推 a016258:土法煉鋼推~ 沒方法 有時間一定要慢慢微 114.42.195.20 10/02 15:39
→ a016258:否則就是一分都沒有... 114.42.195.20 10/02 15:39
※ 編輯: suker 來自: 118.169.80.98 (10/02 20:02)
→ AMANISAPEN:謝謝分享 超累的方法 120.127.36.183 10/03 02:16
※ 引述《AMANISAPEN (老叭ㄅㄨ)》之銘言：
: lim 1-cos(1-cosX) 1
: x→0 ---------------- = _
: x^4 8
: 試問是如何拆解
: 煩請解答 謝謝！
真的不會拆解的
爛方法就羅畢達連用4次微分
1-cos(1-cosx)
lim -------------------
x→0 x^4
屬於0/0 羅畢達
f'(x)
= lim --------------
x->0 4x^3
屬於0/0 使用羅畢達
f''(0) f'''(x)
=lim --------------- =lim --------------------
x->0 12x^2 x->0 24x
f''''(x) f''''(0)
= lim ------------------- = ------------- =3/24 =1/8
x->0 24 24
f(x)=1-cos(1-cosx)
f'(x) = sin(1-cosx) *[sin(x)] =>f'(0)=0
f''(x) = cos(1-cosx)*[sinx]*[sin(x)] + sin(1-cosx) *cosx =>f''(0)=0
= cos(1-cosx)* (sinx)^2 + sin(1-cosx) *cosx
f'''(x) = sin(1-cosx)*(sinx)^3 +2cos(1-cosx)*sinx *cosx
+cos(1-cosx) *sinx*cosx +sin(1-cosx) *(-sinx)
=sin(1-cosx)*(sinx)^3 +cos(1-cosx)*sin2x
+cos(1-cosx) *(1/2)sin2x +sin(1-cosx) *(-sinx)
f'''(0)=0
f''''(x) = cos(1-cosx)*(sinx)^4 + sin(1-cosx)*3*(sinx)^2 cosx
+sin(1-cosx)*sinx*sin2x + cos(1-cosx)*cos2x*2
^^^^^^^^^^^^^^^^^^^^
x=0 =>2
+sin(1-cosx) *(1/2)sin2x*sinx
+cos(1-cosx)*(1/2)*cos2x *2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
x=0 =>1
+cos(1-cosx) *(-sinx)*sinx +sin(1-cosx) *(-cosx)
====> f''''(0) =2+1=3
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