Computation and proof of explicit continued fractions

Sebastien Maulat, ENS de LyonBruno Salvy, INRIA

LIP, Lyon

May 2015

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 1 / 14

Motivation

A long history

Huygens’ planetarium [Huygens, 1682]

De analysis infinitorum [Euler, 1748]

π is irrational [Lambert, 1761]

real roots isolation of P ∈ IR[X ] [Lenstra, 2002, Hallgren, 2007]

A new Stirling series as continued fraction [C. Mortici, 2009]

Renewed interest for numerical evaluation

Handbook of mathematical functions [Abramowitz and Stegun, 1964]

Continued Fractions for numerical analysis [Jones and Thron, 1988]

Continued Fractions with applications [Lorentzen and Waadeland, 1992]

Handbook of Continued Fractions for Special Functions[Cuyt, Peterson, Verdonk, Waadeland and Jones, 2008]

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 2 / 14

Motivation

Convergence

ln(1 + x) =4∑1

(−1)i+1 x i

i+ O(x5)

=1/2 x2 + x

1/6 x2 + x + 1+ O(x5)

=x

1 +x/2

1 +x/6

1 + x/3

+ O(x5)

zone with 10 correct digits, for x ∈ C(with n = 20 and n = 30)

Taylor at order 2n : convergence for |x | < 1,

Pade et order n/n : convergence for 1 + x /∈ IR−.

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 3 / 14

Objects

Corresponding continued fractions

(truncated) C-fraction:

n

Ki=0

aixαi

1:=

a0xα0

1 +a1xα1

1 +. . .

1 +an−1xαn−1

1 + anxαn

, ai ∈ C?, αi ∈ IN∗

Correspondence: {∞

Ki=0

aixαi

1

}'

{ ∞∑i=1

cixi

}([

a0

α0

], . . . ,

[anαn

])straightforward! (c1, . . . , cα0+...+αn)

regular C-fractions : when αi = 1.simple, many formulas and applications. [Stieltjes 1894, Cuyt et alii 2008]

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 4 / 14

Objects

Formulas

Some general classes

[Gauss]

2F1(a, b; c ; z)

2F1(a, b + 1; c + 1; z)= 1 +

∞

Km=1

pmz

1

p2k = − (k + b)(k + c − a)

(2k + c)(2k − 1 + c), p2k+1 = − (k + a)(k + c − b)

(2k + c)(2k + 1 + c)

where 2F1(a, b; c ; z) :=∑

n≥0(a)n(b)n

(c)nzn

n! , and (a)n := a(a + 1) · · · (a + n− 1).

[Khovanskii 1963] similar solution for

(1 + αz)zy ′ + (β + γz)y + δy 2 = εz , y(0) = −β/γ.

We look for C-fractions K∞m=1amz

1 with a2k and a2k+1 rational in k.

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 5 / 14

Objects

Formulas

Some general classes

[Gauss]

2F1(a, b; c ; z)

2F1(a, b + 1; c + 1; z)= 1 +

∞

Km=1

pmz

1

p2k = − (k + b)(k + c − a)

(2k + c)(2k − 1 + c), p2k+1 = − (k + a)(k + c − b)

(2k + c)(2k + 1 + c)

where 2F1(a, b; c ; z) :=∑

n≥0(a)n(b)n

(c)nzn

n! , and (a)n := a(a + 1) · · · (a + n− 1).

[Khovanskii 1963] similar solution for

(1 + αz)zy ′ + (β + γz)y + δy 2 = εz , y(0) = −β/γ.

We look for C-fractions K∞m=1amz

1 with a2k and a2k+1 rational in k.

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 5 / 14

Objects

Proofs

Proofs are performed:

by generalization/specialization,

sometimes indirectly, as in:

exp(x) = limk→∞(1 + x

k

)k.

They rely on:

3 terms linear recurrences,

the invariance of Riccati equations with rational coefficients

y ′(z) = py 2 + qy + r , p, q, r ∈ C(z),

under Mœbius transforms y2(z) :=az

1 + y(z)with a ∈ C?,

and q-analogues of these.

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 6 / 14

Equations as data structures

The equation as data structure

Proposition (Cauchy): the Riccati equation y ′(z) = py 2 + qy + r with y(0) = 0and p, q, r ∈ C(z) admits a unique power series solution y0. A sequence of powerseries (fn)n≥0 tends to y0 iff the remainder valuations satisfy:

val(f ′n − pf 2n − qfn − r)→∞

where val(∑

i≥0 cizi)

:= min{i ≥ 0 | ci 6= 0}.

Aim: one procedure

{y ′(z) = py 2 + qy + r , y(0) = 0} 7→ explicit C-fraction + proof

two steps: guessing the formula, and proving it,

the equation is non-linear,

classical tools are linear.

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 7 / 14

Equations as data structures

The equation as data structure

Proposition (Cauchy): the Riccati equation y ′(z) = py 2 + qy + r with y(0) = 0and p, q, r ∈ C(z) admits a unique power series solution y0. A sequence of powerseries (fn)n≥0 tends to y0 iff the remainder valuations satisfy:

val(f ′n − pf 2n − qfn − r)→∞

where val(∑

i≥0 cizi)

:= min{i ≥ 0 | ci 6= 0}.

Aim: one procedure

{y ′(z) = py 2 + qy + r , y(0) = 0} 7→ explicit C-fraction + proof

two steps: guessing the formula, and proving it,

the equation is non-linear,

classical tools are linear.

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 7 / 14

Equations as data structures

A guess and prove approach

riccati_to_cfrac({exp′(x) = exp(x), exp(0) = 1}, x = 0); exp(x) = 1 +x

1 +. . .

1 +

12(2k+1)

x

1 +

−12(2k+1)

x

. . .

Direct computation : (a0,...,a19)=(1,−12 ,

16 ,−1

6 ,...,1

10 ,−110 ,...,

138 ,−138 ).

Linear algebra : {−n2an+nan+1+n(n+3)an+2=0, (a0,a1,a2)=(1,− 12 ,

16 )},

a2k= 12(2k+1) , a2k+1= −1

2(2k+1) , k>0.

Proof : fn := 1 +Kni=0

aix1 −→n→∞

exp(x)?

. . .

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 8 / 14

Equations as data structures

A guess and prove approach

riccati_to_cfrac({exp′(x) = exp(x), exp(0) = 1}, x = 0); exp(x) = 1 +x

1 +. . .

1 +

12(2k+1)

x

1 +

−12(2k+1)

x

. . .

Direct computation : (a0,...,a19)=(1,−12 ,

16 ,−1

6 ,...,1

10 ,−110 ,...,

138 ,−138 ).

Linear algebra : {−n2an+nan+1+n(n+3)an+2=0, (a0,a1,a2)=(1,− 12 ,

16 )},

a2k= 12(2k+1) , a2k+1= −1

2(2k+1) , k>0.

Proof : fn := 1 +Kni=0

aix1 −→n→∞

exp(x)?

. . .

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 8 / 14

Equations as data structures

D-finiteness

A sequence is D-finite when it satisfies a linear recurrence with polynomialcoefficients. This recurrence is an effective data structure:

(an)n≥0 = 0 is decidable;

a recurrence can be computed for:

(an+1)n≥0 , (an − bn)n≥0 , (anbn)n≥0 ,(∑

i+j=n aibj

)n≥0

. . .

Example (squaring)

> rec:={ u(n+2) = (n+1)*u(n+1) + 2*u(n), u(0)=0, u(1)=1 }:

> gfun:-poltorec( u(n)^2, [rec], [u(n)], c(n) );

3 generators over Q(n) : u(n)2, u(n)u(n + 1), u(n + 1)2

{(8 n + 16

)c (n) +

(−2 n3 − 8 n2 − 14 n − 8

)c (n + 1) +(

−n3 − 5 n2 − 10 n − 8)

c (n + 2) + (n + 1) c (n + 3),

c (0) = 0, c (1) = 1, c (2) = 1}

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 9 / 14

Equations as data structures

D-finiteness

A sequence is D-finite when it satisfies a linear recurrence with polynomialcoefficients. This recurrence is an effective data structure:

(an)n≥0 = 0 is decidable;

a recurrence can be computed for:

(an+1)n≥0 , (an − bn)n≥0 , (anbn)n≥0 ,(∑

i+j=n aibj

)n≥0

. . .

Example (squaring)

> rec:={ u(n+2) = (n+1)*u(n+1) + 2*u(n), u(0)=0, u(1)=1 }:

> gfun:-poltorec( u(n)^2, [rec], [u(n)], c(n) );

3 generators over Q(n) : u(n)2, u(n)u(n + 1), u(n + 1)2

{(8 n + 16

)c (n) +

(−2 n3 − 8 n2 − 14 n − 8

)c (n + 1) +(

−n3 − 5 n2 − 10 n − 8)

c (n + 2) + (n + 1) c (n + 3),

c (0) = 0, c (1) = 1, c (2) = 1}

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 9 / 14

Equations as data structures

D-finiteness of the remainder

Recall that we wanted: val(f ′n − pf 2n − qfn − r)→∞.

for fn = 1 +a0x

1 +. . .

1 +anx

1

and p, q, r rational, we have fn = Pn/Qn with

{Pn = Pn−1 + anx Pn−2 (P−2,P−1) = (1, 0)

Qn = Qn−1 + anx Qn−2 (Q−2,Q−1) = (0, 1)

Lemma

Let Hn := Q2n(f ′n − pf 2

n − qfn − r), then fn → y0(z) ⇐⇒ val(H2n)→∞

Hn+i ∈ Vect(anP ′nQn, anPnQ ′n, Pn+1Q ′n+1, . . .) of finite dimension !

first recurrence for H2n,

Does not conclude

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 10 / 14

Equations as data structures

D-finiteness of the remainder

Recall that we wanted: val(f ′n − pf 2n − qfn − r)→∞.

for fn = 1 +a0x

1 +. . .

1 +anx

1

and p, q, r rational, we have fn = Pn/Qn with

{Pn = Pn−1 + anx Pn−2 (P−2,P−1) = (1, 0)

Qn = Qn−1 + anx Qn−2 (Q−2,Q−1) = (0, 1)

Lemma

Let Hn := Q2n(f ′n − pf 2

n − qfn − r), then fn → y0(z) ⇐⇒ val(H2n)→∞

Hn+i ∈ Vect(anP ′nQn, anPnQ ′n, Pn+1Q ′n+1, . . .) of finite dimension !

first recurrence for H2n,

Does not conclude =⇒ reduction of order . . . DEMO

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 10 / 14

Reduction of order

Principle: Guess and prove again !

Theorem (Dimension of the solutions space)

∀n, p0(n) 6= 0 =⇒ dim{(un)n≥0 , p0(n)un+d + · · ·+ pd(n)un = 0} = d.

Problem (2−n et 2n sont dans un bateau. . . )

Given {2an+2 − 5an+1 + 2an = 0, (a0, a1) = (1, 12 )},

show that limn→∞ an = 0.

guess a “small” recurrence: an+1 + αan = 0?(a0, a1) α = − 1

2 . {2bn+1 − bn = 0, b0 = 1}

it defines a new sequence (bn)n≥0 ,

we prove bn = an, by induction:

(b0, b1) = (a0, a1);2bn+2 − 5bn+1 + 2bn = (2bn+2 − bn+1)− 2(2bn+1 − bn) = 0.

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 11 / 14

Reduction of order

Miracle

all Riccati solutions in [Cuyt et alii, 2008] are covered,

and give a hypergeometric (Hn)n≥0 !

it generalizes to difference, and q-difference equations,=⇒ all explicit formulas in [Cuyt et alii, 2008]

Question: reciprocal ?

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 12 / 14

Classification

Classification

What are the

regular C-fractions

solutions of a Riccati equation with rational coefficients

with a hypergeometric remainder H2n ?

Symbolic study of the constraints. . . (DEMO ?). . . the formula of Khovanskii [Khovanskii 1963] for

(1 + αz)zy ′ + (β + γz)y + δy 2 = εz , y(0) = −β/γ

is the most general one ! (has a caracterization)

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 13 / 14

Classification

Classification

What are the

regular C-fractions

solutions of a Riccati equation with rational coefficients

with a hypergeometric remainder H2n ?

Symbolic study of the constraints. . . (DEMO ?)

. . . the formula of Khovanskii [Khovanskii 1963] for

(1 + αz)zy ′ + (β + γz)y + δy 2 = εz , y(0) = −β/γ

is the most general one ! (has a caracterization)

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 13 / 14

Classification

Classification

What are the

regular C-fractions

solutions of a Riccati equation with rational coefficients

with a hypergeometric remainder H2n ?

Symbolic study of the constraints. . . (DEMO ?). . . the formula of Khovanskii [Khovanskii 1963] for

(1 + αz)zy ′ + (β + γz)y + δy 2 = εz , y(0) = −β/γ

is the most general one ! (has a caracterization)

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 13 / 14

Conclusion

Conclusion

One procedure,

Riccati equation with rational coefficients7→ C-fraction with rational coefficients,proof with a quick and direct computation,

an implementation

prototype gfun:-ContFrac

generalizes by hand (to difference and q-difference equations),

and a classification. . .

to be continued

Maulat, Salvy (LIP, Lyon) Computing and proving continued fractions May 2015 14 / 14