Taylor's Series

2010-08-01 1:33 am

回答 (1)

2010-08-01 4:34 am
✔ 最佳答案
g(x,y,z)=cos(x+y+z)-cosx cosy cosz

g(0, π/2, π/2)= -1

grad(g)
=<sinxcosycosz-sin(x+y+z),cosxsinycosz-sin(x+y+z),cosxcosysinz-sin(x+y+z)>
at P(0, π/2, π/2), grad(g)=<0,0,0>

For P(0,π/2, π/2)
∂^2g/∂x^2=cosxcosycosz-cos(x+y+z)=1
∂^2g/∂x∂y=-sinxsinycosz-cos(x+y+z)=1
∂^2g/∂x∂z=-sinxcosysinz-cos(x+y+z)=1
∂^2g/∂y^2=cosxcosycosz-cos(x+y+z)=1
∂^2g/∂y∂z=-cosxsinysinz-cos(x+y+z)=0
∂^2g/∂z^2=cosxcosycosz-cos(x+y+z)=1
D=
[ 1 1 1 ]
[ 1 1 0 ]
[ 1 0 1 ]

For P(0, π/2, π/2),
g(x,y,z)=g(0, π/2, π/2)+<0,0,0>‧<x-0, y-π/2, z-π/2>
+1/2! <x-0, y-π/2, z-π/2> D <x, y-π/2, z-π/2>^t (M^t= transpose of M)
+....
= -1 + [x^2+(y-π/2)^2+(z-π/2)^2+2x(y-π/2)+2x(z-π/2)]/2! +...


收錄日期: 2021-04-22 00:34:54
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100731000051KK01106

檢視 Wayback Machine 備份