A Mahlers theorem for functions from words to integers Jean- Eric - PowerPoint PPT Presentation

A Mahler’s theorem for functions from words to integers Jean-´ Eric Pin 1 Pedro V. Silva 2 1 LIAFA, CNRS and University Paris Diderot 2 Centro de Matem´ atica, Faculdade de Ciˆ encias, Universidade do Porto, R. Campo Alegre 687, 4169-007 Porto, Portugal. Novembre 2010, LIPN supported by the ESF (European Science Foundation) network AutoMathA LIAFA, CNRS and University Paris Diderot

Outline (1) Mahler’s expansions (2) The p -adic norm (3) Mahler’s theorem (4) Extension to words (5) The p -adic and pro- p topologies (6) An extension of Mahler’s theorem (7) Real motivations LIAFA, CNRS and University Paris Diderot

Part I Mahler’s expansion Mahler’s theorem is the dream of math students: A function is equal to the sum of its Newton series iff it is uniformly continuous. http://en.wikipedia.org/wiki/Mahler’s_theorem LIAFA, CNRS and University Paris Diderot

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Two basic definitions Binomial coefficients � n ( n − 1) ··· ( n − k +1) � n � if 0 � k � n k ! = k 0 otherwise Difference operator Let f : N → Z be a function. We set (∆ f )( n ) = f ( n + 1) − f ( n ) Note that (∆ 2 f )( n )= f ( n + 2) − 2 f ( n + 1) + f ( n ) � n � � (∆ k f )( n )= ( − 1) k f ( n + k ) k 0 � k � n LIAFA, CNRS and University Paris Diderot

Mahler’s expansions For each function f : N → Z , there exists a unique family a k of integers such that, for all n ∈ N , ∞ � n � � f ( n ) = a k k k =0 This family is given by a k = (∆ k f )(0) where ∆ is the difference operator, defined by (∆ f )( n ) = f ( n + 1) − f ( n ) LIAFA, CNRS and University Paris Diderot

Examples Fibonacci sequence: f (0) = f (1) = 1 and f ( n ) = f ( n − 1) + f ( n − 2) for ( n � 2) . Then ∞ � n � � ( − 1) k +1 f ( k ) f ( n ) = k k =0 Let f ( n ) = r n . Then ∞ � n � � ( r − 1) k f ( n ) = k k =0 LIAFA, CNRS and University Paris Diderot

Examples (2) � 0 if n is even The parity function f ( n ) = 1 if n is odd ∞ � n � � ( − 2) k − 1 then f ( n ) = k k> 0 ∞ � n � � Factorial n ! = a k k k =0 where the a k are derangements: number of permutations of k elements with no fixed points: 1 , 0 , 1 , 2 , 9 , 44 , 265 , 1854 , 14833 , 133496 , 1334961 . LIAFA, CNRS and University Paris Diderot

The p -adic valuation Let p be a prime number. The p -adic valuation of a non-zero integer n is k ∈ N | p k divides n � � ν p ( n ) = max By convention, ν p (0) = + ∞ . The p -adic norm of n is the real number | n | p = p − ν p ( n ) Finally, the metric d p can be defined by d p ( u, v ) = | u − v | p LIAFA, CNRS and University Paris Diderot

Examples Let n = 1200 = 2 4 × 3 × 5 2 | n | 2 = 2 − 4 | n | 3 = 3 − 1 | n | 5 = 5 − 2 | n | 7 = 1 LIAFA, CNRS and University Paris Diderot

Examples Let n = 1200 = 2 4 × 3 × 5 2 | n | 2 = 2 − 4 | n | 3 = 3 − 1 | n | 5 = 5 − 2 | n | 7 = 1 Let u = 512 and v = 12 . Then u − v = 500 = 2 2 × 5 3 . Thus d 2 ( u, v ) = 2 − 2 d 5 ( u, v ) = 5 − 3 d p ( u, v ) = p 0 = 1 for p � = 2 , 5 LIAFA, CNRS and University Paris Diderot

Mahler’s theorem Theorem (Mahler) Let f ( n ) = � ∞ � n � k =0 a k be the Mahler’s expansion k of a function f : N → Z . TFCAE: (1) f is uniformly continuous for the p -adic norm, (2) the polynomial functions n → � m � n � k =0 a k k converge uniformly to f , (3) lim k →∞ | a k | p = 0 . �� ∞ � n � �� (2) means that lim m →∞ sup n ∈ N k = m a k p = 0 . � k LIAFA, CNRS and University Paris Diderot

Mahler’s theorem (2) Theorem (Mahler) f is uniformly continuous iff its Mahler’s expansion converges uniformly to f . The most remarkable part of the theorem is the fact that any uniformly continuous function can be approximated by polynomial functions, in contrast to Stone-Weierstrass approximation theorem, which requires much stronger conditions. LIAFA, CNRS and University Paris Diderot

Examples • The Fibonacci function is not uniformly continuous (for any p ). • The factorial function is not uniformly continuous (for any p ). • The function f ( n ) = r n is uniformly continuous iff p | r − 1 since f ( n ) = � ∞ k =0 ( r − 1) k � n � . k � 0 if n is even • If f ( n ) = if n is odd then 1 f ( n ) = � ∞ k> 0 ( − 2) k − 1 � n � and hence f is uniformly k continuous for the p -adic norm iff p = 2 . LIAFA, CNRS and University Paris Diderot

Part II Extension to words Is it possible to obtain similar results for functions from A ∗ to Z ? Questions to be solved: (1) Extend binomial coefficients to words and difference operators to word functions. (2) Find a Mahler expansion for functions from A ∗ to Z . (3) Find a metric on A ∗ which generalizes d p . (4) Extend Mahler’s theorem. LIAFA, CNRS and University Paris Diderot

The free monoid A ∗ An alphabet is a finite set whose elements are letters ( A = { a, b, c } , A = { 0 , 1 } ). Words are finite sequences of letters. The empty word 1 has no letter. Thus 1 , a , bab , aaababb are words on the alphabet { a, b } . The set of all words on the alphabet A is denoted by A ∗ . Words can be concatenated abraca dabra → abracadabra The concatenation product is associative. Further, for any word u , 1 u = u 1 = u . Thus A ∗ is a monoid, in fact the free monoid on A . LIAFA, CNRS and University Paris Diderot

Subwords Let u = a 1 · · · a n and v be two words of A ∗ . Then u is a subword of v if there exist v 0 , . . . , v n ∈ A ∗ such that v = v 0 a 1 v 1 . . . a n v n . For instance, aaba is a subword of aacbdcac . LIAFA, CNRS and University Paris Diderot

Binomial coefficients (see Eilenberg or Lothaire) Given two words u = a 1 a 2 · · · a n and v , the binomial � v � coefficient is the number of times that u u appears as a subword of v . That is, � v � = |{ ( v 0 , . . . , v n ) | v = v 0 a 1 v 1 . . . a n v n }| u = | u | a . If u = a n and � u � If a is a letter, then a v = a m , then � v � � m � = u n LIAFA, CNRS and University Paris Diderot

Pascal triangle Let u, v ∈ A ∗ and a, b ∈ A . Then � u � (1) = 1 , 1 � u � = 0 if | u | � | v | and u � = v , (2) v �� u � if a � = b � ua � vb (3) = � u vb � u � � + if a = b vb v Examples � abab � abab � abab � � � = 2 = 3 = 1 a ab ba LIAFA, CNRS and University Paris Diderot

An exercise Verify that, for every word u , v , � � u � � v       � u � v � uv � � uv � � � 1 1 1 a ab a ab a ab � u � v � uv       � � � 0 1 0 1  = 0 1       b b b      0 0 1 0 0 1 0 0 1 LIAFA, CNRS and University Paris Diderot

Computing the Pascal triangle Let a 1 a 2 · · · a n be a word. The function τ : A ∗ → M n +1 ( Z ) defined by � u � � u u u  � � � � �  1 . . . a 1 a 2 ··· a n a 1 a 1 a 2 a 1 a 2 a 3 � u � u u � � � � 0 1 . . .   a 2 ··· a n a 2 a 2 a 3 � u   u � � � 0 0 1 . . .   a 3 a 3 ··· a n τ ( u ) =   . . . . . ... . . . . .   . . . . .   � u  �  0 0 0 0 . . .  a n  0 0 0 0 . . . 1 is a morphism of monoids. LIAFA, CNRS and University Paris Diderot

Computing the Pascal triangle modulo p The function τ p : A ∗ → M n +1 ( Z /p Z ) defined by τ p ( u ) ≡ τ ( u ) mod p is a morphism of monoids. Further, the unitriangular n × n matrices with entries in Z /p Z form a p -group, that is, a finite group whose number of elements is a power of p . LIAFA, CNRS and University Paris Diderot

Difference operator Let f : A ∗ → Z be a function. For each letter a , we define the difference operator ∆ a by (∆ a f )( u ) = f ( ua ) − f ( u ) One can now define inductively an operator ∆ w for each word w ∈ A ∗ by setting (∆ 1 f )( u ) = f ( u ) , and for each letter a ∈ A , (∆ aw f )( u ) = (∆ a (∆ w f ))( u ) LIAFA, CNRS and University Paris Diderot

Direct definition of ∆ w � w � � ∆ w f ( u ) = ( − 1) | w | + | x | f ( ux ) x 0 � | x | � | w | Example ∆ aab f ( u ) = − f ( u ) + 2 f ( ua ) + f ( ub ) − f ( uaa ) − 2 f ( uab ) + f ( uaab ) LIAFA, CNRS and University Paris Diderot

Mahler’s expansion of word functions Theorem (cf. Lothaire) For each function f : A ∗ → Z , there exists a unique family � f, v � v ∈ A ∗ of integers such that, for all u ∈ A ∗ , � u � � f ( u ) = � f, v � v v ∈ A ∗ This family is given by � v � � ( − 1) | v | + | x | � f, v � = (∆ v f )(1) = f ( x ) x 0 � | x | � | v | LIAFA, CNRS and University Paris Diderot

An example Let f : { 0 , 1 } ∗ → N the function mapping a binary word onto its value: f (010111) = f (10111) = 23 . � f + 1 if the first letter of v is 1 (∆ v f ) = f otherwise � 1 if the first letter of v is 1 (∆ v f )( ε ) = 0 otherwise Thus, if u = 01001 , then � u � u � u � u � u � u � � � � � � f ( u ) = + + + + + = 1 10 11 100 101 1001 2 + 2 + 1 + 1 + 2 + 1 = 9 . LIAFA, CNRS and University Paris Diderot