數(shù)模2013數(shù)值分析ch03b

第三章1函數(shù)逼近—正交多項(xiàng)式內(nèi)容提要2正交多項(xiàng)式正交函數(shù)族與正交多項(xiàng)式Legendre

正交多項(xiàng)式Chebyshev

正交多項(xiàng)式Chebyshev

插值第二類Chebyshev

正交多項(xiàng)式Laguerre

正交多項(xiàng)式Hermite

正交多項(xiàng)式正交函數(shù)族正交函數(shù)定義：設(shè)f(x),g(x)?

C[a,b]，r

(x)是[a,b]上的權(quán)函數(shù)，3若則稱f(x)與g(x)在[a,b]上帶權(quán)r

(x)正交ba(

f

,

g)

=r(

x)

f

(

x)

g(

x)dx

=

0正交函數(shù)族?

C[a,

b]，定義：設(shè)函數(shù)j0(x),j1(x),…,jk(x),…r

(x)是[a,b]上的權(quán)函數(shù)，若則稱{jk(x)}是[a,b]上帶權(quán)r

(x)的正交函數(shù)族正交函數(shù)族40,bjar(

x)ji(

x)j(ji

,j

j

)

=(

x)

dx

=i

?

j

A

?

0,

i

=

j

i若所有Ai

=1

，則稱為標(biāo)準(zhǔn)正交函數(shù)族正交函數(shù)舉例5例：三角函數(shù)系1，

cos

x，sin

x，cos

2x，sin

2x，…在[-p,p]上是帶權(quán)r

(x)=1

的正交函數(shù)族π-π(1, 1)

=dx

=

2π證：π-π(sin

nx,

sin

mx)

=

sin

nx

sin

mx

dx

=

π

dnmπ-π(cos

nx,

cos

mx)

=

cos

nx

cos

mx

dx

=

π

dnmπ-π(cos

nx,

sin

mx)

=

cos

nx

sin

mx

dx

=

0(m,

n

=

1,

2,

3,

…

)(m,

n

=

0,

1,

2,

…

)為[a,b]上帶權(quán)r

(x)正交，n稱j

(x)為n次正交多項(xiàng)式。正交多項(xiàng)式正交多項(xiàng)式定義：設(shè)jn(x)是首項(xiàng)系數(shù)不為0

的n

次多項(xiàng)式，r

(x)是[a,b]上的權(quán)函數(shù)，若60,ba(ji

,j

j

)

=r(

x)ji

(

x)j

j

(

x)

dx

=i

?

j

A

?

0,

i

=

j

i則稱{}nj￥n=0正交多項(xiàng)式性質(zhì)1：設(shè)為[a,b]上帶權(quán)r

(x)正交多項(xiàng)式，Hn

為所有次數(shù)不超過n

的多項(xiàng)式組成的線性空間，則{

j0

(

x),

j1

(

x),

j2

(

x)

,

,

jn

(

x)

}構(gòu)成Hn

的一組基{

}nj￥n=0性質(zhì)2：設(shè)7為[a,b]上帶權(quán)r

(x)正交多項(xiàng)式，則對"p(x)?

Hn-1，有{

}nj￥n=0()bnnar(

x)

p(

x)p(

x),j

(

x)

=j

(

x)

dx

=0正交多項(xiàng)式為[a,b]上帶權(quán)r

(x)正交多項(xiàng)式，其中性質(zhì)3：設(shè){}nj￥n=0且首項(xiàng)系數(shù)均為1，則jn+1

(

x)

=

(

x

-an

)jn

(

x)

-

bnjn-1

(

x)n8n(jn

,jn

)a(jn

,jn

)(j

,j

)n-1

n-1=

(

xjn

,jn

)

,b

=n

=

0,

1,

2,

…證明：板書這就是正交多項(xiàng)式的三項(xiàng)遞推公式！所有首項(xiàng)系數(shù)為1

的正交多項(xiàng)式族都滿足這個(gè)公式，該公式也給出了正交多項(xiàng)式的一個(gè)計(jì)算方法。0-1j

=

0,

j

=

1,正交多項(xiàng)式為[a,b]上帶權(quán)r

(x)正交多項(xiàng)式，9性質(zhì)4：設(shè){}nj￥n=0則jn(x)在(a,b)內(nèi)有n

個(gè)不同的零點(diǎn)證明：板書正交多項(xiàng)式10幾類重要的正交多項(xiàng)式Legendre

多項(xiàng)式Chebyshev

多項(xiàng)式第二類Chebyshev

多項(xiàng)式Laguerre

多項(xiàng)式Hermite

多項(xiàng)式Legendre

多項(xiàng)式2n

n! 2n

(n!)2Pn

(x)的首項(xiàng)xn

的系數(shù)為：2n(2n

-

1)(n

+

1)

=(2n)!21dnnP0

(

x)

=

1,

Pn

(

x)

=

2n

n!

dxn

(

x

-

1)x?

[-1,

1]，n

=

1,

2,

…記號：P0

,P1

,P2

,...2dnnnnn!(2n)!

dxP

(

x)

=(

x

-

1)則Pn

(x)是首項(xiàng)系數(shù)為1

的勒讓德多項(xiàng)式令勒讓德（Legendre）多項(xiàng)式在[-1,

1]

上帶權(quán)r

(x)=1

的正交多項(xiàng)式稱為勒讓德多項(xiàng)式11Legendre

多項(xiàng)式勒讓德多項(xiàng)式的性質(zhì)1-10,

m

?

n(1)

正交性：(Pn

,Pm

)=Pn

(

x)Pm

(

x)

dx

=

2

, m

=

n

2n

+

1(2)

奇偶性：P2n(x)只含偶次冪，P2n+1(x)只含奇次冪，故P

(-x)

=

(-1)n

P

(

x)n

n遞推公式：(n

+1)Pn+1

(x)=(2n

+1)x

Pn

(x)-nPn-1

(x)其中P0(x)=1,

P1(x)=x，n

=1,

2,…Pn(x)在(-1,1)

內(nèi)有n

個(gè)不同的零點(diǎn)12Legendre

多項(xiàng)式132P

(

x)

=

(3x2

-1)

/

23P

(

x)

=

(5

x3

-

3x)

/

24P

(

x)

=

(35

x4

-

30

x2

+

3)

/

85P

(

x)

=

(63x5

-

70

x3

+

15

x)

/

8ex31.m勒讓德多項(xiàng)式的表達(dá)式P0

(

x)

=

1P1

(

x)

=

x14Chebyshev

多項(xiàng)式r(

x)

=11

-

x2Tn

(

x

)

=

cos(n

arccos

x

)切比雪夫多項(xiàng)式的表達(dá)式x?

[-1,

1]，n

=

0,

1,

2,

…令x

=cos

q，則Tn

(x)=cos(nq)，展開后即得4n

2n

n

nn-2

2

4

n-4T

(

x)

=

cos

q

q

sin

q-

C

cos

+

C

cosq

sin

q

+=

xn

-

C

2

xn-2

(1

-

x2

)

+

C

4

xn-4

(1

-

x2

)2

+n

n切比雪夫（Chebyshev）多項(xiàng)式在[-1,

1]

上帶權(quán)r

(x)的正交多項(xiàng)式稱為切比雪夫多項(xiàng)式Chebyshev

多項(xiàng)式切比雪夫多項(xiàng)式的性質(zhì)12Tn

(

x)Tm

(

x)-1

0,(1)

正交性：(Tn

,Tm

)=dx

=

π

/

2,1

-

xm

?

nm

=

n

?

0

π,

m

=

n

=

0(2)

奇偶性：T2n(x)只含偶次冪，T2n+1(x)只含奇次冪，故T

(-x)

=

(-1)nT

(

x)n

n(3)

遞推公式：Tn+1

(x)=2xTn

(x)-Tn-1

(x)其中T0(x)=1,

T1(x)=x，n

=1,

2,…cos(n+1)q

+

cos(n-1)q

=

2cosq

cosnqx

=

cosq15Chebyshev

多項(xiàng)式(4)Tn(x)在(-1,1)

內(nèi)有n

個(gè)不同的零點(diǎn):k2nx

=

cos

2k

-

1

π

(k

=

1,

2,

…

,

n)(5)Tn(x)在[-1,

1]

上有n+1

個(gè)極值點(diǎn):k16n=

cos

kπx(k

=

0,

1,

…

,

n)(6)

Tn(x)的首項(xiàng)系數(shù)為2n-1，且|Tn(x)|

￡

1切比雪夫多項(xiàng)式的性質(zhì)（續(xù)）Chebyshev

多項(xiàng)式(7)令nT

(

x)n2n-1T

(

x)

=n即{T

(x)}為首項(xiàng)系數(shù)為1

的Chebyshev

多項(xiàng)式n-1￡x￡1

-1￡x￡1max

T

(

x)

￡

max

p(

x)定理：記H

n

為所有次數(shù)不超過n

的首項(xiàng)系數(shù)為1的多項(xiàng)式集合，則對

"

p(

x)

?

H

n

有17且n12n-1-1￡x￡1max

T

(

x)

=切比雪夫多項(xiàng)式的性質(zhì)（續(xù)）證明略Chebyshev

多項(xiàng)式n￥￥T

(

x)

￡

p(

x)T

(

x)

是集合

H

中無窮范數(shù)最小的多項(xiàng)式n

n18n"

p(

x)

?

H注：這里的無窮范數(shù)是指C[-1,1]

上的無窮范數(shù)性質(zhì)：設(shè)f(x)?

Hn，且首項(xiàng)系數(shù)為an?0，則f(x)在[-1,1]

上的n-1

次最佳一致逼近多項(xiàng)式為f

(

x)

-

anT

(

x)n證明：留作練習(xí)Chebyshev

多項(xiàng)式例：求f(x)=2x3+x2+2x-1在[-1,1]上的二次最佳一致逼近多項(xiàng)式。解：設(shè)p(x)是f(x)在[-1,1]上的二次最佳一致逼近多項(xiàng)式，則由前面的性質(zhì)可知3

3p(

x)

=

f

(

x)

-

a

T

(

x)3

2334x=

2x

+

x

+

2x

-

1

-

2x

-21972=

x

+x

-

1思考：如何計(jì)算f(x)在區(qū)間[a,b]上的最佳逼近多項(xiàng)式？Chebyshev多項(xiàng)式20T0

(

x)

=

1T1

(

x)

=

x2T

(

x)

=

2x2

-13T

(

x)

=

4

x3

-

3x4T

(

x)

=

8

x4

-

8

x2

+

15T

(

x)

=

16

x5

-

20

x3

+

5

x切比雪夫多項(xiàng)式的表達(dá)式ex32.mChebyshev零點(diǎn)插值以Chebyshev

多項(xiàng)式的零點(diǎn)作為插值節(jié)點(diǎn)進(jìn)行插值好處：所有插值多項(xiàng)式中,總體插值誤差最小kx

=cos

2k

+1

π

2(n+1)定理：設(shè)

f(x)?

Cn+1[-1,

1]，插值節(jié)點(diǎn)

x0

,

x1

,

…

,

xn

為Tn+1

(x)的n+1

個(gè)零點(diǎn)，則121nf

(

n+1)

(

x)￥￥f

(

x)

-

L

(

x)

￡2n

(n

+

1)!用Chebyshev

多項(xiàng)式的零點(diǎn)插值注：此處原ppt

有誤！Chebyshev零點(diǎn)插值若f(x)?

Cn+1[a,b]，怎么辦？x

=

b

-

a

t

+

b

+

a2

2作變量替換插值節(jié)點(diǎn)kx=

b

-

a

cos

2k

+

1

π

+

b

+

a2

2(n

+

1)

2(k

=

0,

1,

…

,

n)插值誤差122nf

(

x)(

n+1)2n+1￥￥(b

-

a)n+1f

(

x)

-

L

(

x)

￡·2n

(n

+

1)!舉例例：(教材64頁，例

4)

求

在

[0,

1]，上的四f

(

x)

=

ex次Chebyshev

插值多項(xiàng)式L4(x)，并估計(jì)誤差。解：板書例：(教材65頁，例5，上機(jī))，插值區(qū)間[-5,5]，試分別用等距節(jié)點(diǎn)和Chebyshev

節(jié)點(diǎn)作10次多項(xiàng)式插值，畫圖比較兩種插值的數(shù)值效果。ex33.m123函數(shù)f

(x)=1

+

x2其他正交多項(xiàng)式24其他正交多項(xiàng)式第二類Chebyshev

多項(xiàng)式Laguerre

多項(xiàng)式Hermite

多項(xiàng)式x?

[-1,

1]，n

=

0,

1,

2,

…n第二類Chebyshev第二類Chebyshev

多項(xiàng)式U

(

x

)

=

sin

((n

+

1)

arccos

x

)1

-

x

2正交，即1-1(Un

,Um

)

=r(

x)Tn

(

x)Tm

(

x)

dx

=π

/

2,0,

m

?

nm

=

n遞推公式：Un+1

(x)=2xUn

(x)-Un-1

(x)其中U0(x)=1,

U1(x)=2x，n

=1,2,…在[-1,

1]

上帶權(quán)r(x)=251

-

x2Laguerre

多項(xiàng)式拉蓋爾（Laguerre）多項(xiàng)式x?

[0,

￥]，n

=

0,

1,

2,

…xndndxn(

xne

-

x

)L

(

x

)

=

e2260(n!)

,￥(

Ln

,

Lm

)

=r(

x)Ln

(

x)Lm

(

x)

dx

=0,

m

?

nm

=

n其中L0(x)=1,

L1(x)=1