Example: bachelor of science

テイラー展開 - math.chs.nihon-u.ac.jp

1 . f (x) x = a . f 00 (a). f (x) = f (a) + f 0 (a)(x a) + (x a)2 + . 2.. |x a| < a ( . ) > 0 f x = a . f . (C ) a = 0 .. 2 .. 1 f 0 (C 1 ) .. f (x) f (a). = f 0 (c) (c x a ). x a . f (x) = f (a) + f 0 (c)(x a). c = a + (x a) (0 1) . f (x) = f (a) + f 0 (a + (x a))(x a) (0 1). x a .. 1. n n f (n) (C n ) . f 00 (a) f (n 1) (a). f (x) = f (a) + f 0 (a)(x a) + (x a)2 + + (x a)n 1. 2 (n 1)! f (n) (a + (x a)). + (x a)n n! . f (n) (a + (x a)). Rn = (x a)n n! . n 1 (n 1 ). f 00 (a) f (n 1) (a). f 0 (a)(x a) + (x a)2 + + (x a)n 1. 2 (n 1)! f (x) Rn .. sin x . x3 x5. sin x = x + + . 3! 5! . f (n) ( x) n Rn = x n! f (n) ( x) sin( x) cos( x) 1 .. |x|n |Rn | . n! . sin x . 2 (0 . ) 2 3 (1 2 ) sin x.

テイラー展開 森 真 1 序 関数f(x) のx = a におけるテイラー展開は f(x) = f(a)+f0(a)(x−a)+ f00(a) 2 (x−a)2 +···と表される.右辺の級数が収束する範囲でのみ成り立つことに注意しなけれ

Tags:

  Math, Nihon

Information

Domain:

Source:

Link to this page:

Please notify us if you found a problem with this document:

Other abuse

Advertisement

Transcription of テイラー展開 - math.chs.nihon-u.ac.jp

1 1 . f (x) x = a . f 00 (a). f (x) = f (a) + f 0 (a)(x a) + (x a)2 + . 2.. |x a| < a ( . ) > 0 f x = a . f . (C ) a = 0 .. 2 .. 1 f 0 (C 1 ) .. f (x) f (a). = f 0 (c) (c x a ). x a . f (x) = f (a) + f 0 (c)(x a). c = a + (x a) (0 1) . f (x) = f (a) + f 0 (a + (x a))(x a) (0 1). x a .. 1. n n f (n) (C n ) . f 00 (a) f (n 1) (a). f (x) = f (a) + f 0 (a)(x a) + (x a)2 + + (x a)n 1. 2 (n 1)! f (n) (a + (x a)). + (x a)n n! . f (n) (a + (x a)). Rn = (x a)n n! . n 1 (n 1 ). f 00 (a) f (n 1) (a). f 0 (a)(x a) + (x a)2 + + (x a)n 1. 2 (n 1)! f (x) Rn .. sin x . x3 x5. sin x = x + + . 3! 5! . f (n) ( x) n Rn = x n! f (n) ( x) sin( x) cos( x) 1 .. |x|n |Rn | . n! . sin x . 2 (0 . ) 2 3 (1 2 ) sin x.

2 X3 0 x x + R2 x + 0x2 + R3. 4 5 (4 . x3. 5 ) x 3! .. 2. 3. 2. 1. -3 -2 -1 1 2 3. -1. -2. -3. 4 4. 2 2. -3 -2 -1 1 2 3 -3 -2 -1 1 2 3. -2 -2. -4 -4. 4. 3. 2. 2. 1. -3 -2 -1 1 2 3 -3 -2 -1 1 2 3. -1. -2. -2. -3. -4. 1: sin x 0 4 . ex . x2 x3. ex = 1 + x + + + . 2 3! . e x n x Rn =. n! e < 3 ex .. 3|x| n Rn |x|. n! 3|x| x m 1 <. |x| m 3|x| < 3m . x . 3. 30. 25. 20. 15. 10. 5. -3 -2 -1 1 2 3. 30 30. 25 25. 20 20. 15 15. 10 10. 5 5. -3 -2 -1 1 2 3 -3 -2 -1 1 2 3. 30 30. 25 25. 20 20. 15 15. 10 10. 5 5. -3 -2 -1 1 2 3 -3 -2 -1 1 2 3. 2: ex 0 4 . log(1 + x) . x2 x3. log(1 + x) = x + + . 2 3.. ( 1)n 1 n Rn = x (1 + x)n 1 x > 0. = 0 Rn x < 0 = 1 . |Rn | .. xn x 0. |Rn | n |x|n x<0. (1+x)n |x| > 1.

3 4. 1. -1 1. -1. -2. -3. 1 1. -1 1 -1 1. -1 -1. -2 -2. -3 -3. 1 1. -1 1 -1 1. -1 -1. -2 -2. -3 -3. 3: log(1 + x) 0 4 . x 0 x = 1.. log 0 . (1 + x)c 2 . c c . (1 + x)c = 1 + x+ x2 + . 1 2. c 1 . (1 + x)a x = 1 .. 3 .. b0 + b1 (x a) + b2 (x a)2 + . 5. Rn 0 . x . b1 , b2 , .. p 1. lim sup n |bn | =. n . |x a| < .. bn+1 . lim . n bn .. 1 + 3x + 22 x2 + 33 x3 + 24 x4 + . 1. 3 . 1 + x2 + x4 + . 1 .. x2 x3. log(1 + x) = x + . 2 3.. x3 x5. x+ + + . 3 5.. x2 x4. + + . 2 4. log(1 + x) . 2 0 3 0. 1 0 0 x x = (log(1 + x)) = x + + . 1+x 2 3. = 1 x + x2 .. Z x x log(1 + x) + log(1 + x) x =. log(1 + t) dt 0. x Z x 2 Z Z x 3. t t = t dt dt + dt . 0 0 2 0 3. x2 x3 x4. = + . 2 2 3 3 4. |x| < 1 .. 6.