Integral D-finite Functions Manuel Kauers Institute for Algebra - PowerPoint PPT Presentation

Integral D-finite Functions Manuel Kauers Institute for Algebra Johannes Kepler University Linz, Austria joint work with Christoph Koutschan 1

Integral D-finite Functions p 0 ( x ) f ( x ) + p 1 ( x ) f ′ ( x ) + · · · + p r ( x ) f ( r ) ( x ) = 0 Manuel Kauers Institute for Algebra Johannes Kepler University Linz, Austria joint work with Christoph Koutschan 1

� Integral D-finite Functions p 0 ( x ) f ( x ) + p 1 ( x ) f ′ ( x ) + · · · + p r ( x ) f ( r ) ( x ) = 0 Manuel Kauers Institute for Algebra Johannes Kepler University Linz, Austria joint work with Christoph Koutschan 1

Z · · · · · · Q Integral D-finite Functions p 0 ( x ) f ( x ) + p 1 ( x ) f ′ ( x ) + · · · + p r ( x ) f ( r ) ( x ) = 0 Manuel Kauers Institute for Algebra Johannes Kepler University Linz, Austria joint work with Christoph Koutschan 1

k [ x ] · · · · · · k ( x ) Z · · · · · · Q Integral D-finite Functions p 0 ( x ) f ( x ) + p 1 ( x ) f ′ ( x ) + · · · + p r ( x ) f ( r ) ( x ) = 0 Manuel Kauers Institute for Algebra Johannes Kepler University Linz, Austria joint work with Christoph Koutschan 1

O k [ x ] · · · · · · k ( x ) k [ x ] · · · · · · k ( x ) Z · · · · · · Q Integral D-finite Functions p 0 ( x ) f ( x ) + p 1 ( x ) f ′ ( x ) + · · · + p r ( x ) f ( r ) ( x ) = 0 Manuel Kauers Institute for Algebra Johannes Kepler University Linz, Austria joint work with Christoph Koutschan 1

??? · · · · · · D-finite functions O k [ x ] · · · · · · k ( x ) k [ x ] · · · · · · k ( x ) Z · · · · · · Q Integral D-finite Functions p 0 ( x ) f ( x ) + p 1 ( x ) f ′ ( x ) + · · · + p r ( x ) f ( r ) ( x ) = 0 Manuel Kauers Institute for Algebra Johannes Kepler University Linz, Austria joint work with Christoph Koutschan 1

• An element of Q is called integral if its denominator is 1 or − 1 . 2

• An element of Q is called integral if its denominator is 1 or − 1 . • An element of k ( x ) is called integral if its denominator is in k . 2

• An element of Q is called integral if its denominator is 1 or − 1 . • An element of k ( x ) is called integral if its denominator is in k . Note: This is the case if and only if the element has no poles. 2

• An element of Q is called integral if its denominator is 1 or − 1 . • An element of k ( x ) is called integral if its denominator is in k . Note: This is the case if and only if the element has no poles. • An element of k ( x ) is called integral if its monic (!) minimal polynomial belongs to k [ x ][ y ] . 2

• An element of Q is called integral if its denominator is 1 or − 1 . • An element of k ( x ) is called integral if its denominator is in k . Note: This is the case if and only if the element has no poles. • An element of k ( x ) is called integral if its monic (!) minimal polynomial belongs to k [ x ][ y ] . Note: This is the case if and only if the corresponding function has no poles on any of its branches. 2

• An element of Q is called integral if its denominator is 1 or − 1 . • An element of k ( x ) is called integral if its denominator is in k . Note: This is the case if and only if the element has no poles. • An element of k ( x ) is called integral if its monic (!) minimal polynomial belongs to k [ x ][ y ] . Note: This is the case if and only if the corresponding function has no poles on any of its branches. √ √ √ x 2 are integral but 3 Example: x and 1/x is not. 2

• An element of Q is called integral if its denominator is 1 or − 1 . • An element of k ( x ) is called integral if its denominator is in k . Note: This is the case if and only if the element has no poles. • An element of k ( x ) is called integral if its monic (!) minimal polynomial belongs to k [ x ][ y ] . Note: This is the case if and only if the corresponding function has no poles on any of its branches. √ √ √ x 2 are integral but 3 Example: x and 1/x is not. M = y 2 − x 2

• An element of Q is called integral if its denominator is 1 or − 1 . • An element of k ( x ) is called integral if its denominator is in k . Note: This is the case if and only if the element has no poles. • An element of k ( x ) is called integral if its monic (!) minimal polynomial belongs to k [ x ][ y ] . Note: This is the case if and only if the corresponding function has no poles on any of its branches. √ √ √ x 2 are integral but 3 Example: x and 1/x is not. M = y 2 − x M = y 3 − x 2 2

• An element of Q is called integral if its denominator is 1 or − 1 . • An element of k ( x ) is called integral if its denominator is in k . Note: This is the case if and only if the element has no poles. • An element of k ( x ) is called integral if its monic (!) minimal polynomial belongs to k [ x ][ y ] . Note: This is the case if and only if the corresponding function has no poles on any of its branches. √ √ √ x 2 are integral but 3 Example: x and 1/x is not. M = y 2 − x M = y 3 − x 2 M = y 2 − 1 x 2

• An element of Q is called integral if its denominator is 1 or − 1 . • An element of k ( x ) is called integral if its denominator is in k . Note: This is the case if and only if the element has no poles. • An element of k ( x ) is called integral if its monic (!) minimal polynomial belongs to k [ x ][ y ] . Note: This is the case if and only if the corresponding function has no poles on any of its branches. √ √ √ x 2 are integral but 3 Example: x and 1/x is not. M = y 2 − x M = y 3 − x 2 M = xy 2 − 1 2

√ � 3 x 2 � Consider the field K = k ( x ) . 3

√ � 3 x 2 � Consider the field K = k ( x ) . √ √ 3 3 x 2 ) 2 . x 2 , ( It is the k ( x ) -vector space generated by 1, 3

√ � 3 x 2 � Consider the field K = k ( x ) . √ √ 3 3 x 2 ) 2 . x 2 , ( It is the k ( x ) -vector space generated by 1, Let O be the set of all integral elements of K . 3

√ � 3 x 2 � Consider the field K = k ( x ) . √ √ 3 3 x 2 ) 2 . x 2 , ( It is the k ( x ) -vector space generated by 1, Let O be the set of all integral elements of K . √ √ 3 3 x 2 ) 2 . x 2 , ( This is a k [ x ] -module, but it is not generated by 1, 3

√ � 3 x 2 � Consider the field K = k ( x ) . √ √ 3 3 x 2 ) 2 . x 2 , ( It is the k ( x ) -vector space generated by 1, Let O be the set of all integral elements of K . √ √ 3 3 x 2 ) 2 . x 2 , ( This is a k [ x ] -module, but it is not generated by 1, √ √ √ x 4 ∈ O \ x 2 + k [ x ]( In fact, 1 3 3 3 � x 2 ) 2 � k [ x ] + k [ x ] . x 3

√ � 3 x 2 � Consider the field K = k ( x ) . √ √ 3 3 x 2 ) 2 . x 2 , ( It is the k ( x ) -vector space generated by 1, Let O be the set of all integral elements of K . √ √ 3 3 x 2 ) 2 . x 2 , ( This is a k [ x ] -module, but it is not generated by 1, √ √ √ x 4 ∈ O \ x 2 + k [ x ]( In fact, 1 3 3 3 � x 2 ) 2 � k [ x ] + k [ x ] . x √ √ 3 1 3 � x 2 , x 4 � It can be shown that 1, is a module basis of O . x 3

√ � 3 x 2 � Consider the field K = k ( x ) . √ √ 3 3 x 2 ) 2 . x 2 , ( It is the k ( x ) -vector space generated by 1, Let O be the set of all integral elements of K . √ √ 3 3 x 2 ) 2 . x 2 , ( This is a k [ x ] -module, but it is not generated by 1, √ √ √ x 4 ∈ O \ x 2 + k [ x ]( In fact, 1 3 3 3 � x 2 ) 2 � k [ x ] + k [ x ] . x √ √ 3 1 3 � x 2 , x 4 � It can be shown that 1, is a module basis of O . x Such a basis is called an integral basis for K . 3

Classical Problem: Given an irreducible polynomial M ∈ k ( x )[ y ] , find an integral basis for K = k ( x )[ y ] / � M � , i.e., a k [ x ] -module basis for the set O of all integral elements of K . 4

Classical Problem: Given an irreducible polynomial M ∈ k ( x )[ y ] , find an integral basis for K = k ( x )[ y ] / � M � , i.e., a k [ x ] -module basis for the set O of all integral elements of K . Classical Algorithms: • Trager’s algorithm • van Hoeij’s algorithm 4

Classical Problem: Given an irreducible polynomial M ∈ k ( x )[ y ] , find an integral basis for K = k ( x )[ y ] / � M � , i.e., a k [ x ] -module basis for the set O of all integral elements of K . Classical Algorithms: • Trager’s algorithm – based on ideal arithmetic • van Hoeij’s algorithm 4

Classical Problem: Given an irreducible polynomial M ∈ k ( x )[ y ] , find an integral basis for K = k ( x )[ y ] / � M � , i.e., a k [ x ] -module basis for the set O of all integral elements of K . Classical Algorithms: • Trager’s algorithm – based on ideal arithmetic • van Hoeij’s algorithm – based on series arithmetic 4

Classical Problem: Given an irreducible polynomial M ∈ k ( x )[ y ] , find an integral basis for K = k ( x )[ y ] / � M � , i.e., a k [ x ] -module basis for the set O of all integral elements of K . Classical Algorithms: • Trager’s algorithm – based on ideal arithmetic • van Hoeij’s algorithm – based on series arithmetic 4