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DERET TAYLOR DAN DERET MACLAURIN
Misalkan f Misalkan f dapat dapat diturunkan untuk semua ordo pada selang (a − R ,a + R ) dan misalkan ∞
n = c + c (x − a ) + c (x − a )2 + c (x − a )3 + c (x − a )4 + ... o 1 2 3 4
∑ c n (x − a )
f (x ) =
n =0 konvergen pada selang (a − R ,a + R ) , maka f ( a ) = co 2
3
f ' ( x) = c1 + 2c2 ( x − a) + 3c3 ( x − a) + 4c4 ( x − a ) + ... ⇒ f ' ( a ) = 1!c1 ⇒ c1 = 2
f ' ' ( x ) = 2.1.c2 + 3.2.c3 ( x − a ) + 4.3.c4 ( x − a ) + ... ⇒ f ' ' ( a ) = 2!c2 ⇒ c2 = f ' ' ' ( x) = 3.2.1.c3 + 4.3.2.c4 ( x − a ) + ... ⇒ f ' ' ' ( a ) = 3!c3 ⇒ c3 = f
(4)
( x ) = 4.3.2.c4 ( x − a ) + ... ⇒ f ( 4 ) ( a ) = 4!c3 ⇒ c4 =
∴ cn =
f
(n)
2!
f ' ' ' ( a )
3!
(a)
4!
n! f
∴ f ( x ) = ∑ ∞
∑ n = 0 ∞
∑ n =0
( n)
(a )
n!
n =0
f (x ) =
1! f ' ' ( a )
(a)
∞
f (x ) =
f
( 4)
f ' (a )
( x − a )
n
f (n )(a ) (x − a )n disebut deret Taylor. Jika a = 0 , maka n !
f (n )(0) n x disebut deret Maclaurin. n !
Contoh: 1. Ekspansikan f (x ) = ln x dalam deret Taylor di x = 1 (a (a = = 1), kemudian tentukan selang konvergensinya 2. Ekspansikan f (x ) = sin x , f (x ) = cos x , f (x ) = e x , f (x ) = ln(1 + x ) , dan f (x ) = (1 + x )m dalam deret Maclaurin, kemudian tentukan selang konvergensinya. Penyelesaian: 1. f (x ) = ln x
f (1) = ln1 = 0 1 1 f ' (1) ⇒ f ' (1) = = 1 ⇒ c 1 = =1 x 1 1! 1 1 f ' ' (1) 1 ⇒ f ' ' (1) = − = −1 ⇒ c 2 = =− f ' ' (x ) = − 1 2! 2 x 2 f ' (x ) =
f ' ' ' (x ) =
2 2 f ' ' ' (1) 2 1 ⇒ f ' ' ' (1) = = 2 ⇒ c 3 = = = 3 3 3 ! 3 ! 3 x 1
1 (−1)n +1 c n n = lim n + 1 = 1 x − 1 < lim = lim n →∞ c n +1 n →∞ (−1)n + 2 1 n →∞ n n + 1 x − 1 < 1 ⇔ −1 < x − 1 < 1 ⇔ 0 < x < 2, harus di cek untuk x = 0 dan 2 . ∞
x = 0 ⇒ f (0) = ∑ (−1) n =1
∞
n +1
1 n
∞
( −1) =∑ (−1) n
2 n +1
n =1
∞
1 n
∞
=−∑
1
n =1 n
divergen
1 1 x = 2 ⇒ f (2) = ∑ (−1)n +1 (1)n = ∑ (−1)n +1 konvergen n n n =1 n =1 1 1 1 ln x = (x − 1) − (x − 1)2 + (x − 1)3 − (x − 1)4 + ... konvergen pada 2 3 4 selang 0 < x ≤ 2
2. ⊗ f (x ) = sin x f (0) = sin 0 = 0 f ' (0) =1 1! f ' ' (0) =0 f ' ' (x ) = − sin x ⇒ f ' ' (0) = − sin 0 = 0 ⇒ c 2 = 2! f ' ' ' (0) 1 = − f ' ' ' (x ) = − cos x ⇒ f ' ' ' (0) = − cos 0 = −1 ⇒ c 3 = 3! 3! f (4)(0) 0 ( 4 ) ( 4 ) = =0 f (x ) = sin x ⇒ f (0) = sin 0 = 0 ⇒ c 4 = 4! 4! f ' (x ) = cos x ⇒ f ' (0) = cos 0 = 1 ⇒ c 1 =
1 3 1 5 1 7 ∴ f (x ) = x − x + x − x + ... = 3! 5! 7!
x < lim n →∞
cn cn +1
( −1) n +1 = lim n →∞
( −1)
n+ 2
∞
1 (−1)n +1 x 2n −1 (2n − 1)! n =1
∑
1 ( 2n − 1)! 1
= lim n →∞
( 2n + 1)! ( 2n − 1)!
= lim 2n( 2n + 1) = ∞ n →∞
( 2( n + 1) − 1)!
1 3 1 5 1 7 x + x − x + ... konvergen untuk semua x 3! 5! 7!
sin x = x − ⊗ f (x ) = cos x
f (0 ) = cos 0 = 1 f ' (0 ) = 0 1! f ' ' (0 ) 1 f ' ' ( x ) = − cos x ⇒ f ' ' (0 ) = − cos 0 = − 1 ⇒ c 2 = = − 2! 2! f ' ' ' (0 ) f ' ' ' ( x ) = sin x ⇒ f ' ' ' (0 ) = sin 0 = 0 ⇒ c 3 = = 0 3! f (4 )(0 ) 1 f (4 )( x ) = cos x ⇒ f (4 )(0 ) = cos 0 = 1 ⇒ c 4 = 0 = 4! 4! f ' ( x ) = − sin x ⇒ f ' (0 ) = − sin 0 = 0 ⇒ c 1 =
1 2 1 4 1 6 x + x − x + ... = ∴ f ( x ) = 1 − 2! 4! 6!
∞
∑ n = 0
1 ( −1)n x 2n (2n )!
1 (−1)n c n (2n + 2)! (2n )! x < lim = lim = lim = lim (2n + 1)(2n + 2) = ∞ 1 c ( 2 n )! + n 1 n →∞ n +1 n →∞ (−1) n →∞ n →∞ (2(n + 1))! cos x = 1 −
1 2 1 4 1 6 x + x − x + ... konvergen untuk semua x 2! 4! 6!
x
⊗ f (x ) = e
f (0) = e 0 = 1 f ' (0) =1 f ' (x ) = e x ⇒ f ' (0) = e 0 = 1 ⇒ c 1 = 1! f ' ' (0) 1 = f ' ' (x ) = e x ⇒ f ' ' (0) = e 0 = 1 ⇒ c 2 = 2! 2! f ' ' ' (0) 1 = f ' ' ' (x ) = e x ⇒ f ' ' ' (0) = e 0 = 1 ⇒ c 3 = 3! 3! f (4)(0) 1 ( 4 ) x ( 4 ) 0 = f (x ) = e ⇒ f (0) = e = 1 ⇒ c 4 = 4! 4! 1 2 1 3 1 4 ∴ f (x ) = 1 + x + x + x + x + ... = 2! 3! 4!
∞
∑ n = 0
1 n x n !
1 c n (n + 1)! n ! x < lim = lim = lim = lim (n + 1) = ∞ 1 n →∞ c n +1 n →∞ n →∞ n ! n →∞ (n + 1)! 1 1 1 4 e x = 1 + x + x 2 + x 3 + x + ... konvergen untuk semua x 2! 3! 4! ⊗ f (x ) = ln(x + 1)
Dari soal 1, ekspansi f (y ) = ln y dalam deret Taylor di y = 1 . ln y = (y − 1) −
selang 0 < y ≤ 2 . Misalkan y = x + 1 , jika y = 1 , maka x = 0 , sehingga: ln(x + 1) = x −
1 2 1 3 1 4 x + x − x + ... konvergen pada selang 2 3 4
0 < x + 1 ≤ 2 atau − 1 < x ≤ 1
x
⊗ f (x ) = e
f (0) = e 0 = 1 f ' (0) =1 f ' (x ) = e x ⇒ f ' (0) = e 0 = 1 ⇒ c 1 = 1! f ' ' (0) 1 = f ' ' (x ) = e x ⇒ f ' ' (0) = e 0 = 1 ⇒ c 2 = 2! 2! f ' ' ' (0) 1 = f ' ' ' (x ) = e x ⇒ f ' ' ' (0) = e 0 = 1 ⇒ c 3 = 3! 3! f (4)(0) 1 ( 4 ) x ( 4 ) 0 = f (x ) = e ⇒ f (0) = e = 1 ⇒ c 4 = 4! 4! 1 2 1 3 1 4 ∴ f (x ) = 1 + x + x + x + x + ... = 2! 3! 4!
∞
∑ n = 0
1 n x n !
1 c n (n + 1)! n ! x < lim = lim = lim = lim (n + 1) = ∞ 1 c n ! n →∞ n +1 n →∞ n →∞ n →∞ (n + 1)!
1 1 1 4 e x = 1 + x + x 2 + x 3 + x + ... konvergen untuk semua x 2! 3! 4! come back
