fungsi bessel

4
10.4 Buktikanlah (a) J 1/ 2 ( x ) = 2 πx sin x , (b) J 1/ 2 ( x ) = 2 πx cos x Pembahasan : (a) Persamaan umumnya : J n ( x ) = r=0 ( 1) r ( x / 2 ) n +2r r!Γ ( n+ r +1) J 1/ 2 ( x ) = r=0 ( 1 ) r ( x / 2) 1/ 2+2 r r!Γ( 1 2 + r +1) = r=0 ( 1 ) r ( x / 2) 1/ 2+2 r r!Γ ( r +3 / 2 ) = (1 ) 0 ( x / 2 ) 1/ 2+2( 0) 0 (0 +3/ 2) + (1 ) 1 ( x / 2) 1/ 2+2 (1) 1 ( 1+3 / 2 ) + ( 1) 2 ( x / 2) 1 /2+ 2(2 ) 2 ( 2+3 / 2) += ( x / 2) 1/ 2 Γ ( 3/ 2) (x / 2) 5 /2 Γ ( 5 / 2) + (x / 2) 9/ 2 4 Γ ( 9 / 2) += ( x / 2 ) 1/ 2 ( 1 / 2) π ( x / 2) 5/ 2 ( 3 / 2)(1 / 2) π + ( x / 2) 9/ 2 4 ( 5 / 2) (3 / 2 )( 1/ 2) π = ( x / 2 ) 1/ 2 ( 1 / 2) π ( 1(x / 2) 2 ( 3 / 4 ) + ( x / 2) 4 ( 15 / 2) ) = ( x / 2 ) 1/ 2 ( 1 / 2) π ( 1x 2 2 2 (3 / 4 ) + x 4 2 4 ( 15 / 2) ) = ( x / 2 ) 1/ 2 ( 1 / 2) π ( 1x 2 3 + ( x / 2) 3 120 )

Upload: diana-petra

Post on 21-Jan-2016

33 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: Fungsi Bessel

10.4 Buktikanlah (a) J1 /2 (x )=√ 2πx

sin x , (b) J−1 /2 ( x )=√ 2πx

cos x

Pembahasan :

(a)Persamaan umumnya :

Jn ( x )=∑r=0

∞ (−1 )r(x /2)n+2 r

r ! Γ (n+r+1)

J1 /2 (x ) = ∑r=0

∞ (−1 )r( x /2)1/2+2 r

r ! Γ (12+r+1)

= ∑r=0

∞ (−1 )r( x /2)1/2+2 r

r ! Γ (r+3/2)

= (−1 )0(x /2)1 /2+2(0)

0! Γ (0+3 /2)+

(−1 )1(x /2)1/2+2(1)

1 ! Γ (1+3 /2)+

(−1 )2(x /2)1/2+2(2)

2 ! Γ (2+3 /2)+…

= (x /2)1/2

Γ (3/2)−

(x /2)5/2

Γ (5 /2)+

(x /2)9 /2

4 Γ (9 /2)+…

= (x /2)1/2

(1/2)√π−

(x /2)5/2

(3 /2)(1/2)√ π+

(x /2)9 /2

4 (5/2 )(3/2)(1/2)√ π−…

= (x /2)1/2

(1/2)√π (1−(x /2)2

(3/4)+

(x /2)4

(15 /2)−…)

= (x /2)1/2

(1/2)√π (1− x2

22

(3 /4)+

x4

24

(15/2)−…)

= (x /2)1/2

(1/2)√π (1− x2

3+(x /2)3

120−…)

= (x /2)1/2

(1/2)√π (1− x2

3 !+

(x /2)3

5 !−…)

Page 2: Fungsi Bessel

= (x /2)1/2

(1/2)√πsin xx

= √ 2πx

sin x

(b)J−1 /2 ( x ) = ∑r=0

∞ (−1 )r( x /2)−1 /2+2r

r ! Γ (−12

+r+1)

= ∑r=0

∞ (−1 )r( x2 )−1 /2+2 r

r ! Γ (r+1 /2)

= (−1 )0(x /2)−1/2+2(0)

0 ! Γ (0+1/2)+

(−1 )1(x /2)−1 /2+2 (1 )

1 ! Γ (1+1/2)+

(−1 )2(x /2)−1 /2+2(2)

2 ! Γ (2+1/2)+…

= (x /2)−1/2

Γ (1/2)−

(x /2)3 /2

Γ (3 /2)+

(x /2)7/2

4 Γ (5 /2)+…

= (x /2)−1/2

√π−

(x /2)3 /2

(1/2)√π+

(x /2)7 /2

4 (3/2)(1/2)√π−…

= (x /2)−1/2

√π (1−(x /2)2

1/2+

(x /2)4

3−…)

= (x /2)−1/2

√π (1−

x2

22

1 /2+

x4

24

3−…)

= (x /2)−1/2

√π (1− x2

2+

(x /2)3

48−…)

= (x /2)−1/2

√π (1− x2

2 !+(x /2)3

4 !−…)

= (x /2)−1/2

√πcos xx

= √ 2πx

cos x

Page 3: Fungsi Bessel

10.7. Tunjukkanlah bahwa (a) J1 /2 (x )=√ 2πx ( sin x−x cos x

x )

(b) J−1 /2 ( x )=−√ 2πx ( x sin x−cos x

x )Pembahasan :

Dari Soal 10.6 (b) didapatkan persamaan :

Jn−1 (x )+Jn+1 ( x )=2nxJ n( x)

Dan dari soal 10.4. diatas diketahui bahwa

J1 /2 (x ) = √ 2πx

sin x ; J−1 /2 ( x ) = √ 2πx

cos x

dengan mengambil n = ½ maka diperoleh

(a) Jn+1 ( x ) = 2nxJ n ( x )−J n−1(x)

J1 /2+1 ( x ) = 2(1/2)x

J 1/2 ( x )−J1 /2−1(x )

J3 /2 ( x ) = 1xJ 1/2 ( x )−J−1/2( x)

= 1x √ 2πx

sin x−√ 2πx

cos x

= √ 2πx (sin x

x−cos x )

= √ 2πx (sin x−x cos x

x )

Page 4: Fungsi Bessel

(b) dengan mengambil n = - ½ maka diperoleh

Jn−1 (x ) = 2nxJ n ( x )−J n+1(x )

J−1 /2−1 ( x ) = 2(−1/2)x

J−1 /2 (x )−J−1/2+1(x )

J3 /2 ( x ) = −1xJ−1 /2 ( x )−J 1/2(x)

= −1x √ 2

πxcos x−√ 2

πxsin x

= −√ 2πx ( cos x

x−sin x )

= −√ 2πx ( cos x−x sin x

x )