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MATEMAATILINE ANALÜÜS II - Tallinna Tehnikaülikool

MATEMAATILINE ANALÜÜS II - Tallinna Tehnikaülikool

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2.11. ORTOGONAALSED POLÜNOOMID 119<br />

Järgmisena leiame valemi (2.11.7) abil polünoomid P 2 (x) ja P 3 (x) :<br />

〈<br />

Q 2 (x) = x 2 − x 2 ,<br />

= x 2 − √ 1 ∫1<br />

2<br />

−1<br />

〉<br />

1<br />

√ · 2<br />

〈<br />

1<br />

√ − x 2 , 1 〉<br />

2 2 x√ 6 · 1<br />

2 x√ 6 =<br />

x 2 dx<br />

√ − 1 ∫ 1<br />

2 2 x√ 6<br />

−1<br />

= x 2 − 1 3 ,<br />

∫ 1 (<br />

‖Q 2 (x)‖ = √ x 2 − 1 ) 2<br />

dx = 2 √<br />

10,<br />

3 15<br />

P 2 (x) =<br />

−1<br />

1<br />

2 x3√ 6dx =<br />

(<br />

x 2 − 1 ) ( ) 2<br />

/ 10 =<br />

3 15√ 1 (<br />

3x 2 − 1 ) √ 10;<br />

4<br />

〈 〉 〈<br />

Q 3 (x) = x 3 − x 3 1 1<br />

, √ √2 − x 3 , 1 1<br />

6〉<br />

2 2 x√ 2 x√ 6−<br />

〈<br />

− x 3 , 1 (<br />

3x 2 − 1 ) √ 〉 1 (<br />

10 3x 2 − 1 ) √ 10 =<br />

4<br />

4<br />

− 3 4<br />

= x 3 − 1 √<br />

2<br />

∫ 1<br />

−1<br />

x 3 dx<br />

√ − x√ ∫<br />

6 1<br />

2 2 −1<br />

x 4√ 6<br />

dx−<br />

2<br />

(<br />

x 2 − 1 ) √10<br />

∫ 1<br />

3x 3 (<br />

x 2 − 1 √10dx<br />

= x<br />

3<br />

−1 4 3) 3 − 3 5 x,<br />

‖Q 3 (x)‖ =<br />

P 3 (x) =<br />

√ ∫ 1<br />

−1<br />

(<br />

x 3 − 3 ) 2<br />

5 x dx = 2<br />

35<br />

(x 3 − 3 5 x )<br />

/<br />

( 2<br />

35√<br />

14<br />

)<br />

√<br />

14,<br />

= 5√ 14<br />

4<br />

(x 3 − 3 5 x )<br />

.<br />

Analoogiliselt saab leida järgmised polünoomid P k (x) (k ≥ 4) .<br />

Legendre’i ortonormaalsed polünoomid P n (x) (n ∈ N 0 ) avalduvad Rodrigues’i<br />

valemi abil<br />

( d<br />

n (<br />

P n (x) = x 2<br />

dx n − 1 ) )<br />

n<br />

/<br />

d n (<br />

∥ x 2<br />

dx n − 1 ) n<br />

∥ . (2.11.8)<br />

Veenduge, et Rodrigues’i valem kehtib n = 0; 1; 2; 3 korral.<br />

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