′P)2=∑i≠j1x-ci1x-cj=∑i≠j[1/(ci-cj)x-ci+1/(cj-ci)x-cj]=2∑i≠j1ci-cj1x-ci.(2)
We putt=1/x. We have
(4.32)(P′P)′+(P′P)2=2t∑i≠j1ci-cj11-cit=2t∑i≠j1ci-cj∑k=0∞(cit)k=∑k=0∞(2∑i≠jcikci-cj)tk+1=∑k=1∞∑i≠jcik-cjkci-cjtk+1.
as
(4.33)∑i≠jcik-cjkci-cj=∑i≠j∑μ+ν=k-1,0≤μ≤k-1ciμcjν=∑μ+ν=k-1,0≤μ≤k-1[σμσν-∑i=1mciμ+ν]=[∑μ+ν=k-1,0≤μ≤k-1σμσν]-kσk-1.
Thus
(4.34)(P′P)′+(P′P)2=∑k=1∞[(∑μ=0k-1σμσk-1-μ)-kσk-1]tl=∑l=2∞[(∑μ=0l-2σμσl-2-μ)-(l-1)σl-2]tl.


### 4.3. Research ofD0

Letr=N/D1D22D32D42 be a rational fraction and θ∈ℂ(x) where θ′+θ2=r.We put(4.35)θ=S+D0′D0,
where S=E(θ)+∑cpolesofrθc and D0 monic polynomial of degree m. We have(4.36)θ′+θ2=S′+S2+(D0′D0)′+(D0′D0)2+2SD0′D0,θ′+θ2=r⟺(D0′D0)′+(D0′D0)2+2SD0′D0=RB
with B=D1D2D3D4 and R polynomials as r-S′-S2=R/B.Case 1 (dor=2ν≥0).
We havedoS=ν and doR/B=ν-1. Taylors expansion of (1/xν)S, (1/xν-1)(R/B) and x(D0′/D0) at order μ.
We obtain the Taylors expansion of(1/xν-1)[(R/B)-2S(D0′/D0)] at order μ.
The Taylors expansion equals the Taylors expansion of(1/xν-1)[(D0′/D0)′+(D0′/D0)2]. Let t=1/x. In Taylors expansion of (1/xν-1)[(P′/P)′+(P′/P)2] (see Proposition 4.2) the coefficient of tν+1 is: σ02-σ0=m2-m. For l≥2, coefficient of tl+ν-1 use: σ0,…,σl-2.
Besides, we have(4.37)xD0′D0=∑i=1m11-cit=∑i=1m∑k=0∞ciktk=∑k=0∞σktk.
Thus, Taylors expansion of
(4.38)Φ=1xν-1[(D0′D0)′+(D0′D0)2]-1xν-1[RB-2SD0′D0]
and the coefficient of tk is
(4.39)-2ασk+polynomialatσ0,…,σk-1,Φ=0 determine σk by recurrence at k=ν.Case 2 (dor<0).
The coefficient oftk in Taylors expansion of
(4.40)Ψ=x2[(D0′D0)′+(D0′D0)2]-x2RB+2xS(xD0′D0)
is
(4.41)2mσk-(k+1)σk+(2lim⁡x→∞xS)σk+polynomialatσ0,…,σk-1.
By recurrence σk at condition
(4.42)k≠2lim⁡x→∞xS+2m-1.
We have limx→∞xS+m=α∞2+(1/2) with α∞2=limx→∞x2r+(1/4).
Fork≠2α∞, to be σk at function of σ0,…,σk-1.
If2α∞∈ℕ then, the coefficient of t2α∞ is nonzero implied that there is no solution, or the coeffcient of t2α∞ is zero implied that σ2α∞ is arbitrary and σk; k>2α∞ depend in a unique way.
D0 is polynomial determined by σ0,…,σ2m-1 by determinant formulae; the problem of nonunique suite (σk)k∈ℕ is put if 2α∞ is positive integer equal to or less than 2m-1, then 4limx→∞x2r+1 is square of integer and
(4.43)lim⁡x→∞x2r≤m2-m.Example 4.3.
In this example we consider the Riccati differential equation (1.2) where
(4.44)r=-1+z2-(1/4)x2;z∈R*.
We have D1=1, D2=1, D3=x and E2(θ)=-1.
We can assumeE(θ)=i,
(4.45)lim⁡x→0x2r+14=z2=α02,θ=i+α0+(1/2)x+D0′D0,2i(α0+(1/2)+doD0)=0. Thus m=doD0=-α0-(1/2)∈ℕ.
Thusα0=-m-(1/2), r=-1+((m+(1/2))2-(1/4))/x2=-1+(m2+m)/x2 and θ=i-(m/x)+(D0′/D0). θ is a solution of Riccati equation if and only if D0 verifies
(4.46)(D0′D0)′+(D0′D0)2+2(i-mx)D0′D0=2imx.
Thus Φ=0 where
(4.47)Φ=x[(D0′D0)′+(D0′D0)2]+2(i-mx)(xD0′D0)-2im.
Taylors expansion of Φ=0 at t=1/x(4.48)Φ=∑k=1∞[(∑μ=1k-1σμσk-1-μ)-kσk-1-2mσk-1+2iσk]tk.
As σ0=m. Thus
(4.49)Φ=0⟺∀k≥1,(∑μ=0k-1σμσk-1-μ)-kσk-1-2mσk-1+2iσk=0.
For k=1, σ02-σ0-2mσ0+2iσ1=0 equivalent to 2iσ1=m2+m.
For all,k≥2, 2iσk=kσk-1-∑μ=1k-2σμσk-1-μ. For example
(4.50)m=3:{2iσ1=12,2iσ2=2σ1,2iσ3=3σ2-σ12,2iσ4=4σ3-2σ1σ2,2iσ5=5σ4-2σ1σ3-σ22.
Thus, the polynomial D0 is partner to
(4.51)|σ0σ1σ21σ1σ2σ3xσ2σ3σ4x2σ3σ4σ5x3|=|3-6i-61-6i-6-9ix-6-9i-54x2-9i-5499ix3|=135x3+810ix2-2025x-2025i.
Thus
(4.52)D0=x3+6ix2-15x-15i.


## 4.1. Presentation ofD0 as Determinant

Proposition 4.1.
Letc1,…,cm be complex constants. We put
(4.1)P(x)=(x-c1)⋯(x-cm)=xm-p1xm-1+p2xm-2-⋯+(-1)mpm.
For j∈ℕ, we put σj=c1j+⋯+cmj and
(4.2)Δ=|1c1⋅⋅c1m-11c2⋅⋅c2m-1⋅⋅⋅⋅1cm⋅⋅cmm-1|.(1)
One has(4.3)Δ2=|σ0σ1⋅⋅σm-1σ1σ2⋅⋅σm⋅⋅⋅⋅σm-1σm⋅⋅σ2m-2|.(2)
One has(4.4)Δ2P(x)=|σ0⋅⋅σm-11⋅⋅x⋅⋅⋅⋅⋅⋅σm⋅⋅σ2m-1xm|.Proof.
Δ,|σ0σ1··σm-1σ1σ2··σm····σm-1σm··σ2m-2| are polynomials at c1,…,cm with real coefficients. Thus, Δ2P(x) and |σ0··σm-11··x······σm··σ2m-1xm| are polynomials at c1,…,cm,x with real coefficients. Thus, to have both identities we can assume c1,…,cm,x are reals. Furthermore, we can restrict to the open of Zariski c1,…,cm,x distinct real nonzero.
We have for allj=1,…,m,cjm=p1cjm-1-p2cjm-2-⋯-(-1)mpm.
We put(4.5)Δ(x)=|1c1⋅⋅c1m-1c1m1c2⋅⋅c2m-1c2m⋅⋅⋅⋅⋅1cm⋅⋅cmm-1cmm1x⋅⋅xm-1xm|,Δ(x)=|1c1⋅⋅c1m-101c2⋅⋅c2m-10⋅⋅⋅⋅⋅⋅1cm⋅⋅cmm-101x⋅⋅xm-1P(x)|=ΔP(x).
We put(4.6)vj=(c1j-1..cmj-1),forj=1,…,m.
The scalar product
(4.7)〈vj,vk〉=c1j+k-2+⋯+cmj+k-2=σj+k-2.
The  Gram matrix of v1,…,vm is
(4.8)G=(σ0σ1⋅⋅σm-1σ1σ2⋅⋅σm⋅⋅⋅⋅σm-1σm⋅⋅σ2m-2)⋅
Thus
(4.9)Δ2=|σ0σ1..σm-1σ1σ2..σm....σm-1σm..σ2m-2|.
We put
(4.10)vj(x)=(c1j-1⋅⋅cmj-1xj-1),j=1,…,m+1.
We notice
(4.11)〈vj(x),vk(x)〉=〈vj,vk〉+xj+k-2=σj+k-2+xj+k-2.
The jth column of Gram of matrix which defines Δ(x) is
(4.12)(σj-1σj⋅⋅σj+m-1)+xj-1(1x⋅⋅xm).
Thus
(4.13)Δ2(x)=|σ0σ1⋅⋅σmσ1σ2⋅⋅σm+1⋅⋅⋅⋅σmσm+1⋅⋅σ2m|+|1σ1⋅⋅σmxσ2⋅⋅σm+1⋅⋅⋅⋅xmσm+1⋅⋅σ2m|+⋯+|σ0⋅σm-1xmσ1⋅σmxm+1⋅⋅⋅⋅σm⋅σ2m-1x2m|=∑i,jxi+j-2cofactori,j(M),
where
(4.14)M=(σ0σ1⋅⋅σmσ1σ2⋅⋅σm+1⋅⋅⋅⋅σmσm+1⋅⋅σ2m).
We obtain
(4.15)Δ2(x)=Δ2P2(x)=(1x⋯xm)Com(M)(1x⋅⋅xm)
The cofactor(m+1,m+1) of M is Δ2. The adjoint of M is nonzero. We prove that adjoint of M of rank 1.  det(M)=0. Because: σm=p1σm-1-p2σm-2-⋯-(-1)mpmσ0 and for all k≥m;σk=p1σk-1-p2σk-2-⋯-(-1)mpmσk-m. Thus, the (m+1)th column of M is is a linear combination of other column.
We consider the matrix(4.16)E1,1=(10⋅00⋅⋅⋅0⋅⋅0),(4.17)M+E1,1=(σ0+1σ1⋅⋅σmσ1σ2⋅⋅σm+1⋅⋅⋅⋅σmσm+1⋅⋅σ2m),det⁡(M+E1,1)=det⁡(M)+|1σ1⋅⋅σm0σ2⋅⋅σm+1⋅⋅⋅⋅0σm+1⋅⋅σ2m|=|σ2σ3⋅⋅σm+1σ3σ4⋅⋅σm+2⋅⋅⋅⋅σm+1σm+2⋅⋅σ2m|⋅
It is the determinant of Gram of
(4.18)(c1⋅⋅cm),…,(c1m⋅⋅cmm).
Thus
(4.19)det⁡(M+E1,1)=|c1⋅⋅⋅c1m⋅⋅⋅⋅⋅⋅⋅⋅cm⋅⋅⋅cmm|2⋅
But
(4.20)|c1⋅⋅⋅c1m⋅⋅⋅⋅⋅⋅⋅⋅cm⋅⋅⋅cmm|=(-1)m-1pm|c1⋅⋅⋅c1m-11⋅⋅⋅⋅1⋅⋅⋅⋅cm⋅⋅⋅cmm-11|=pmΔ.
Thus det(M+E1,1)=pm2Δ2. Or pm=c1⋯cm is scalar nonzero, M+E1,1 is invertible matrix.
WithM symmetric matrix, Com(M) symmetric matrix, we have
(4.21)MCom(M)=Com(M)M=det⁡(M)I=0,Ibeingmatriceidentity.
Thus
(4.22)(M+E1,1)Com(M)=E1,1Com(M).
Or M+E1,1 is invertible and Com(M)≠0, Com(M) of rank 1.
Thus(4.23)cofactori,j(M)=cofactori,m+1(M)cofactorm+1,m+1(M)cofactorm+1,j(M).
For i=1,…,m+1,j=1,…,m+1, we put: Ci,j=cofactori,j(M),
(4.24)Ci,jΔ2=Ci,m+1Δ2Cm+1,jΔ2.
Thus:
(4.25)P(x)2=∑1≤i,j≤m+1Ci,jΔ2xi+j-2=∑1≤i,j≤m+1Ci,m+1Δ2Cj,m+1Δ2xi+j-2=(∑1≤i≤m+1Ci,m+1Δ2xi-1)2.
Or P is monic polynomial and Cm+1,m+1/Δ2=1(4.26)P(x)=∑iCi,m+1Δ2xi-1=1Δ2|σ0⋅⋅⋅σm-11σ1⋅⋅⋅σmx⋅⋅⋅⋅⋅⋅σm⋅⋅⋅σ2m-1xm|.
We obtain
(4.27)Δ2P(x)=|σ0⋅⋅⋅σm-11σ1⋅⋅⋅σmx⋅⋅⋅⋅⋅⋅σm⋅⋅⋅σ2m-1xm|⋅


## 4.2. Taylor Expansion ofP′/P

Proposition 4.2.
Letc1,…,cm be pairwise distinct constants.
(1)
We put, forj=0,…,m: σj=c1j+⋯+cmj and P(x)=(x-c1)⋯(x-cm)(4.28)(P′P)′+(P′P)2=2∑i≠j1ci-cj1x-ci.(2)
We putt=1/x. Taylor expansion on the neighborhood of infinity:
(a)
(4.29)(P′P)′+(P′P)2=∑l=2∞[(∑μ=0l-2σμσl-2-μ)-(l-1)σl-2]tl,(b)
(4.30)xP′P=∑l=0∞σltl.Proof.
We haveP′/P=∑i=1m1/(x-ci).
(1)
(4.31)(P′P)′+(P′P)2=∑i≠j1x-ci1x-cj=∑i≠j[1/(ci-cj)x-ci+1/(cj-ci)x-cj]=2∑i≠j1ci-cj1x-ci.(2)
We putt=1/x. We have
(4.32)(P′P)′+(P′P)2=2t∑i≠j1ci-cj11-cit=2t∑i≠j1ci-cj∑k=0∞(cit)k=∑k=0∞(2∑i≠jcikci-cj)tk+1=∑k=1∞∑i≠jcik-cjkci-cjtk+1.
as
(4.33)∑i≠jcik-cjkci-cj=∑i≠j∑μ+ν=k-1,0≤μ≤k-1ciμcjν=∑μ+ν=k-1,0≤μ≤k-1[σμσν-∑i=1mciμ+ν]=[∑μ+ν=k-1,0≤μ≤k-1σμσν]-kσk-1.
Thus
(4.34)(P′P)′+(P′P)2=∑k=1∞[(∑μ=0k-1σμσk-1-μ)-kσk-1]tl=∑l=2∞[(∑μ=0l-2σμσl-2-μ)-(l-1)σl-2]tl.


## 4.3. Research ofD0

Letr=N/D1D22D32D42 be a rational fraction and θ∈ℂ(x) where θ′+θ2=r.We put(4.35)θ=S+D0′D0,
where S=E(θ)+∑cpolesofrθc and D0 monic polynomial of degree m. We have(4.36)θ′+θ2=S′+S2+(D0′D0)′+(D0′D0)2+2SD0′D0,θ′+θ2=r⟺(D0′D0)′+(D0′D0)2+2SD0′D0=RB
with B=D1D2D3D4 and R polynomials as r-S′-S2=R/B.Case 1 (dor=2ν≥0).
We havedoS=ν and doR/B=ν-1. Taylors expansion of (1/xν)S, (1/xν-1)(R/B) and x(D0′/D0) at order μ.
We obtain the Taylors expansion of(1/xν-1)[(R/B)-2S(D0′/D0)] at order μ.
The Taylors expansion equals the Taylors expansion of(1/xν-1)[(D0′/D0)′+(D0′/D0)2]. Let t=1/x. In Taylors expansion of (1/xν-1)[(P′/P)′+(P′/P)2] (see Proposition 4.2) the coefficient of tν+1 is: σ02-σ0=m2-m. For l≥2, coefficient of tl+ν-1 use: σ0,…,σl-2.
Besides, we have(4.37)xD0′D0=∑i=1m11-cit=∑i=1m∑k=0∞ciktk=∑k=0∞σktk.
Thus, Taylors expansion of
(4.38)Φ=1xν-1[(D0′D0)′+(D0′D0)2]-1xν-1[RB-2SD0′D0]
and the coefficient of tk is
(4.39)-2ασk+polynomialatσ0,…,σk-1,Φ=0 determine σk by recurrence at k=ν.Case 2 (dor<0).
The coefficient oftk in Taylors expansion of
(4.40)Ψ=x2[(D0′D0)′+(D0′D0)2]-x2RB+2xS(xD0′D0)
is
(4.41)2mσk-(k+1)σk+(2lim⁡x→∞xS)σk+polynomialatσ0,…,σk-1.
By recurrence σk at condition
(4.42)k≠2lim⁡x→∞xS+2m-1.
We have limx→∞xS+m=α∞2+(1/2) with α∞2=limx→∞x2r+(1/4).
Fork≠2α∞, to be σk at function of σ0,…,σk-1.
If2α∞∈ℕ then, the coefficient of t2α∞ is nonzero implied that there is no solution, or the coeffcient of t2α∞ is zero implied that σ2α∞ is arbitrary and σk; k>2α∞ depend in a unique way.
D0 is polynomial determined by σ0,…,σ2m-1 by determinant formulae; the problem of nonunique suite (σk)k∈ℕ is put if 2α∞ is positive integer equal to or less than 2m-1, then 4limx→∞x2r+1 is square of integer and
(4.43)lim⁡x→∞x2r≤m2-m.Example 4.3.
In this example we consider the Riccati differential equation (1.2) where
(4.44)r=-1+z2-(1/4)x2;z∈R*.
We have D1=1, D2=1, D3=x and E2(θ)=-1.
We can assumeE(θ)=i,
(4.45)lim⁡x→0x2r+14=z2=α02,θ=i+α0+(1/2)x+D0′D0,2i(α0+(1/2)+doD0)=0. Thus m=doD0=-α0-(1/2)∈ℕ.
Thusα0=-m-(1/2), r=-1+((m+(1/2))2-(1/4))/x2=-1+(m2+m)/x2 and θ=i-(m/x)+(D0′/D0). θ is a solution of Riccati equation if and only if D0 verifies
(4.46)(D0′D0)′+(D0′D0)2+2(i-mx)D0′D0=2imx.
Thus Φ=0 where
(4.47)Φ=x[(D0′D0)′+(D0′D0)2]+2(i-mx)(xD0′D0)-2im.
Taylors expansion of Φ=0 at t=1/x(4.48)Φ=∑k=1∞[(∑μ=1k-1σμσk-1-μ)-kσk-1-2mσk-1+2iσk]tk.
As σ0=m. Thus
(4.49)Φ=0⟺∀k≥1,(∑μ=0k-1σμσk-1-μ)-kσk-1-2mσk-1+2iσk=0.
For k=1, σ02-σ0-2mσ0+2iσ1=0 equivalent to 2iσ1=m2+m.
For all,k≥2, 2iσk=kσk-1-∑μ=1k-2σμσk-1-μ. For example
(4.50)m=3:{2iσ1=12,2iσ2=2σ1,2iσ3=3σ2-σ12,2iσ4=4σ3-2σ1σ2,2iσ5=5σ4-2σ1σ3-σ22.
Thus, the polynomial D0 is partner to
(4.51)|σ0σ1σ21σ1σ2σ3xσ2σ3σ4x2σ3σ4σ5x3|=|3-6i-61-6i-6-9ix-6-9i-54x2-9i-5499ix3|=135x3+810ix2-2025x-2025i.
Thus
(4.52)D0=x3+6ix2-15x-15i.


## 5. Method of Last Minor

Letr=N/D1D22D32D42 be a rational fraction and θ∈ℂ(x) as: θ′+θ2=r.θ solution of (1.2) if and only if D0 verifies(5.1)BD0′′+2SBD0′=RD0.
We choose a complex number c not pole of r and we use the expression of polynomials following the powers of x-c.For the sake of simplicity, in the following we assume thatc=0. Constant coefficient constant of B is nonzero.Denote byak,bk, and rk coefficients of xk, in A,B, and R, respectively, k∈ℕ. If k∈ℤ then ak=bk=rk=0.


### 5.1. Casedor<0

We putS=A/B.IfdoB=1 then r are pole unique, it is simple or double. If c simple pole of r then r of degree −1 and (1.2) has no rational solution.Ifc double pole of r then A=αc+(1/2), r=(αc2-(1/4))/(x-c)2 and R=0.Thus,(5.1)(5.2)D0′′D0′=-2(αc+(1/2))x-c.
The solutions(5.3)D0′=(-2αc)(x-c)-2αc-1,D0=(x-c)-2αc+β,
where β nonzero constant.We have, an infinity of the other rational solution when-2αc∈ℕ*, with the same parity of S.We assumedoB≥2and we consider D0 a polynomial of degree m≥1 that verifies (5.1).Proposition 5.1.
One assumesdor<0,{doB≥2do=m≥1 and D0 verifies (5.1). (1)
One has(5.4)do(BD0′′+2AD0′)≤doB+m-2,doR≤doB-2.(2)
One hasrdoB-2=2adoB-1m+m(m-1).(3)
One has(5.5)adoB-1=12-m+α∞,rdoB-2=m(2α∞-m).Proof.
(1)do(BD0′′+2AD0′)≤sup(do(BD0′′),do(2AD0′)). If doD0=1 then
(5.6)sup⁡(do(BD0′′),do(2AD0′))=do(2AD0′)=doA≤doB-1=doB+doD0-2.
If doD0≥2 then
(5.7)do(BD0′′)=doB+m-2,do(2AD0′)=doA+m-1≤doB+m-2.
(2) We havedo(RD0)≤doB+m-2 and coefficients of xdoB+m-2 in (5.1).
(3) Accordingly, in view oflimx→∞xθ=α∞+(1/2).We put(5.8)v(x)=-R(1xx2⋅⋅xm)+2A(012x⋅⋅mxm-1)+B(002⋅⋅m(m-1)xm-2)=(v1(x)v2(x)v3(x)⋅⋅vm+1(x)).
The kth element of v(x) is(5.9)vk(x)=B(k-1)(k-2)xk-3+2A(k-1)xk-2-Rxk-1,vk(x) of degree equal to or less than doB+k-3 and xdoB+k-3 of coefficient(5.10)(k-1)(k-2)+2adoB-1(k-1)-rdoB-2=(k-m-1)(rdoB-2m+k-1)=(k-m-1)(2adoB-1+m+k-2).
For k≤m, the coefficient of xdoB+k-3 is zero if and only if (5.11)2adoB-1=-(m+k-2)∈{-(m-1),-m,…,-2(m-1)},vm+1 of degree equal to or less than doB+m-3.Fork≥3 the coefficient of more low degree of vk is coefficient of xk-3 equal b0(k-1)(k-2).LetV matrix of m+1 row and kth row is row lk of coefficients of vk(x) in basis of ℂdoB+m-3[X].l3,…,lm+1 is linearly independent system.Equation (5.1) gives(5.12)D0=d0+d1x+⋯+xm,
obtained(5.13)d0v1(x)+⋯+dm-1vm(x)+vm+1(x)=0,d0l1+⋯+dm-1lm+lm+1=0.
Thus, the matrix V is rank m-1,m or m+1. Accordingly, existence of D0 correspondent is linearly dependent in row lm+1 of l1,…,lm.Letc1,…,cm roots of D0 distinct.We put, forj∈ℕ: σj=c1j+⋯+cmj and (5.14)Δ2=|σ0σ1⋅⋅σm-1σ1σ2⋅⋅σm⋅⋅⋅⋅σm-1σm⋅⋅σ2m-2|.The m column vector (σ0··σm),…,(σm-1··σ2m-1) of m+1 column is linearly independent.The equation of hyperplaneH engendered by vectors (σ0··σm),…,(σm-1··σ2m-1) is(5.15)|σ0⋅⋅⋅σm-1x1σ1⋅⋅⋅σmx2⋅⋅⋅⋅⋅⋅σm⋅⋅⋅σ2m-1xm+1|=0.
The coefficient of xm+1 is Δ2≠0 and H as equation(5.16)xm+1=λ1x1+⋯+λmxm,
where λ1⋯λm scalar satisfying(5.17)(σm,…,σ2m-1)=λ1(σ0,…,σm-1)+⋯+λm(σm-1,…,σ2m-2).
The derivative of(5.18)|σ0⋅⋅⋅σm-11σ1⋅⋅⋅σmx⋅⋅⋅⋅⋅⋅σm⋅⋅⋅σ2m-1xm|
is derived from last column. Equation (5.1) is equivalent to:(5.19)|σ0⋅⋅⋅σm-11σ1⋅⋅⋅σmv1(x)⋅⋅⋅⋅⋅⋅σm⋅⋅⋅σ2m-1vm+1(x)|=0.
Thus, for all x, v(x)∈H. With application of Taylor, this is equivalent to columns of matrix V in hyperplane H,(5.20)lm+1=λ1l1+⋯+λmlm,|σ0⋅⋅⋅σm-11σ1⋅⋅⋅σmx⋅⋅⋅⋅⋅⋅σm⋅⋅⋅σ2m-1xm|=|σ0⋅⋅⋅σm-11σ1⋅⋅⋅σmx⋅⋅⋅σm-1⋅⋅⋅σ2m-2xm-10⋅⋅⋅0xm-λ1-λ2x-⋯-λmxm-1|=Δ2[xm-λ1-λ2x-⋯-λmxm-1].
The polynomial D0 is partner to (5.21)|σ0⋅⋅⋅σm-11σ1⋅⋅⋅σmx⋅⋅⋅⋅⋅⋅σm⋅⋅⋅σ2m-1xm|.
Thus(5.22)D0(x)=xm-λ1-λ2x-⋯-λmxm-1.
For k=1,…,m+1, we put(5.23)vk(x)=ρk(x)+xdoB-2wk,
where doρk(x)≤doB-3.The matrixW partner to (w1·wm) is triangular where kth entry diagonal equal (k-1-m)(2adoB-1+m+k-2). The coefficient is nonzero except possibly for a single value of k. Thus, rank of (w1·wm) is m or m-1.For rank(V)=m, (5.1) as solution is equivalent to l1,…,lm which is linearly independent.Hence, we assume rank(V)=m-1. In that case l1 and l2 are linear combinations of l3,…,lm+1. One of both combinations has to express lm+1. Thus we have two situations.Situation 1
l2 linear combination of l3,…,lm+1 where coefficient of lm+1 is nonzero. By replacement of lm+1, we obtain l1 a linear combination of l2,…,lm.
Thus,w1 is linear combination of w2,…,wm. (w2,…,wm) libre system where dowk=k-1 for 0k=2,…,m. Thus: w1=0.
If(5.24)l1=∑j=2mαjlj,
where αj is constant, then
(5.25)w1=∑j=2mαjwj.
As w1=0, thus for all j=2,…,m, αj=0 equivalent to l1=0. Thus R=0.First Way of HavingD0:
(5.26)D0′′D0′=-2AB-2A/B has to be the sum of simple elements of the shape μc/(x-c) where μc is greater or equal to 1.
ThusD4=D2=D1=1 and D3=B.
For allc root of D3,-2αc=μc+1 where αc2=limx→c(x-c)2r+(1/4).Thus
(5.27)αc<0,4[lim⁡x→c(x-c)2r+14]=(μc+1)2.
Thus
(5.28)D0′=m∏c∈Root(D3)(x-c)μc.D0 is primitive nonzero in roots of D3.Second Way of HavingD0:
(5.29)lm+1=∑j=2mλjlj,
where λj is constant (j=2,…,m). The relations which are linearly dependent between lm+1 and l1,…,lm are
(5.30)lm+1=λl1+∑j=2mλjlj.
Thus we have an infinity of solutions:
(5.31)D0=xm-λ-∑j=2mλjxj-1,
where λ is arbitrarily constant.Situation 2
l1 is linear combination of l3,…,lm+1 where coefficient of lm+1 is nonzero and l2 is linear combination of l3,…,lm. Thus, w2 is linear combination of w3,…,wm. Thus, w2=0. w1,w3,…,wm is libre system.
If(5.32)l2=∑j=3mαjlj,
then
(5.33)w2=∑j=3mαjwj.
As w2=0. Thus, for all j=3,…,m, αj=0 equivalent to l2=0 and 2A-xR=0.First Way of HavingD0:
(5.34)2AxD0=2AD0′+BD0′′,2A(D0-xD0′)=xBD0′′.
We put Q=D0-xD0′.  Q′/Q=-2A/B,
(5.35)D4=D2=D1=1,Q=(1-m)∏c∈Root(D3)(x-c)μc.
Thus D0=Cx with C′=-Q/x2; C rational function.
The coefficient ofx in Q is zero. Q′(0)=0 (A(0)=0; 2A=xR).Second Way of HavingD0
(5.36)lm+1=λ1l1+∑j=3mλjlj,
thus
(5.37)D0=xm-∑j=1mλjxj-1.λ2 is arbitrarily constant.


### 5.2. Casedor=2ν≥0

We putE=E(θ) and S=E+(A/B). We have(5.38)do(BD0′′)≤doB+m-2,do(EB+A)D0′=doB+m-1+ν.
Thus(5.39)do(RD0)=doB+m-1+ν.
Thus, doR=doB+ν-1 and rdoB+ν-1=2αm where α is dominant coefficient of E.We put(5.40)v(x)=-R(1xx2⋅⋅xm)+2(EB+A)(012x⋅⋅mxm-1)+B(002⋅⋅m(m-1)xm-2)=(v1(x)v2(x)v3(x)⋅⋅vm+1(x))⋅
The kth element of v(x) is(5.41)vk(x)=B(k-1)(k-2)xk-3+2(EB+A)(k-1)xk-2-Rxk-1.vk(x) is of degree equal to or less than doB+ν+k-2 where coefficient of xdoB+ν+k-2 equal 2α(k-1-m). For k=1,…,m, the kth element of v(x) is of degree: doR+k-1. The (m+1)th element of v(x) is of degree equal to or less than doR+m-1.Proposition 5.2.
One putsvk(x)=ρk(x)+xdoRwk(x) where doρk(x)<doR.(1)
(5.42)dowk(x)=k-1fork=1,…,m,dowm+1(x)≤m-1,(2)
(w1,…,wm) libre system and
(5.43)wm+1(x)=λ1w1(x)+⋯+λmwm(x),
where λ1,…,λm constants.(3)
(5.44)D0=xm-λ1-λ2x-⋯-λm-1xm-1
is solution if and only if
(5.45)lm+1=λ1l1+⋯+λmlm.Example 5.3.
In this example we consider the Riccati differential equation (1.2) where
(5.46)r=1(x+1)4-5(x+1)3+74(x+1)2+1x+1+x2+2,D1=D2=D3=1,D4=(x+1)2,do(r)=2,ν=1. We have
(5.47)N1=1-5(x+1),F=x2+2+1-5(x+1)(x+1)4+14(x+1)2.Study at (−1)
Laurent series development at −1:(5.48)(F)1/2=ε-1(1(x+1)2-52(x+1)+⋯);ε-1=±1.Study at Infinity
We haveE(r)/x2=1+t2 where t=1/x,
(5.49)(1+2t2)1/2=1+t2+o(t2).
Thus sν+1=1 and E(θ)=αx, where α2=1,
(5.50)θ=αx+ε-1(1(x+1)2-52(x+1))+1(x+1)+D0′D0,doD0=αsν+1-ν2+5ε-12-1=αsν+1+5ε-12-32.
Case {α=-1ɛ-1=-1 and {α=1ɛ-1=-1 are to be rejected because we obtain negative values of D0.  (i)
Ifα=1 and ε-1=1 then: doD0=2.(ii)
Ifα=-1 and ε-1=1 then: doD0=0.Case 1 (α=1 and ɛ-1=1).
We have the following(5.51)θ=x+1(x+1)2-32(x+1)+D0′D0,
where doD0=2. Research of coefficients of D0,
(5.52)S=x+1(x+1)2-32(x+1)=x3+2x2-(1/2)x-(1/2)(x+1)2r-S′-S2=R/B, where R=4x2+4x and B=(1+x)2(5.53)v(x)=-R(1xx2)+2SB(012x)+B(002)=(-4x-4x2-1-x-2x32+2x+4x3),V=(0-4-40-1-10-22204),l3=-2l2. Thus D0=x2+2x,
(5.54)θ=x+1(x+1)2-32(x+1)+2(x+1)x2+2x=x+1(x+1)2-32(x+1)+1x+1x+2.Case 2 (α=-1 and ε-1=1).
We obtain the rational fraction(5.55)θ=-x+1(x+1)2-32(x+1).
It is not a solution because the sum of this fraction with the solution already found is not the logarithmic derivative of a rational fraction.Example 5.4.
In this example we consider the Riccati differential equation (1.2) where
(5.56)r=116+1(x-1)8-4(x-1)5-296(x-1)4-89(x-1)3-6427(x-1)2-15218(x-1)+30(x+2)2-1081(x+2),D1=D2=1,D3=x+2 and D4=(x-1)4. We have
(5.57)N1=(2418x-2454)(x-1)3+(x+2)2,δ=-14(x+2)2,F=116+(2418x-2454)(x-1)3+(x+2)2(x-1)8(x+2)2+14(x+2)2.
Laurent series development at 1:
(5.58)(F)1/2=ε1(1(x-1)4-2x-1+⋯);ε1=±1.
Laurent series development at −2:
(5.59)(F)1/2=ε-2(112(x+2)+⋯);ɛ-2=±1.
We have
(5.60)12(D2D3D4)′D2D3D4=12(x+2)+2x-1.
Thus
(5.61)θ=E(θ)+ε1[1(x-1)4-2x-1]+ε-2[112(x+2)]+12(x+2)+2x-1+D0′D0,
where
(5.62)E2(θ)=E(r)=116,doD0=-1E(θ)+2ε1-11ε-22-52.
Cases {ε1=1ε-2=1 and {ε1=-1ε-2=1 are to be rejected because we obtain negative values of D0.Case 1.
Ifε1=1,ε-2=-1,E(θ)=1/4 then doD0=1,
(5.63)θ=14+1(x-1)4-5x+2+D0′D0.
Research of coefficients of D0,
(5.64)S=14+1(x-1)4-5x+2,r-(S′+S2)=R/B, where R=(1/2)x4-10x3+19x2-18x+(5/2), B=(x-1)4(x+2),
(5.65)v(x)=-R(1x)+2SB(01)=(-12x4+10x3-19x2+18x-52-x4+20x3-38x2+36x-1),V=(-5218-1910-12-536-3820-5),l2=2l1. Thus, D0=x-2,
(5.66)θ1=14+1(x-1)4-5x+2+1x-2.Case 2.
Ifε1=1, ε-2=-1, E(θ)=-1/4 then: doD0=9,
(5.67)θ=-14+1(x-1)4-5x+2+D0′D0.
It is not a solution because the sum of this fraction with the solution already found is not the logarithmic derivative of a rational fraction.Case 3.
Ifε1=-1, ε-2=-1, E(θ)=-1/4 then doD0=5,
(5.68)θ3=-14-1(x-1)4-5x+2+4x-1+D0′D0.
Research of coefficients of D0,
(5.69)S=-14-1(x-1)4-5x+2+4x-1,R=-x4+36x3-137x2+168x-88,B=(x-1)4(x+2),V=(-37-168137-36520000-32-1512-266180-3720004-78-162-368219-363200012-138-2092-474254-33100024-212-95-584285-281200040-300-1392-698312-21).
We consider Vp the minor 6×6 obtained by column vectors 1,2,6,7,8, and 9. In ℤ/5ℤ(5.70)Vp=(-220000-222000-12-1-10002-121000102-200-222-1)detVp=1. Accordingly, row 6 is not linear combination of other row and So no solution D0.


## 5.1. Casedor<0

We putS=A/B.IfdoB=1 then r are pole unique, it is simple or double. If c simple pole of r then r of degree −1 and (1.2) has no rational solution.Ifc double pole of r then A=αc+(1/2), r=(αc2-(1/4))/(x-c)2 and R=0.Thus,(5.1)(5.2)D0′′D0′=-2(αc+(1/2))x-c.
The solutions(5.3)D0′=(-2αc)(x-c)-2αc-1,D0=(x-c)-2αc+β,
where β nonzero constant.We have, an infinity of the other rational solution when-2αc∈ℕ*, with the same parity of S.We assumedoB≥2and we consider D0 a polynomial of degree m≥1 that verifies (5.1).Proposition 5.1.
One assumesdor<0,{doB≥2do=m≥1 and D0 verifies (5.1). (1)
One has(5.4)do(BD0′′+2AD0′)≤doB+m-2,doR≤doB-2.(2)
One hasrdoB-2=2adoB-1m+m(m-1).(3)
One has(5.5)adoB-1=12-m+α∞,rdoB-2=m(2α∞-m).Proof.
(1)do(BD0′′+2AD0′)≤sup(do(BD0′′),do(2AD0′)). If doD0=1 then
(5.6)sup⁡(do(BD0′′),do(2AD0′))=do(2AD0′)=doA≤doB-1=doB+doD0-2.
If doD0≥2 then
(5.7)do(BD0′′)=doB+m-2,do(2AD0′)=doA+m-1≤doB+m-2.
(2) We havedo(RD0)≤doB+m-2 and coefficients of xdoB+m-2 in (5.1).
(3) Accordingly, in view oflimx→∞xθ=α∞+(1/2).We put(5.8)v(x)=-R(1xx2⋅⋅xm)+2A(012x⋅⋅mxm-1)+B(002⋅⋅m(m-1)xm-2)=(v1(x)v2(x)v3(x)⋅⋅vm+1(x)).
The kth element of v(x) is(5.9)vk(x)=B(k-1)(k-2)xk-3+2A(k-1)xk-2-Rxk-1,vk(x) of degree equal to or less than doB+k-3 and xdoB+k-3 of coefficient(5.10)(k-1)(k-2)+2adoB-1(k-1)-rdoB-2=(k-m-1)(rdoB-2m+k-1)=(k-m-1)(2adoB-1+m+k-2).
For k≤m, the coefficient of xdoB+k-3 is zero if and only if (5.11)2adoB-1=-(m+k-2)∈{-(m-1),-m,…,-2(m-1)},vm+1 of degree equal to or less than doB+m-3.Fork≥3 the coefficient of more low degree of vk is coefficient of xk-3 equal b0(k-1)(k-2).LetV matrix of m+1 row and kth row is row lk of coefficients of vk(x) in basis of ℂdoB+m-3[X].l3,…,lm+1 is linearly independent system.Equation (5.1) gives(5.12)D0=d0+d1x+⋯+xm,
obtained(5.13)d0v1(x)+⋯+dm-1vm(x)+vm+1(x)=0,d0l1+⋯+dm-1lm+lm+1=0.
Thus, the matrix V is rank m-1,m or m+1. Accordingly, existence of D0 correspondent is linearly dependent in row lm+1 of l1,…,lm.Letc1,…,cm roots of D0 distinct.We put, forj∈ℕ: σj=c1j+⋯+cmj and (5.14)Δ2=|σ0σ1⋅⋅σm-1σ1σ2⋅⋅σm⋅⋅⋅⋅σm-1σm⋅⋅σ2m-2|.The m column vector (σ0··σm),…,(σm-1··σ2m-1) of m+1 column is linearly independent.The equation of hyperplaneH engendered by vectors (σ0··σm),…,(σm-1··σ2m-1) is(5.15)|σ0⋅⋅⋅σm-1x1σ1⋅⋅⋅σmx2⋅⋅⋅⋅⋅⋅σm⋅⋅⋅σ2m-1xm+1|=0.
The coefficient of xm+1 is Δ2≠0 and H as equation(5.16)xm+1=λ1x1+⋯+λmxm,
where λ1⋯λm scalar satisfying(5.17)(σm,…,σ2m-1)=λ1(σ0,…,σm-1)+⋯+λm(σm-1,…,σ2m-2).
The derivative of(5.18)|σ0⋅⋅⋅σm-11σ1⋅⋅⋅σmx⋅⋅⋅⋅⋅⋅σm⋅⋅⋅σ2m-1xm|
is derived from last column. Equation (5.1) is equivalent to:(5.19)|σ0⋅⋅⋅σm-11σ1⋅⋅⋅σmv1(x)⋅⋅⋅⋅⋅⋅σm⋅⋅⋅σ2m-1vm+1(x)|=0.
Thus, for all x, v(x)∈H. With application of Taylor, this is equivalent to columns of matrix V in hyperplane H,(5.20)lm+1=λ1l1+⋯+λmlm,|σ0⋅⋅⋅σm-11σ1⋅⋅⋅σmx⋅⋅⋅⋅⋅⋅σm⋅⋅⋅σ2m-1xm|=|σ0⋅⋅⋅σm-11σ1⋅⋅⋅σmx⋅⋅⋅σm-1⋅⋅⋅σ2m-2xm-10⋅⋅⋅0xm-λ1-λ2x-⋯-λmxm-1|=Δ2[xm-λ1-λ2x-⋯-λmxm-1].
The polynomial D0 is partner to (5.21)|σ0⋅⋅⋅σm-11σ1⋅⋅⋅σmx⋅⋅⋅⋅⋅⋅σm⋅⋅⋅σ2m-1xm|.
Thus(5.22)D0(x)=xm-λ1-λ2x-⋯-λmxm-1.
For k=1,…,m+1, we put(5.23)vk(x)=ρk(x)+xdoB-2wk,
where doρk(x)≤doB-3.The matrixW partner to (w1·wm) is triangular where kth entry diagonal equal (k-1-m)(2adoB-1+m+k-2). The coefficient is nonzero except possibly for a single value of k. Thus, rank of (w1·wm) is m or m-1.For rank(V)=m, (5.1) as solution is equivalent to l1,…,lm which is linearly independent.Hence, we assume rank(V)=m-1. In that case l1 and l2 are linear combinations of l3,…,lm+1. One of both combinations has to express lm+1. Thus we have two situations.Situation 1
l2 linear combination of l3,…,lm+1 where coefficient of lm+1 is nonzero. By replacement of lm+1, we obtain l1 a linear combination of l2,…,lm.
Thus,w1 is linear combination of w2,…,wm. (w2,…,wm) libre system where dowk=k-1 for 0k=2,…,m. Thus: w1=0.
If(5.24)l1=∑j=2mαjlj,
where αj is constant, then
(5.25)w1=∑j=2mαjwj.
As w1=0, thus for all j=2,…,m, αj=0 equivalent to l1=0. Thus R=0.First Way of HavingD0:
(5.26)D0′′D0′=-2AB-2A/B has to be the sum of simple elements of the shape μc/(x-c) where μc is greater or equal to 1.
ThusD4=D2=D1=1 and D3=B.
For allc root of D3,-2αc=μc+1 where αc2=limx→c(x-c)2r+(1/4).Thus
(5.27)αc<0,4[lim⁡x→c(x-c)2r+14]=(μc+1)2.
Thus
(5.28)D0′=m∏c∈Root(D3)(x-c)μc.D0 is primitive nonzero in roots of D3.Second Way of HavingD0:
(5.29)lm+1=∑j=2mλjlj,
where λj is constant (j=2,…,m). The relations which are linearly dependent between lm+1 and l1,…,lm are
(5.30)lm+1=λl1+∑j=2mλjlj.
Thus we have an infinity of solutions:
(5.31)D0=xm-λ-∑j=2mλjxj-1,
where λ is arbitrarily constant.Situation 2
l1 is linear combination of l3,…,lm+1 where coefficient of lm+1 is nonzero and l2 is linear combination of l3,…,lm. Thus, w2 is linear combination of w3,…,wm. Thus, w2=0. w1,w3,…,wm is libre system.
If(5.32)l2=∑j=3mαjlj,
then
(5.33)w2=∑j=3mαjwj.
As w2=0. Thus, for all j=3,…,m, αj=0 equivalent to l2=0 and 2A-xR=0.First Way of HavingD0:
(5.34)2AxD0=2AD0′+BD0′′,2A(D0-xD0′)=xBD0′′.
We put Q=D0-xD0′.  Q′/Q=-2A/B,
(5.35)D4=D2=D1=1,Q=(1-m)∏c∈Root(D3)(x-c)μc.
Thus D0=Cx with C′=-Q/x2; C rational function.
The coefficient ofx in Q is zero. Q′(0)=0 (A(0)=0; 2A=xR).Second Way of HavingD0
(5.36)lm+1=λ1l1+∑j=3mλjlj,
thus
(5.37)D0=xm-∑j=1mλjxj-1.λ2 is arbitrarily constant.


## 5.2. Casedor=2ν≥0

We putE=E(θ) and S=E+(A/B). We have(5.38)do(BD0′′)≤doB+m-2,do(EB+A)D0′=doB+m-1+ν.
Thus(5.39)do(RD0)=doB+m-1+ν.
Thus, doR=doB+ν-1 and rdoB+ν-1=2αm where α is dominant coefficient of E.We put(5.40)v(x)=-R(1xx2⋅⋅xm)+2(EB+A)(012x⋅⋅mxm-1)+B(002⋅⋅m(m-1)xm-2)=(v1(x)v2(x)v3(x)⋅⋅vm+1(x))⋅
The kth element of v(x) is(5.41)vk(x)=B(k-1)(k-2)xk-3+2(EB+A)(k-1)xk-2-Rxk-1.vk(x) is of degree equal to or less than doB+ν+k-2 where coefficient of xdoB+ν+k-2 equal 2α(k-1-m). For k=1,…,m, the kth element of v(x) is of degree: doR+k-1. The (m+1)th element of v(x) is of degree equal to or less than doR+m-1.Proposition 5.2.
One putsvk(x)=ρk(x)+xdoRwk(x) where doρk(x)<doR.(1)
(5.42)dowk(x)=k-1fork=1,…,m,dowm+1(x)≤m-1,(2)
(w1,…,wm) libre system and
(5.43)wm+1(x)=λ1w1(x)+⋯+λmwm(x),
where λ1,…,λm constants.(3)
(5.44)D0=xm-λ1-λ2x-⋯-λm-1xm-1
is solution if and only if
(5.45)lm+1=λ1l1+⋯+λmlm.Example 5.3.
In this example we consider the Riccati differential equation (1.2) where
(5.46)r=1(x+1)4-5(x+1)3+74(x+1)2+1x+1+x2+2,D1=D2=D3=1,D4=(x+1)2,do(r)=2,ν=1. We have
(5.47)N1=1-5(x+1),F=x2+2+1-5(x+1)(x+1)4+14(x+1)2.Study at (−1)
Laurent series development at −1:(5.48)(F)1/2=ε-1(1(x+1)2-52(x+1)+⋯);ε-1=±1.Study at Infinity
We haveE(r)/x2=1+t2 where t=1/x,
(5.49)(1+2t2)1/2=1+t2+o(t2).
Thus sν+1=1 and E(θ)=αx, where α2=1,
(5.50)θ=αx+ε-1(1(x+1)2-52(x+1))+1(x+1)+D0′D0,doD0=αsν+1-ν2+5ε-12-1=αsν+1+5ε-12-32.
Case {α=-1ɛ-1=-1 and {α=1ɛ-1=-1 are to be rejected because we obtain negative values of D0.  (i)
Ifα=1 and ε-1=1 then: doD0=2.(ii)
Ifα=-1 and ε-1=1 then: doD0=0.Case 1 (α=1 and ɛ-1=1).
We have the following(5.51)θ=x+1(x+1)2-32(x+1)+D0′D0,
where doD0=2. Research of coefficients of D0,
(5.52)S=x+1(x+1)2-32(x+1)=x3+2x2-(1/2)x-(1/2)(x+1)2r-S′-S2=R/B, where R=4x2+4x and B=(1+x)2(5.53)v(x)=-R(1xx2)+2SB(012x)+B(002)=(-4x-4x2-1-x-2x32+2x+4x3),V=(0-4-40-1-10-22204),l3=-2l2. Thus D0=x2+2x,
(5.54)θ=x+1(x+1)2-32(x+1)+2(x+1)x2+2x=x+1(x+1)2-32(x+1)+1x+1x+2.Case 2 (α=-1 and ε-1=1).
We obtain the rational fraction(5.55)θ=-x+1(x+1)2-32(x+1).
It is not a solution because the sum of this fraction with the solution already found is not the logarithmic derivative of a rational fraction.Example 5.4.
In this example we consider the Riccati differential equation (1.2) where
(5.56)r=116+1(x-1)8-4(x-1)5-296(x-1)4-89(x-1)3-6427(x-1)2-15218(x-1)+30(x+2)2-1081(x+2),D1=D2=1,D3=x+2 and D4=(x-1)4. We have
(5.57)N1=(2418x-2454)(x-1)3+(x+2)2,δ=-14(x+2)2,F=116+(2418x-2454)(x-1)3+(x+2)2(x-1)8(x+2)2+14(x+2)2.
Laurent series development at 1:
(5.58)(F)1/2=ε1(1(x-1)4-2x-1+⋯);ε1=±1.
Laurent series development at −2:
(5.59)(F)1/2=ε-2(112(x+2)+⋯);ɛ-2=±1.
We have
(5.60)12(D2D3D4)′D2D3D4=12(x+2)+2x-1.
Thus
(5.61)θ=E(θ)+ε1[1(x-1)4-2x-1]+ε-2[112(x+2)]+12(x+2)+2x-1+D0′D0,
where
(5.62)E2(θ)=E(r)=116,doD0=-1E(θ)+2ε1-11ε-22-52.
Cases {ε1=1ε-2=1 and {ε1=-1ε-2=1 are to be rejected because we obtain negative values of D0.Case 1.
Ifε1=1,ε-2=-1,E(θ)=1/4 then doD0=1,
(5.63)θ=14+1(x-1)4-5x+2+D0′D0.
Research of coefficients of D0,
(5.64)S=14+1(x-1)4-5x+2,r-(S′+S2)=R/B, where R=(1/2)x4-10x3+19x2-18x+(5/2), B=(x-1)4(x+2),
(5.65)v(x)=-R(1x)+2SB(01)=(-12x4+10x3-19x2+18x-52-x4+20x3-38x2+36x-1),V=(-5218-1910-12-536-3820-5),l2=2l1. Thus, D0=x-2,
(5.66)θ1=14+1(x-1)4-5x+2+1x-2.Case 2.
Ifε1=1, ε-2=-1, E(θ)=-1/4 then: doD0=9,
(5.67)θ=-14+1(x-1)4-5x+2+D0′D0.
It is not a solution because the sum of this fraction with the solution already found is not the logarithmic derivative of a rational fraction.Case 3.
Ifε1=-1, ε-2=-1, E(θ)=-1/4 then doD0=5,
(5.68)θ3=-14-1(x-1)4-5x+2+4x-1+D0′D0.
Research of coefficients of D0,
(5.69)S=-14-1(x-1)4-5x+2+4x-1,R=-x4+36x3-137x2+168x-88,B=(x-1)4(x+2),V=(-37-168137-36520000-32-1512-266180-3720004-78-162-368219-363200012-138-2092-474254-33100024-212-95-584285-281200040-300-1392-698312-21).
We consider Vp the minor 6×6 obtained by column vectors 1,2,6,7,8, and 9. In ℤ/5ℤ(5.70)Vp=(-220000-222000-12-1-10002-121000102-200-222-1)detVp=1. Accordingly, row 6 is not linear combination of other row and So no solution D0.



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*Source: 852656-2011-06-18.xml*