Download Michel Waldschmidt Abstract Applications of linear - IMJ-PRG

Abstract
December 2016
Linear recurrence sequences are ubiquitous. They occur in
biology, economics, computer science (analysis of algorithms),
digital signal processing.
Linear recurrence sequences,
exponential polynomials
and Diophantine approximation
We give a survey of this subject, together with connections
with linear combinations of powers and with exponential
polynomials, with an emphasis on arithmetic questions.
Michel Waldschmidt
Institut de Mathématiques de Jussieu — Paris VI
This lecture will include new results, arising from a joint work
with Claude Levesque, involving families of Diophantine
equations, with explicit examples related to some units of
L. Bernstein and H. Hasse.
http://webusers.imj-prg.fr/~michel.waldschmidt/
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Applications of linear recurrence sequences
2 / 119
Leonardo Pisano (Fibonacci)
Combinatorics
Elimination
Fibonacci sequence (Fn )n 0 ,
Symmetric functions
Leonardo Pisano (Fibonacci)
(1170–1250)
0, 1, 1, 2, 3, 5, 8, 13, 21,
Hypergeometric series
34, 55, 89, 144, 233, . . .
Language
is defined by
Communication, shift registers
F0 = 0, F1 = 1,
Finite di↵erence equations
Fn+2 = Fn+1 + Fn
Logic
for n
0.
http://oeis.org/A000045
Approximation
Pseudo–random sequences
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4 / 119
Fibonacci rabbits
Fibonacci squares
Fibonacci considered the growth of a rabbit population.
A newly born pair of rabbits,
a male and a female, are put
in a field. Rabbits are able to
mate at the age of one month
so that at the end of its
second month a female can
produce another pair of
rabbits ; rabbits never die and
a mating pair always produces
one new pair (one male, one female) every month from the
second month on. The puzzle that Fibonacci posed was : how
many pairs will there be in one year ?
Answer : F12 = 144.
http://mathforum.org/dr.math/faq/faq.golden.ratio.html
5 / 119
6 / 119
Fibonacci numbers in nature
Geometric construction of the
Fibonacci sequence
8
5
3
2
1
1
Ammonite (Nautilus shape)
2
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8 / 119
Reflections of a ray of light
Levels of energy of an electron of an atom of
hydrogen
Consider three parallel sheets of glass and a ray of light which
crosses the first sheet. Each time it touches one of the sheets,
it can cross it or reflect on it.
An atom of hydrogen can have three levels of energy, 0 at the
ground level when it does not move, 1 or 2. At each step, it
alternatively gains and looses some level of energy, either 1 or
2, without going sub 0 nor above 2. Let `n be the number of
di↵erent possible scenarios for this electron after n steps.
Denote by pn the number of di↵erent paths with the ray going
out of the system after n reflections.
p0 = 1,
We have `0 = 1 (initial state
level 0)
`1 = 2 : state 1 or 2, scenarios
01 or 02.
`2 = 3 : scenarios 010, 021 or
020.
`3 = 5 : scenarios 0101, 0102,
0212, 0201 or 0202.
In general, `n = Fn+2 .
p1 = 2,
p2 = 3,
p3 = 5.
In general, pn = Fn+2 .
9 / 119
Rhythmic patterns
10 / 119
Fibonacci sequence and the Golden ratio
The Fibonacci sequence appears in Indian mathematics, in
connection with Sanskrit prosody. Several Indian scholars,
Pingala (200 BC), Virahanka (c. 700 AD), Gopāla (c. 1135),
and the Jain scholar Hemachandra (c. 1150) studied rhythmic
patterns that are formed from one-beat notes (or short
syllables, ti in Morse Alphabet) : • and two-beat notes (or
long syllables, ta ta in Morse) : ⌅⌅.
1 beat, 1 pattern :
•
2 beats, 2 patterns : •• and ⌅⌅
3 beats, 3 patterns : • • •, •⌅⌅ and ⌅⌅•
4 beats, 5 patterns :
For n
0, the Fibonacci number Fn is the nearest integer to
1
p
5
where
n
is the Golden Ratio :
http://oeis.org/A001622
p
1+ 5
Fn+1
=
= lim
= 1.6180339887499 . . .
n!1 Fn
2
which satisfies
=1+
• • ••, ⌅⌅ • •, •⌅⌅•, • • ⌅⌅, ⌅⌅⌅⌅
,
1
·
n beats, Fn+1 patterns
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Binet’s formula
For n
n
(1 +
p
(
p
)
Formula of A. De Moivre (1718, 1730), Daniel Bernoulli
(1726), L. Euler (1728, 1765), J.P.M. Binet (1843) : for n
p !n
p !n
1
1+ 5
1
1
5
p
Fn = p
.
2
2
5
5
Jacques Philippe Marie Binet
(1843)
0,
Fn =
=
The so–called Binet Formula
n
5
5)n (1
p
2n 5
p
5)n
,
p
1+ 5
=
,
2
X 2 X 1 = (X
Abraham de
Moivre
(1667–1754)
1
=
1
p
2
)(X +
5
Daniel
Bernoulli
(1700–1782)
Leonhard
Euler
(1707–1783)
0,
Jacques P.M.
Binet
(1786–1856)
,
1
).
13 / 119
14 / 119
Fn is the nearest integer to
Generating series
Generating series of the Fibonacci
A single series encodes all the Fibonacci sequence :
X
Fn X n = X + X 2 + 2X 3 + 3X 4 + 5X 5 + · · · + Fn X n + · · ·
1
p n.
sequence
5
n 0
Fact : this series is the Taylor expansion of a rational fraction :
X
X
Fn X n =
·
1 X X2
n 0
X
X
X 2 in the right hand side
X + X 2 + 2X 3 + 3X 4 + · · · + Fn X n + · · · =
Proof : the product
(X + X 2 + 2X 3 + 3X 4 + 5X 5 + 8X 6 + · · · )(1
Remark. The denominator 1
of
X 2)
is X 2 f (X 1 ), where f (X) = X 2
polynomial of the Golden ratio .
is a telescoping series
X
1
X
X X2
1 is the irreducible
X + X 2 + 2X 3 + 3X 4 + 5X 5 + 8X 6 + · · ·
X2
X 3 2X 4 3X 5 5X 6 · · ·
X3
X 4 2X 5 3X 6 · · ·
= X.
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16 / 119
Fibonacci and powers of matrices
Characteristic polynomial
The Fibonacci linear recurrence relation Fn+2 = Fn+1 + Fn for
n 0 can be written
✓
◆ ✓
◆✓
◆
Fn+1
0 1
Fn
=
.
Fn+2
1 1
Fn+1
The characteristic polynomial of the matrix
✓
◆
0 1
A=
1 1
By induction one deduces, for n 0,
✓
◆ ✓
◆n ✓ ◆
Fn
0 1
0
=
.
Fn+1
1 1
1
is
det(XI
An equivalent formula is, for n 1,
✓
◆n ✓
◆
0 1
Fn 1 Fn
=
.
1 1
Fn Fn+1
A) = det
✓
X
1
1 X 1
◆
= X2
X
which is the irreducible polynomial of the Golden ratio
1,
.
17 / 119
Fibonacci sequence and the Golden ratio
(continued)
Fibonacci sequence and Hilbert’s 10th problem
Yuri Matiyasevich (1970) showed that there is a polynomial P
in n, m, and a number of other variables x, y, z, . . . having the
property that n = F2m i↵ there exist integers x, y, z, . . . such
that P (n, m, x, y, z, . . . ) = 0.
For n 1, n 2 Z[ ] = Z + Z is a linear combination of 1
and with integer coefficients, namely
n
= Fn
1
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This completed the proof of
the impossibility of the tenth
of Hilbert’s problems (does
there exist a general method
for solving Diophantine
equations ?) thanks to the
previous work of Hilary
Putnam, Julia Robinson and
Martin Davis.
+ Fn
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The Fibonacci Quarterly
Lucas sequence
http://oeis.org/000032
The Lucas sequence (Ln )n 0 satisfies the same recurrence
relation as the Fibonacci sequence, namely
for n
Ln+2 = Ln+1 + Ln
The Fibonacci sequence
satisfies a lot of very
interesting properties. Four
times a year, the Fibonacci
Quarterly publishes an issue
with new properties which
have been discovered.
0,
only the initial values are di↵erent :
L0 = 2, L1 = 1.
The sequence of Lucas numbers starts with
2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, . . .
A closed form involving the Golden ratio
Ln =
n
+(
from which it follows that for n
to n .
)
n
is
,
2, Ln is the nearest integer
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François Édouard Anatole Lucas (1842 - 1891)
22 / 119
Generating series of the Lucas sequence
Edouard Lucas is best known
for his results in number
theory. He studied the
Fibonacci sequence and
devised the test for Mersenne
primes still used today.
The generating series of the Lucas sequence
X
Ln X n = 2 + X + 3X 2 + 4X 3 + · · · + Ln X n + · · ·
n 0
is nothing else than
2 X
·
1 X X2
http://www-history.mcs.st-andrews.ac.uk/history/
Mathematicians/Lucas.html
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The Lucas sequence and power of matrices
Perrin sequence
The Perrin sequence (also called skiponacci sequence) is the
linear recurrence sequence (Pn )n 0 defined by
From the linear recurrence relation Ln+2 = Ln+1 + Ln one
deduces, (as we did for the Fibonacci sequence), for n 0,
✓
◆ ✓
◆✓
◆
Ln+1
0 1
Ln
=
,
Ln+2
1 1
Ln+1
hence
✓
Ln
Ln+1
◆
=
✓
0 1
1 1
http://oeis.org/A001608
for n
Pn+3 = Pn+1 + Pn
0,
with the initial conditions
P0 = 3, P1 = 0, P2 = 2.
◆n ✓ ◆
2
.
1
It starts with
Any one of the three sequences (Fn )n 0 , (Ln )n 0 , ( n )n
can be written as a linear combination of the two others.
3, 0, 2, 3, 2, 5, 5, 7, 10, 12, 17, 22, 29, 39, 51, 68, . . .
0
François Olivier Raoul Perrin (1841-1910) :
https://en.wikipedia.org/wiki/Perrin_number
25 / 119
Plastic (or silver) constant
https://oeis.org/A060006
The ratio of successive terms in the Perrin sequence
approaches the plastic number %, which is the minimal
Pisot–Vijayaraghavan number, real root of
x
3
x
26 / 119
Perrin sequence and the plastic constant
Decompose the polynomial X 3
factors over C
X3
1
1 = (X
%)(X
1 into irreducible
⇢)(X
⇢)
and over R
which has a value of approximately 1.324718.
X3
This constant is equal to
p
p
p
3
3
108 + 12 69 + 108
%=
6
X
X
p
12 69
X
1 = (X
%)(X 2 + %X + % 1 ).
Hence ⇢ and ⇢ are the roots of X 2 + %X + % 1 . Then, for
n 0,
P n = %n + ⇢n + ⇢n .
·
It follows that, for n
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0, Pn is the nearest integer to %n .
28 / 119
Generating series of the Perrin sequence
Perrin sequence and power of matrices
The generating series of the Perrin sequence
X
Pn X n = 3 + 2X 2 + 3X 3 + 2X 4 + · · · + Pn X n + · · ·
From
Pn+3 = Pn+1 + Pn
we deduce
n 0
0
is nothing else than
3 X2
·
1 X2 X3
Hence
The denominator 1 X 2 X 3 is X 3 f (X 1 ) where
f (X) = X 3 X 1 is the irreducible polynomial of %.
1 0
10
1
Pn+1
0 1 0
Pn
@Pn+2 A = @0 0 1A @Pn+1 A .
Pn+3
1 1 0
Pn+2
0
1 0
1n 0 1
Pn
0 1 0
3
@Pn+1 A = @0 0 1A @0A .
Pn+2
1 1 0
2
29 / 119
Characteristic polynomial
The characteristic polynomial of
0
0
A = @0
1
is
det(XI
0
X
@
A) = det 0
1
30 / 119
Perrin’s remark
the matrix
1
1 0
0 1A
1 0
1
X
1
1
0
1A = X 3
X
X
R. Perrin L’intermédiaire des mathématiciens, Query 1484, v.6,
76–77 (1899).
1,
The website www.Perrin088.org maintained by Richard Turk is
devoted to Perrin numbers. See OEISA113788.
which is the irreducible polynomial of the plastic constant %.
31 / 119
32 / 119
Perrin pseudoprimes
https://oeis.org/A013998
Padovan sequence
https://oeis.org/A000931
The Padovan sequence (pn )n
If p is prime, then p divides Pp .
0
satisfies the same recurrence
pn+3 = pn+1 + pn
The smallest composite n such that n divides Pn is
5212 = 271441.
The number P271441 has 33 150 decimal digits (the number c
which satisfies 10c = %271441 is c = 271441(log %)/(log 10)).
as the Perrin sequence but has di↵erent initial values :
p0 = 1,
p1 = p2 = 0.
It starts with
Also for the composite number n = 904631 = 7 ⇥ 13 ⇥ 9941,
the number n divides Pn .
1, 0, 0, 1, 0, 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, . . .
Jon Grantham has proved that there are infinitely many Perrin
pseudoprimes.
Richard Padovan
http://mathworld.wolfram.com/LinearRecurrenceEquation.html
33 / 119
Generating series and power of matrices
1 + X 3 + X 5 + · · · + pn X n + · · · =
For n
34 / 119
Padovan triangles
1 X2
·
1 X2 X3
pn = pn
0,
0
1n 0
1
0 1 0
pn pn+2 pn+1
@0 0 1A = @pn+1 pn+3 pn+2 A .
1 1 0
pn+2 pn+4 pn+3
2
+ pn
3
pn
1
= pn
3
+ pn
4
pn
2
= pn
4
+ pn
5
Hence
pn
pn = pn
35 / 119
1
+ pn
pn
1
= pn
5
5
36 / 119
Padovan triangles
Padovan triangles vs Fibonacci squares
C 2 curve, not C 3
C 1 curve, not C 2
37 / 119
Padovan, Euler, Zagier and Brown
38 / 119
Narayana sequence
https://oeis.org/A000930
Narayana sequence is defined by the recurrence relation
For n 0, the number of compositions s = (s1 , . . . , sk ) with
si 2 {2, 3} and s1 + · · · + sk = n is pn+3 . This is (an upper
bound for) the dimension of the space spanned by the multiple
zeta values of weight n of Euler and Zagier.
Cn+3 = Cn+2 + Cn
with the initial values C0 = 2, C1 = 3, C2 = 4.
It starts with
2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, . . .
Leonhard Euler
Don Zagier
Real root of x3 x2 1
s
s
p
p
93
3 93
3 29 + 3
3 29
+
+1
2
2
= 1.465571231876768 . . .
3
Francis Brown
39 / 119
40 / 119
Generating series and power of matrices
2 + 3X + 4X 2 + 6X 3 + · · · + Cn X n + · · · =
For n
Narayana’s cows
2 + X + X2
·
1 X X3
Narayana was an Indian mathematician in the 14th century
who proposed the following problem :
A cow produces one calf every year. Beginning in its fourth
year each calf produces one calf at the beginning of each year.
How many calves are there altogether after, for example, 17
years ?
7,
0
1n 0
0 1 0
Cn
@0 0 1 A = @C n
1 0 1
Cn
6
5
4
Cn
Cn
Cn
7
6
5
1
Cn 5
Cn 4 A .
Cn 3
41 / 119
Music :
42 / 119
Narayana’s cows
http://www.pogus.com/21033.html
In working this out, Tom Johnson found a way to translate
this into a composition called Narayana’s Cows.
Music : Tom Johnson
Saxophones : Daniel Kientzy
http://webusers.imj-prg.fr/~michel.waldschmidt/
43 / 119
Year
1 2
3
4
5
6
7
8
9
10
11
12
13
14
Original
Cow
1 1
1
1
1
1
1
1
1
1
1
1
1
1
Second
generation
1 2
3
4
5
6
7
8
9
10
11
12
13
14
Third
generation
0 0
0
1
3
6
10
15
21
28
36
45
55
66
Fourth
generation
0 0
0
0
0
0
1
4
10
20
35
56
84
120
Fifth
generation
0 0
0
0
0
0
0
0
0
1
5
15
35
70
Sixth
generation
0 0
0
0
0
0
0
0
0
0
0
0
1
6
Total
2 3
4
6
9
13
19
28
41
60
88
129
189
277
44 / 119
Jean-Paul Allouche and Tom Johnson
Cows, music and morphisms
Jean-Paul Allouche and Tom Johnson
• Narayana’s Cows and Delayed Morphisms
In 3èmes Journées d’Informatique Musicale (JIM ’96), Ile de
Tatihou, Les Cahiers du GREYC (1996 no. 4), pages 2-7, May
1996.
http://kalvos.org/johness1.html
• Finite automata and morphisms in assisted musical
composition,
Journal of New Music Research, no. 24 (1995), 97 – 108.
http://www.tandfonline.com/doi/abs/10.1080/
09298219508570676
http://web.archive.org/web/19990128092059/www.swets.
nl/jnmr/vol24_2.html
http://webusers.imj-prg.fr/~jean-paul.allouche/
bibliorecente.html
http://www.math.jussieu.fr/~allouche/johnson1.pdf
45 / 119
Linear recurrence sequences : definitions
Linear recurrence sequences : examples
A linear recurrence sequence is a sequence of numbers
u = (u0 , u1 , u2 , . . . ) for which there exist a positive integer d
together with numbers a1 , . . . , ad with ad 6= 0 such that, for
n 0,
(?)
un+d = a1 un+d
1
• Constant sequence : un = u0 .
Linear recurrence sequence of order 1 : un+1 = un .
Characteristic polynomial : f (X) = X 1.
Generating series :
X
+ · · · + ad u n .
Here, a number means an element of a field K of zero
characteristic.
Given a = (a1 , . . . , ad ) 2 Kd , the set Ea of linear recurrence
sequences u = (un )n 0 satisfying (?) is a K–vector subspace
of dimension d of the space KN of all sequences .
The characteristic (or companion) polynomial of the linear
recurrence is
f (X) = X d
a1 X d
1
···
46 / 119
n 0
Xn =
1
1
X
·
• Geometric progression : un = u0 n .
Linear recurrence sequence of order 1 : un = un 1 .
Characteristic polynomial f (X) = X
.
Generating series :
X
u0
u0 n X n =
·
1
X
n 0
ad .
47 / 119
48 / 119
Linear recurrence sequences : examples
Linear recurrence sequences : examples
• un = n. Linear recurrence sequence of order 2 :
n + 2 = 2(n + 1)
• un = f (n), f polynomial of degree d. Linear recurrence
sequence of order d + 1.
n.
Characteristic polynomial
Proof. The sequences
f (X) = X 2
2X + 1 = (X
1)2 .
(f (n))n 0 ,
Generating series
X
nX n =
n 0
1
··· ,
(f (n + 1))n 0 ,
(f (n + k))n
are K–linearly independent in KN for k = d
dependent for k = d .
1
·
2X + X 2
0
1 and linearly
A basis of the space of polynomials of degree d is given by the
d + 1 polynomials
Power of matrices :
✓
◆n ✓
◆
0 1
n+1
n
=
.
1 2
n
n+1
f (X), f (X + 1), . . . , f (X + d).
49 / 119
Linear sequences which are ultimately recurrent
Order of a linear recurrence sequence
If u = (un )n 0 satisfies the linear recurrence, the characteristic
polynomial of which is f , then, for any monic polynomial
g 2 K[X] with g(0) 6= 0, this sequence u also satisfies the
linear recurrence, the characteristic polynomial of which is f g.
Example : for g(X) = X
with 6= 0, from
The sequence
(1, 0, 0, . . . )
is not a linear recurrence sequence.
The condition
(?)
un+1 = un
is satisfied only for n
50 / 119
un+d
a1 un+d
1
···
ad u n = 0
we deduce
1.
un+d+1
The relation
un+2 = un+1 + 0un
with d = 2, ad = 0 does not fulfill the requirement ad 6= 0.
a1 un+d · · · ad un+1
(un+d a1 un+d 1 · · · ad un ) = 0.
The order of a linear recurrence sequence is the smallest d
such that (?) holds for all n 0.
51 / 119
52 / 119
Generating series of a linear recurrence sequence
Let u = (un )n
(?)
0
be a linear recurrence sequence
un+d = a1 un+d
1
+ · · · + ad u n
Assume
for
n
0
un+d = a1 un+d
with characteristic polynomial
f (X) = X d
Denote by f
Then
1
···
1
X
ad .
1
X
1
)=1
a1 X
···
1
+ · · · + ad u n
un X n =
n=0
the reciprocal polynomial of f :
f (X) = X d f (X
Then
a1 X d
Generating series of a linear recurrence sequence
for
n
r(X)
·
f (X)
Proof. Comparing the coefficients of X n for n
ad X d .
f (X)
r(X)
un X =
,
f
(X)
n=0
n
1
X
0.
d shows that
un X n
n=0
is a polynomial of degree less than d
where r is a polynomial of degree less than d determined by
the initial values of u.
53 / 119
Taylor coefficients of rational functions
54 / 119
Matrix notation for a linear recurrence sequence
Conversely, the coefficients the Taylor expansion of any
rational fraction a(X)/b(X) with deg a < deg b and b(0) 6= 0
satisfies the recurrence relation with characteristic polynomial
f 2 K[X] given by f (X) = b (X).
The linear recurrence sequence
(?)
un+d = a1 un+d
1
+ · · · + ad u n
for
n
0
can be written
Therefore a sequence u = (un )n 0 satisfies the recurrence
relation (?) with characteristic polynomial f 2 K[X] if and
only if
1
X
r(X)
un X n =
,
f (X)
n=0
0
1
0
0
1
0
un+1
B0
0
1
Bun+2 C B
B
C B ..
.
..
..
B .. C = B .
.
@ . A B
@0
0
0
un+d
ad ad 1 ad 2
where r is a polynomial of degree less than d determined by
the initial values of u.
55 / 119
···
···
..
.
···
···
1
1
0 0
un
C
0CB
C
.. C B un+1 C .
B .. C
.C
C@ . A
1A
un+d 1
a1
56 / 119
Matrix notation for a linear recurrence sequence
Powers of matrices
Un+1 = AUn
with
0
B
B
Un = B
@
1
un
un+1
..
.
un+d
1
C
C
C,
A
The determinant of Id X
A) is nothing but
0
0
1
0
B0
0
1
B
B ..
.
..
..
A=B.
.
B
@0
0
0
ad ad 1 ad 2
f (X) = X d
···
···
..
.
···
···
Let A = (aij )1i,jd 2 GLd⇥d (K) be a d ⇥ d matrix with
coefficients in K and nonzero determinant. For n 0, define
1
0
0C
C
.. C .
.C
C
1A
a1
(n)
An = aij
1i,jd
.
(n)
Then each of the d2 sequences aij n 0 , (1  i, j  d) is a
linear recurrence sequence. The roots of the characteristic
polynomial of these linear recurrences are the eigenvalues of A.
A (the characteristic polynomial of
a1 X d
1
···
In particular the sequence Tr(An ) n 0 satisfies the linear
recurrence, the characteristic polynomial of which is the
characteristic polynomial of the matrix A.
ad ,
the characteristic polynomial of the linear recurrence sequence.
By induction
Un = An U0 .
57 / 119
Conversely :
58 / 119
Linear recurrence sequences : simple roots
A basis of Ea over K is obtained by attributing to the initial
values u0 , . . . , ud 1 the values given by the canonical basis of
Kd .
Given in K⇥ , a necessary and sufficient condition for a
sequence ( n )n 0 to satisfy (?) is that is a root of the
characteristic polynomial
Given a linear recurrence sequence u 2 KN , there exist an
integer d 1 and a matrix A 2 GLd (K) such that, for each
n 0,
(n)
un = a11 .
f (X) = X d
The characteristic polynomial of A is the characteristic
polynomial of the linear recurrence sequence.
a1 X d
1
If this polynomial has d distinct roots
Everest G., van der Poorten A., Shparlinski I., Ward T. –
Recurrence sequences, Mathematical Surveys and Monographs (AMS,
2003), volume 104.
f (X) = (X
1 ) · · · (X
···
ad .
1, . . . ,
d ),
d
i
in K,
6=
j,
then a basis of Ea over K is given by the d sequences (
i = 1, . . . , d.
59 / 119
n
i )n 0 ,
60 / 119
Linear recurrence sequences : double roots
Linear recurrence sequences : multiple roots
The characteristic polynomial of the linear recurrence
2
un = 2 u n 1
un 2 is X 2 2 X + 2 = (X
)2 with a
double root .
The sequence (n
n
n
n
)n
0
In general, when the characteristic polynomial splits as
satisfies
= 2 (n
1)n
n 1
2
(n
2)
n 2
Xd
a1 X d
.
1
···
ad =
Ỳ
(X
i)
ti
,
i=1
a basis of Ea is given by the d sequences
A basis of Ea for a1 = 2 , a2 =
sequences ( n )n 0 , (n n )n 0 .
2
is given by the two
(nk
n
i )n 0 ,
0  k  ti
1,
1  i  `.
Given 2 K⇥ , a necessary and sufficient condition for the
sequence n n to satisfy the linear recurrence relation (?) is
that is a root of multiplicity 2 of f (X).
61 / 119
Polynomial combinations of powers
62 / 119
Sum of polynomial combinations of powers
The sum and the product of any two linear recurrence
sequences are linear recurrence sequences.
If u1 and u2 are two linear recurrence sequences of
characteristic polynomials f1 and f2 respectively, then u1 + u2
satisfies the linear recurrence, the characteristic polynomial of
which is
f1 f2
·
gcd(f1 , f2 )
The set [a Ea of all linear recurrence sequences with
coefficients in K is a sub–K–algebra of KN .
Given polynomials p1 , . . . , p` in K[X] and elements
in K⇥ , the sequence
p1 (n)
n
1
+ · · · + p` (n)
1, . . . ,
`
n
` n 0
is a linear recurrence sequence.
Fact : any linear recurrence sequence is of this form.
63 / 119
64 / 119
Product of polynomial combinations of powers
Linear recurrence sequences and
Brahmagupta–Pell–Fermat Equation
If the characteristic polynomials of the two linear recurrence
sequences u1 and u2 are respectively
f1 (T ) =
Ỳ
0
(T
j)
tj
and f2 (T ) =
j=1
`
Y
(T
Let d be a positive integer, not a square. The solutions
(x, y) 2 Z ⇥ Z of the Brahmagupta–Pell–Fermat Equation
0 t0k
k) ,
x2
k=1
form a sequence (xn , yn )n2Z defined by
p
p
xn + dyn = (x1 + dy1 )n .
then u1 u2 satisfies the linear recurrence, the characteristic
polynomial of which is
0
`
Ỳ Y
(T
dy 2 = ±1
From
0 tj +t0k 1
.
j k)
2xn = (x1 +
j=1 k=1
p
dy1 )n + (x1
p
dy1 )n
we deduce that (xn )n 0 is a linear recurrence sequence. Same
for yn , and also for n  0.
65 / 119
Doubly infinite linear recurrence sequences
Discrete version of linear di↵erential equations
A sequence u 2 KN can be viewed as a linear map N ! K.
Define the discrete derivative D by
A sequence (un )n2Z indexed by Z is a linear recurrence
sequence if it satisfies
(?)
un+d = a1 un+d
1
66 / 119
Du : N !
K
n 7 ! un+1 un .
+ · · · + ad u n .
for all n 2 Z.
A sequence u 2 KN is a linear recurrence sequence if and only
if there exists Q 2 K[T ] with Q(1) 6= 1 such that
Recall ad 6= 0.
Q(D)u = 0.
Linear recurrence sequences are a discrete version of linear
di↵erential equations with constant coefficients.
Such a sequence is determined by d consecutive values.
The condition Q(1) 6= 0 reflects ad 6= 0 – otherwise one gets ultimately
recurrent sequences.
67 / 119
68 / 119
97th Indian Science Congress, 2010
A. P. J. Abdul Kalam (1931-2015)
Public Lecture during the 97th Indian Science Congress,
Thiruvananthapuram – 4 January 2010 Thiruvananthapuram
A.K. Agarwal
Invited by Ashok Agrawal to
the 97th Indian Science
Congress in
Thiruvananthapuram
(Trivandrum, Kerala), January
3-7, 2010.
• Lecture on Number Theory
Challenges of 21st Century
Basic research is vital for
enhancing national and
international competitiveness
http://www.abdulkalam.com/kalam/theme/jsp/guest/
content-display.jsp
69 / 119
Kerala 2010
Sudhir Ghorpade
70 / 119
KSOM 2010
Jugal K. Verma
January 8, 2010, Calicut = Kozhikode (Kerala) The Kerala
School of Mathematics (KSoM)
Ambar Vijayatkumar
January 9-10, 2010, Cochin = Kochi (Kerala) Department of
Mathematics, Cochin University of Science and Technology
CUSAT
A. J. Parameswaran, Director of the Kerala School of
Mathematical Science (KSOM) in Kozikhode (Calicut)
71 / 119
72 / 119
KSOM 2010
A dynamical system
Work on dynamical systems by A. J. Parameswaran and S.G.
Dani
Let V be a finite dimensional vector space over a field of zero
characteristic, H an hyperplane of V , f : V ! V an
endomorphism (linear map) and x an element in V .
Theorem. If there exist infinitely many n 1 such that
f n (x) 2 H, then there is an (infinite) arithmetic progression of
n for which it is so.
A. J. Parameswaran
S.G. Dani
73 / 119
Skolem – Mahler – Lech Theorem
Wolfgang M. Schmidt
Theorem (Skolem 1934 – Mahler 1935 – Lech 1953). Given a
linear recurrence sequence, the set of indices n 0 such that
un = 0 is a finite union of arithmetic progressions.
Linear recurrence sequence :
un+d = a1 un+d
1
+ · · · + ad u n ,
n
74 / 119
0
Thue – Siegel – Roth – Schmidt,
Schmidt’s Subspace Theorem. The generalized S–unit
Theorem
Let K be a field of characteristic zero, let G be a finitely
multiplicative subgroup of the multiplicative group
K⇥ = K \ {0} and let n 2. Then the equation
(ad 6= 0).
Characteristic polynomial :
Xd
a1 X d
1
un =
···
ad =
j 1
X̀ tX
Ỳ
u1 + u2 + · · · + un = 1,
(X
j)
where the values of the unknowns u1 , u2 , · · · , un are in G for
which no nontrivial subsum
X
ui
;=
6 I ⇢ {1, . . . , n}
tj
j=1
cij ni
n
j.
i2I
j=1 i=0
vanishes, has only finitely many solutions.
75 / 119
76 / 119
Schmidt’s subspace Theorem
Balu’s 60’s Birthday, 2011
December 15 - 20, 2011 : HRI : International Meeting on
Number Theory 2011 celebrating the 60th Birthday of
Professor R. Balasubramanian.
Pietro Corvaja
M. Manickam, director of KSOM.
Wolfgang M. Schmidt
Pietro Corvaja
December 16, 2011 : lecture on Families of Thue-Mahler
equations.
Umberto Zannier
77 / 119
Joint work with Claude Levesque
78 / 119
http://www.ramanujanmathsociety.org/publications/rms-lecture-notes-series
http://arxiv.org/abs/1505.06653
Solving simultaneously Thue
Diophantine equations :
almost totally imaginary case
Proceedings of the
International Meeting on
Number Theory HRI 2011, in
honor of R. Balasubramanian.
Ramanujan Mathematical Society, Lecture Notes Series 23,
Highly composite : papers in number theory, (2016), 137–156.
Editors Kumar Murty, Ravindranathan Thangadurai.
http://www.ramanujanmathsociety.org/publications/
rms-lecture-notes-series
79 / 119
80 / 119
KSOM 2013
Reference
Workshop number theory and dynamical systems in KSOM
(Director M. Manickam) in February 2013
M. Waldschmidt. Diophantine approximation with
applications to dynamical systems. Proceedings of the
International Conference on Pure and Applied Mathematics
ICPAM–LAE 2013, South Pacific Journal of Pure and Applied
Mathematics, vol. 1, No 2 (2014), 1–18.
Yann Bugeaud
Pietro Corvaja
S.G. Dani
81 / 119
Skolem – Mahler – Lech Theorem
A dynamical system
Theorem (Skolem 1934 – Mahler 1935 – Lech 1953). Given a
linear recurrence sequence, the set of indices n 0 such that
un = 0 is a finite union of arithmetic progressions.
Thoralf Albert Skolem
(1887 – 1963)
Kurt Mahler
(1903 – 1988)
82 / 119
Let V be a finite dimensional vector space over a field of zero
characteristic, W a subspace of V , f : V ! V an
endomorphism (linear map) and x an element in V .
Christer Lech
Corollary of the Skolem – Mahler – Lech Theorem. The
set of n 0 such that f n (x) 2 W is a finite union of
arithmetic progressions.
By induction, it suffices to consider the case where W = H is
an hyperplane of V .
An arithmetic progression is a set of positive integers of the
form {n0 , n0 + k, n0 + 2k, . . .}. Here, we allow k = 0.
83 / 119
84 / 119
Idea of the proof of the corollary
Theorem of Cayley – Hamilton
Arthur Cayley
(1821 – 1895)
Choose a basis of V . The endomorphism f is given by a
square d ⇥ d matrix A, where d is the dimension of V .
Consider the characteristic polynomial of A, say
Xd
ad 1 X d
1
···
a1 X
Sir William Rowan Hamilton
(1805 – 1865)
a0 .
By the Theorem of Cayley – Hamilton,
Ad = ad 1 Ad
1
Hence, for n
+ · · · + a1 A + a0 I d
0,
An+d = ad 1 An+d
where Id is the identity d ⇥ d matrix.
1
+ · · · + a1 An+1 + a0 An .
It follows that each entry aij (n), 1  i, j  d, satisfies a linear
recurrence relation, the same for all i, j.
85 / 119
Hyperplane membership
86 / 119
Remark on the theorem of Skolem–Mahler–Lech
Let b1 x1 + · · · + bd xd = 0 be an equation of the hyperplane H
in the selected basis of V . Let t b denote the 1 ⇥ d matrix
(b1 , . . . , bd ) (transpose of a column matrix b). Using the
notation v for the d ⇥ 1 (column) matrix given by the
coordinates of an element v in V , the condition v 2 H can be
written t b v = 0.
T.A. Skolem treated the case K = Q of in 1934
K. Mahler the case K = Q, the algebraic closure of Q, in 1935
Let x be an element in V and x the d ⇥ 1 (column) matrix
given by its coordinates. The condition f n (x) 2 H can now be
written
t
bAn x = 0.
The general case was settled by C. Lech in 1953.
The entry un of the 1 ⇥ 1 matrix t bAn x satisfies a linear
recurrence relation, hence, the Skolem – Mahler – Lech
Theorem applies.
87 / 119
88 / 119
Finite characteristic
Polynomial-linear recurrence relation
C. Lech pointed out in 1953 that such a result may not hold if
the characteristic of K is positive : he gave as an example the
sequence un = (1 + x)n xn 1, a third-order linear
recurrence over the field of rational functions in one variable
over the field Fp with p elements, where un = 0 for
n 2 {1, p, p2 , p3 , . . .}. A substitute is provided by a result of
Harm Derksen (2007), who proved that the zero set in
characteristic p is a p–automatic sequence. Further results by
Boris Adamczewski and Jason Bell.
A generalization of the Theorem of Skolem–Mahler–Lech has
been achieved by Jason P. Bell, Stanley Burris and Karen Yeats
who prove that the same conclusion holds if the sequence
(un )n 0 satisfies a polynomial-linear recurrence relation
un =
d
X
Pi (n)un
i
i=1
where d is a positive integer and P1 , . . . , Pd are polynomials
with coefficient in the field K of zero characteristic, provided
that Pd (x) is a nonzero constant.
Harm Derksen
Boris Adamczewski
Jason Bell
89 / 119
Algebraic maps, algebraic groups
Open problem
One main open problem related with Theorem of
Skolem–Mahler–Lech is that it is not e↵ective : explicit upper
bounds for the number of arithmetic progressions, depending
only on the order d of the linear recurrence sequence, are
known (W.M. Schmidt, U. Zannier), but no upper bound for
the arithmetic progressions themselves is known. A related
open problem raised by T.A. Skolem and C. Pisot is :
Given an integer linear recurrence sequence, is the
truth of the statement “xn 6= 0 for all n” decidable
in finite time ?
There are also analogues of the Theorem of
Skolem–Mahler–Lech for algebraic maps on varieties (Jason
Bell).
A version of the Skolem–Mahler–Lech Theorem for any
algebraic group is due to Umberto Zannier.
Jason Bell
90 / 119
T. Tao, E↵ective Skolem Mahler Lech theorem. In “Structure and
Randomness : pages from year one of a mathematical blog”, American
Mathematical Society (2008), 298 pages.
Umberto Zannier
http://terrytao.wordpress.com/2007/05/25/open-question-effective-skolem-mahler-lech-theorem/
91 / 119
92 / 119
Zeros of linear recurrence sequences
Zeros of linear recurrence sequences
Jean Berstel et Maurice Mignotte. – Deux propriétés
décidables des suites récurrentes linéaires Bulletin de la
S.M.F., tome 104 (1976), p. 175-184.
Maurice Mignotte Propriétés arithmétiques des suites
récurrentes linéaires. Besançon, 1989
http://pmb.univ-fcomte.fr/1989/Mignotte.pdf
http://www.numdam.org/item?id=BSMF_1976__104__175_0
Given a linear recurrence sequence with integer coefficients ;
are there finitely or infinitely many zeroes ?
E. Bavenco↵e and J-P. Bézivin Une famille remarquable de
suites recurrentes lineaires. – Monatshefte für Mathematik,
(1995) 120 3, 189–203
Philippe Robba. – Zéros de suites récurrentes linéaires. Groupe
de travail d’analyse ultramétrique (1977-1978) Volume : 5,
page 1-5.
Karim Samake. – Suites récurrentes linéaires, problème
d’e↵ectivité. Inst. de Recherche Math. Avancée, 1996 - 62
pages
L. Cerlienco, M. Mignotte, F. Piras. Suites récurrentes
linéaires. Propriétés algébriques et arithmétiques.
L’Enseignement Mathématique 33 (1987).
93 / 119
Reference
94 / 119
Berstel’s sequence
Everest, Graham ; van der Poorten, Alf ;
Shparlinski, Igor ; Ward, Tom – Recurrence
sequences, Mathematical Surveys and Monographs (AMS,
2003), volume 104.
1290 references.
0, 0, 1, 2, 0,
4, 0, 16, 16,
http://oeis.org/A007420
32,
64, 64, 256, 0,
768, . . .
b0 = b1 = 0, b2 = 1,
bn+3 = 2bn+2 4bn+1 + 4bn
for n 0.
Graham Everest
Linear recurrence sequence of
order 3 with exactly 6 zeros :
n = 0, 1, 4, 6, 13, 52.
Alf van der Poorten
Jean Berstel
Igor Shparlinski
http://www-igm.univ-mlv.fr/~berstel/
Tom Ward
95 / 119
96 / 119
Ternary linear recurrences
Edgard Bavenco↵e and Jean-Paul Bézivin
Berstel’s sequence is a linear recurrence sequence of order 3
with 6 zeroes.
Let n
2. The sequence with initial values
u0 = 1, u1 = · · · = un
Frits Beukers (1991) : up to
trivial transformation, any
other linear recurrence of
order 3 with finitely many
zeroes has at most 5 zeros.
1
=0
satisfying the recurrence relation of order n with characteristic
polynomial
X n+1 ( 2)n 1 X + ( 2)n
X +2
has at least
n(n + 1)
1
2
zeroes.
Frits Beukers
97 / 119
Edgard Bavenco↵e and Jean-Paul Bézivin
Berstel’s sequence
For n = 3 one obtains Berstel’s sequence which happens to
have an extra zero.
X 4 + 4X
X +2
8
= X3
2X 2 + 4X
98 / 119
0, 0, 1, 2, 0,
4, 0, 16, 16,
32,
b0 = b1 = 0, b2 = 1, bn+3 = 2bn+2
4.
64, 64, 256, 0,
4bn+1 + 4bn for n
768, . . .
0.
The equation bm = ±bn has
exactly 21 solutions (m, n)
with m 6= n.
Edgard Bavenco↵e
Jean-Paul Bézivin
Maurice Mignotte
99 / 119
The equation bn = ±2r 3s has
exactly 44 solutions (n, r, s).
100 / 119
Joint work with Claude Levesque
Families of binary forms
Consider a binary form F0 (X, Y ) 2 C[X, Y ] which satisfies
F0 (1, 0) = 1. We write it as
Linear recurrence sequences
and twisted binary forms.
Proceedings of the
International Conference on
Pure and Applied
Mathematics
ICPAM-GOROKA 2014.
South Pacific Journal of Pure
and Applied Mathematics.
d
F0 (X, Y ) = X + a1 X
d 1
Fn (X, Y ) =
(X
↵i Y ).
i=1
d
Y
↵i ✏ni Y ),
(X
i=1
Xd
http://webusers.imj-prg.fr/~michel.waldschmidt//articles/
pdf/ProcConfPNG2014.pdf
Therefore
101 / 119
Families of Diophantine equations
U1 (n)X d 1 Y + · · · + ( 1)d Ud (n)Y d .
Uh (0) = ( 1)h ah
(1  h  d).
102 / 119
Families of Diophantine equations
With Claude Levesque, we considered some families of
diophantine equations
Each of the sequences Uh (n)
coefficients of the relation
Fn (x, y) = m
obtained in the same way from a given irreducible form
F (X, Y ) with coefficients in Z, when ✏1 , . . . , ✏d are algebraic
units and when the algebraic numbers ↵1 ✏1 , . . . , ↵d ✏d are
Galois conjugates with d 3.
Theorem. Let K be a number field of degree d 3, S a finite
set of places of K containing the places at infinity. Denote by
OS the ring of S–integers of K and by OS⇥ the group of
S–units of K. Assume ↵1 , . . . , ↵d , ✏1 , . . . , ✏d belong to K⇥
Then there are only finitely many (x, y, n) in OS ⇥ OS ⇥ Z
satisfying
xy 6= 0 and Card{↵1 ✏n1 , . . . , ↵d ✏nd }
=
d
Y
Let ✏1 , . . . , ✏d be d nonzero complex numbers not necessarily
distinct. Twisting F0 by the powers ✏n1 , . . . , ✏nd (n 2 Z) boils
down to considering the family of binary forms
which we write as
Fn (x, y) 2 OS⇥ ,
Y + · · · + ad Y
d
Fn (X, Y ) = X d
103 / 119
coming from the
U1 (n)X d 1 Y + · · · + ( 1)d Ud (n)Y d
is a linear recurrence sequence.
For example, for n 2 Z,
U1 (n) =
d
X
↵i ✏ni ,
Ud (n) =
i=1
n2Z
d
Y
↵i ✏ni .
i=1
For 1  h  d, the sequence Uh (n)
combination of the sequences
(✏i1 · · · ✏ih )n
3.
n2Z
,
n2Z
is a linear
(1  i1 < · · · < ih  d).
104 / 119
Some units of Bernstein and Hasse
Let t and s be two positive integers, D an integer
c 2 { 1, +1}. Let ! > 1 satisfy
st
Helmut Hasse (1898-1979)
1, and
D > 0, s 1, t 1,
c 2 { 1, +1}, ! > 0,
st
! = D + c,
! st = Dst + c,
where it is assumed that Q(!) is of degree st.
Consider
↵ = D !, ✏ = Dt ! t .
L. Bernstein and H. Hasse noticed that ↵ and ✏ are units of
degree st and s respectively, and showed that these units can
be obtained from the Jacobi–Perron algorithm. H.-J. Stender
proved that for s = t = 2, {↵, ✏} is a fundamental system of
units of the quartic field Q(!).
↵=D
!,
✏ = Dt
!t.
(↵
D)st = ( 1)st (Dst + c).
105 / 119
Diophantine equations associated with some units
of Bernstein and Hasse
The irreducible polynomial of ↵ is F0 (X, 1), with
F0 (X, Y ) = (X
DY )st
( 1)st (Dst + c)Y st .
For n 2 Z, the binary form Fn (X, Y ), obtained by twisting
F0 (X, Y ) with the powers ✏n of ✏, is the homogeneous version
of the irreducible polynomial Fn (X, 1) of ↵✏n . So Fn depends
of the parameters n, D, s, t and c.
Theorem (LW). Suppose st 3. There exists an e↵ectively
computable constant , depending only on D, s and t, with
the following property. Let m, a, x, y be rational integers
satisfying m 2, xy 6= 0, [Q(↵✏a ) : Q] = st and
Then
|Fn (x, y)|  m.
max{log |x|, log |y|, |n|}   log m.
106 / 119
Hankel determinants
To test an arbitrary sequence u = (un )n 0 of elements of a
field K for the property of being a linear recurrence sequence,
consider the Hankel determinants
N,d (u)
The sum
f (z) =
1
X
un z n
n=0
Hermann Hankel
(1839–1873)
107 / 119
= det (ud+i+j )0i,jN .
represents a rational function
if and only if for some d,
N,d (u) = 0 for all
sufficiently large N
108 / 119
Hankel determinants
Perfect powers in the Fibonacci sequence
Alan Haynes, Wadim Zudilin. – Hankel determinants of zeta
values
(Submitted on 7 Oct 2015)
Yann Bugeaud, Maurice Mignotte, Samir Siksek (2004) : The
only perfect powers (squares, cubes, etc.) in the Fibonacci
sequence are 1, 8 and 144.
Abstract: We study the asymptotics of Hankel determinants
constructed using the values ⇣(an + b) of the Riemann zeta
function at positive integers in an arithmetic progression. Our
principal result is a Diophantine application of the
asymptotics.
Y. Bugeaud
Alan Haynes
M. Mignotte
S. Siksek
Wadim Zudilin
109 / 119
Powers in recurrence sequences
110 / 119
Bases of the space of linear recurrence sequences
Given a1 , . . . , ad with ad 6= 0, consider the vector space of
linear recurrence sequences satisfying, for n 0,
(?)
M. A. Bennett, Powers in
recurrence sequences : Pell
equations, Trans. Amer.
Math. Soc. 357 (2005),
1675-1691.
un+d = a1 un+d
1
+ · · · + ad u n .
Assuming the characteristic polynomial
f (X) = X d
a1 X d
1
···
ad
of the recurrence splits completely in K,
Mike Bennett
f (X) =
http://www.math.ubc.ca/~bennett/paper31.pdf
Ỳ
(X
j)
ti
j=1
we have two bases. The first one given by the initial conditions
(u0 , . . . , ud 1 ), and the second one is given by the sequences
(ni
111 / 119
n
j )n 0 ,
0  i  tj
1, 1  j  `.
112 / 119
Change of basis
Exponential polynomials
The sequence of derivatives of an exponential polynomial
evaluated at one point satisfies a linear recurrence relation.
Let p1 (z), . . . , p` (z) be nonzero polynomials of C[z] of degrees
smaller than t1 , . . . , t` respectively. Let 1 , . . . , ` be distinct
complex numbers. Suppose that the function
The matrix of change of bases is
0 1
M1
B .. C
M =@ . A
M`
F (z) = p1 (z)e
where
0
1
B
B
B0
B
B
Mj = B
B0
B.
B ..
@
0
2
j
j
1
2
1
···
j
tj 1
j
tj
j
···
d 1
j
C
C
C
C
C
3 C.
C
C
C
A
...
tj 1
1
tj 2
j
tj
1
tj 1
j
...
d 1
1
d 2
j
tj 1
2
tj 3
j
tj
2
tj 2
j
...
..
.
d 1
2
d
j
···
d 1
tj 1
0
..
.
1
..
.
...
..
.
0
0
...
..
.
1
..
.
tj
tj 1
j
..
.
1
d tj
j
1z
+ · · · + p` (z)e
`z
is not identically 0. Then its vanishing order at a point z0 is
smaller than or equal to t1 + · · · + t` 1.
In other terms, when the complex numbers j are distinct, the
determinant
✓ ◆a
d
z i e j z z=0
0itj 1, 1j`
dz
0ad 1
is di↵erent from 0. This is no surprise that we come across the
determinant of the matrix M .
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The matrix M
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Interpolation
The determinant of M is
det M =
Y
(
j
i)
ti tj
.
1i<j`
For 1  j  `, 0  i  tj 1, 0  k  d
entry of the matrix M is
✓ ◆i
✓ ◆
1
d
k
k
T
=
i! dT
i
T=
1, the (sj + i, k)
k i
j .
di f
( j ) = ⌘ij ,
dz i
j
The matrix M is associated with the linear system of d
equations in d unknowns which amounts to finding a
polynomial f 2 K[z] of degree < d for which the d numbers
di f
( j ),
(1  j  `, 0  i  tj
dz i
take prescribed values.
Let j (1  j  `) be distinct elements in K, tj (1  j  `)
be positive integers, ⌘ij (1  j  `, 0  i  tj 1) be
elements in K. Set d = t1 + · · · + t` . There exists a unique
polynomial f 2 K[z] of degree < d satisfying
(1  j  `, 0  i  tj
1).
1)
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Truncated Taylor expansion
Explicit solution to the interpolation problem
Let g 2 K(z), let z0 2 K and let t
pole of g. We set
Tg,z0 ,t (z) =
t 1 i
X
dg
i=0
dz i
For j = 1, . . . , `, define
1. Assume z0 is not a
(z0 )
(z
hj (z) =
z0 )i
·
i!
Y ✓z
1k`
k6=j
j
k
k
◆t k
tj 1
and
pj (z) =
X
⌘ij
(z
j)
i!
i=0
i
·
Then the solution f of the interpolation problem
In other words, Tg,z0 ,t is the unique polynomial in K[z] of
degree < t such that there exists r(z) 2 K(z) having no pole
at z0 with
g(z) = Tg,z0 ,t (z) + (z z0 )t r(z).
di f
( j ) = ⌘ij ,
dz i
(1  j  `, 0  i  tj
1).
is given by
Notice that if g is a polynomial of degree < t, then g = Tg,z0 ,t
for any z0 2 K.
f=
X̀
j=1
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December 2016
Linear recurrence sequences,
exponential polynomials
and Diophantine approximation
Michel Waldschmidt
Institut de Mathématiques de Jussieu — Paris VI
http://webusers.imj-prg.fr/~michel.waldschmidt/
119 / 119
h j T pj ,
hj
j ,tj
.
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