Orthogonal polynomials and - - PowerPoint PPT Presentation

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Orthogonal polynomials and - - PowerPoint PPT Presentation

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


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  • E.N.S.A.I.T. (Roubaix)

Orthogonal polynomials

and

formal computing

Pierre L. Douillet

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  • aut : aut f x 21 f x 1 f x

n n 1 2 3

a linear operator

we start from a linear operator : an eigenfunction is a polynomial when usual values : chebyschev legendre gegenbauer ...

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  • aut f x

1 µ x

  • x

fac x f x µ x 1x 2

1 2

fac x 1x 2

1 2

f |g c

1 1 f t g t µ t dt

f |aut g c

1 1 f t g t fac t dt

a selfadjoint operator

Rodrigue’s formula identifying : and a scalar product and we have

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  • ck,n

Fn x cj,n x n2j x 21 F

n 1 F n n n Fn 0

cj1,n cj,n n2j1 n2j 2 2n2j2 j1

ck,n 1 4

k

2n 2n2k nk/2 ! n! k! n2k ! n/2 ! c0,n

a closed formula for the Fn (1)

Define by Use Obtain Conclude

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  • Fn x

cj,n x n2j cd Fn x n Hyp n 2 , 1n 2 , 1 2 n , 1 x 2 Hyp a, b , c , z z j j!

j1 k0

ak bk ck

a closed formula for the Fn (2)

Using the hypergeometric function

  • but ... you must quite rewrite hypergeom

from scratch to deal with the polynomial case.

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  • 1 T4 ÷ cd T4

x 4 Hyp 2, 3 2 , 3 , 1 x 2 x 4x 2 1 8 2 x 3 x 2 x 21 x 21 f x x f x 16f x

taking the bad branch !

We have but Maple gives i.e. the other solution of

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  • S x, z zn Fn x

2 Un 1 1 arctan z 1x 2 1xz ÷ z 1x 2 Un 1 n1 1 12xzz 2

normalizing (shorter version)

What choice for the leading coefficient ? Orthonormality introduces square roots ... One criterion is simplifying the generating function For (gegenbauer) leads to while gives

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  • An, k

x n2k ck,n/c0,n Oper n,k an,k xbn1,k cn2,k a n1 n ; b c 2n 2n2 K 2x 2 n1 n 2n2k k n2k1 n2k2

Oper An, k An, k1Kn, k1 An, kKn, k

Oper k An, k 0

the "A=B" method

define as

  • btain

and check conclude

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  • c0,n 1

n n1 Fn1 x 2n2 2n Fn 2n2 2n Fn1,k 0 Fn 1 1

c0,n n 2n n/2 ! ! 2n n /2 !

n Fn1 x 2n Fn nFn1,k 0

recurrences over the Fn

trying , we obtain normalizing by ,we obtain and

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  • F

m

Sz 1xz 12xzz 2

  • 1

1 SySy µ dx 1

2 2zy 1zy S2

1 1 S y Sz µ dx y

1y 22y 3z 1y 2 2 1yz 2

S2 1 y dy y 1yz 1y 2

F

m|Fn

m

decomposition of

when , we have thus (obvious)

  • thus
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  • F

m

S3

1 1 S y S z µ dx

S3 1 yz dy dz yz 1yz 1yz 1y 2 1z 2

F

m|F n

m n min m,n F

m|F n

m m n 1 ! n! n ! m n m n

norm of

Put (here again ) compute conclude general (where and is even)

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  • P

n P |P P|P ; P 0 ; dg P n

the Markov-Bernstein problem

maximize the norm of

  • ver the

normalized polynomials of degree at most i.e. maximize

  • btain asymptotic results on that max
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  • An

Fj|Fk Bn F

j |F k

n µ det µ An Bn T Tjj 1 Tj2,j F

j |F j4

F

j2|F j4

Cn Tn µ An Bn

tTn

n µ det Cn n n n n 1/

A 5-band matrix

define , and find the roots of define by , and otherwise then is [-2,0,+2]-diagonal and define by

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  • n n1 n3 n4

n n n1 n n2 n4 1 1 1 2 1 16 1 56 722 1 144 10242

  • 3-terms recurrence

we obtain a recurrence looking as for parity reasons, recurrence thus

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  • n j cj n

c0 n 1 c1 n n n n 2 n 2 4 1 cj n n 4j µ n 4

  • n

ˆ cj n

asymptotics (1)

define

  • ; by recurrence,

since , put and take the leading term in of coefficients

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  • ˆ

cj n 1 j 1 4 42j m1 m 1 4

ˆ n 1

  • 4 1

Hyp 1 , 2, 5 4 , 1 16

  • asymptotics (2)

we have and therefore

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  • 1

10 89 247 ν

1 1 /8 Hyp 1 , 3/2, 2 , /16 cos

1 2

2 2k1 2

legendre ( )

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  • BesselJ

3 4 , 1 2

  • 5 2 2k 2

remarks

the Bessel function the same formula hold for the other parity when ,

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  • Aut

remaining work

  • ther families of orthogonal polynomials

the second derivative generating functions direct derivation from

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