S ebastien Maulat, ENS de Lyon Bruno Salvy,...
Transcript of S ebastien Maulat, ENS de Lyon Bruno Salvy,...
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