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E.N.S.A.I.T. (Roubaix) Orthogonal polynomials and formal computing Pierre L. Douillet we start from a linear operator : aut : aut f x 2


  1. E.N.S.A.I.T. (Roubaix) Orthogonal polynomials � ������� � ����������� and ���������� formal computing Pierre L. Douillet

  2. � we start from a linear operator : aut : aut f � x 2 � 1 f �� x � � � 1 f � x � an eigenfunction is a polynomial when � � n n � � �������� � usual values : ������������ ���������� � � 0 � � 1 � � 2 � � 3 chebyschev legendre gegenbauer ... a linear operator

  3. � Rodrigue’s formula � fac x f � x 1 � aut f x � µ x � x � identifying : � � 1 � � 1 µ x � 1 � x 2 fac x � 1 � x 2 2 2 and �������� ������������ � a scalar product ���������� � c � � 1 f | g � 1 f t g t µ t d t � and we have � c � � 1 f � t g � t fac t d t � 1 f |aut g a selfadjoint operator

  4. F n x � � c j , n x n � 2 j c k , n � Define by x 2 � 1 F �� n � � � 1 F � n � n n � � F n � 0 � Use n � 2 j � 1 n � 2 j c j � 1, n � � c j , n � Obtain 2 2 n � 2 j � 2 � � j � 1 �������� ������������ ���������� � Conclude k � � 2 n n � k � � /2 ! n ! c k , n � � 1 c 0, n � � 2 n � 2 k k ! n � 2 k ! n � � /2 ! 4 a closed formula for the Fn (1)

  5. � Using the hypergeometric function j � 1 z j a � k b � k j ! � Hyp a , b , c , z � c � k k � 0 � c j , n x n � 2 j � F n x � � �������� � n , 1 � n ������������ � n , 1 cd F n x n Hyp , 1 � � ���������� x 2 2 2 2 � but ... you must quite rewrite hypergeom from scratch to deal with the polynomial case. a closed formula for the Fn (2)

  6. � 1 � T 4 ÷ cd T 4 � � We have � 2, � 3 , � 3 , 1 � x 4 � x 2 � 1 x 4 Hyp x 2 2 8 x 3 � x x 2 � 1 � but Maple gives � 2 � �������� 2 ������������ ���������� i.e. the other solution of x 2 � 1 f �� x � x f � x � 16 f x taking the bad branch !

  7. � What choice for the leading coefficient ? � Orthonormality introduces square roots ... � One criterion is simplifying the generating S x , z � � z n F n x function �������� � For � � 2 (gegenbauer) ������������ arctan z 1 � x 2 U n 1 � 1 ÷ z 1 � x 2 ���������� leads to 1 � xz 1 U n 1 � n � 1 while gives 1 � 2 xz � z 2 normalizing (shorter version)

  8. x n � 2 k c k , n / c 0, n A n , k � define as � obtain Oper � n , k � a � n , k � xb � n � 1, k � c � n � 2, k a � n � 1 n � � ; b � c � 2 n � � 2 n � � � 2 �������� K � � 2 x 2 n � 1 n � � 2 n � 2 k � � k ������������ and n � 2 k � 1 n � 2 k � 2 ���������� Oper A n , k � A n , k � 1 K n , k � 1 � A n , k K n , k � check Oper � k A n , k � 0 � conclude the "A=B" method

  9. c 0, n � 1 � trying , we obtain n n � 1 � � F n � 1 � x 2 n � 2 � � 2 n � � F n � 2 n � 2 � � 2 n � � F n � 1, k � 0 F n 1 � 1 � normalizing by ,we obtain �������� ������������ n � � /2 ! � ! 2 n � � n c 0, n � and ���������� 2 � � n � � n � /2 � ! n � � F n � 1 � x 2 n � � F n � nF n � 1, k � 0 recurrences over the Fn

  10. 1 � xz S z � � when � � 0 , we have 1 � 2 xz � z 2 2 � zy � 1 S y S y µ d x � 1 � 1 � � thus (obvious) 1 � zy 2 �������� S 2 � � 1 � y 2 � 2 y 3 z ������������ � 1 � 1 S � y S z µ d x � y � ���������� 1 � y 2 2 1 � yz 2 y 1 F � m | F n � m d y � thus S 2 y 1 � yz 1 � y 2 decomposition of F � m

  11. S 3 � � � 1 � 1 S � y S � z µ d x � Put (here again � � 0 ) yz 1 � yz 1 � compute d y d z � S 3 yz 1 � yz 1 � y 2 1 � z 2 F � m | F � � m n min m , n � conclude �������� n ������������ � m m � � n � � � ! n ! ���������� F � m | F � � general n � � n ! � � 1 m � n m � n (where and is even) norm of F � m

  12. P � � maximize the norm of over the n normalized polynomials of degree at most i.e. P � | P � ; P � 0 ; dg P � n maximize P | P �������� � obtain asymptotic results on that max ������������ ���������� the Markov-Bernstein problem

  13. F � j | F � A n � F j | F k B n � � define , k � det µ A n � B n and find the roots of � n µ F � j | F � j � 4 T T jj � 1 T j � 2, j � � � define by , F � j � 2 | F � �������� j � 4 ������������ t T n C n � T n µ A n � B n and otherwise then 0 ���������� � det C n is [-2,0,+2]-diagonal and � n µ � n � � � n � n 1/ � � define by � n A 5-band matrix

  14. � we obtain a recurrence looking as � n � � � � � � � n � 1 � � � � � � n � 3 � � � n � 4 � for parity reasons, � n � � n � n � 1 � recurrence � n � � � � � � � n � 2 � � � n � 4 �������� 1 1 ������������ ���������� 1 � 2 � 1 � 16 � � thus 1 � 56 � � 72 � 2 1 � 144 � � 1024 � 2 � � 3-terms recurrence

  15. � n � � � � j c j n � define c 0 n � 1 � ; by recurrence, c 1 n � � n n � � n � 2 n � � � 2 4 � � 1 �������� ������������ µ � n 4 c j n � � n 4 j ���������� � since , put and take � n c j n the leading term in of coefficients ˆ asymptotics (1)

  16. � � 1 � 1 j � 4 c j n � � we have ˆ 4 2 j � m � 1 � m � � � 1 4 �������� � and therefore ������������ ���������� 1 , 2, 5 � � , � 1 ˆ Hyp � � n � � 1 � � 4 � � 1 4 16 asymptotics (2)

  17. 1 10 89 247 ν �������� ������������ ���������� 1 1 � � /8 Hyp 1 , 3/2, 2 , � � /16 � cos 2 � � � � 2 2 k � 1 2 legendre ( ) � � 1

  18. � � 3 , 1 BesselJ � the Bessel function � 4 2 � the same formula hold for the other parity �������� ������������ � � 5 � � � 2 2 k 2 � when , ���������� remarks

  19. � other families of orthogonal polynomials � the second derivative � generating functions �������� � direct derivation from Aut ������������ ���������� remaining work

  20. �������� ������������ ����������

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