数学问题,taylorw公式的展开,有个地方不会化简了....

[Crazy_hand]

//taylor 展开
lncosx = ln[1+(cosx-1)] = (cosx-1)-(1/2)(cosx-1)^2+(1/3)(cosx-1)^3 +o(x^6)

因为 cosx-1 的taylor展开= -(1/2)x^2+ (1/24)x^4 - (1/720)x^6 + o(x^6)
所以
ln cos =[(-1/2)x^2+ (1/21)x^4 - (1/720)x^6)] 
    (1)          -(1/2)[ (-1/2)x^2+ (1/24)x^4 - (1/720)x^6) ]^2 
                 +(1/3)[ (-1/2)x^2+ (1/24)x^4 - (1/720)x^6) ]^3
                   +o(x^6)
       = [ (-1/2)x^2+ (1/21)x^4 - (1/720)x^6) ] 
    (1)         -(1/2)[ (1/4)x^4-(1/24)x^6 ]
                    +(1/3)[-(1/8)x^6)]
                       +o(x^6)
------
从(1)开始是怎样化简的

[bfhtian]

[Quote=引用楼主 Crazy_hand 的帖子:]
C/C++ code
//taylor 展开
lncosx = ln[1+(cosx-1)] = (cosx-1)-(1/2)(cosx-1)^2+(1/3)(cosx-1)^3 +o(x^6) 1式

因为 cosx-1 的taylor展开= -(1/2)x^2+ (1/24)x^4 - (1/720)x^6 + o(x^6) 2式
所以
ln cos =[(-1/2)x^2+ (1/21)x^4 - (1/720)x^6)] //这里直接将2式代入1式
(1) -(1/2)[ (-1/2)x^2+ (1/24)x^4 - (1/720)x^6) ]^2
+(1/3)[ (-1/2)x^2+ (1/24)x^4 - (1/720)x^6) ]^3
+o(x^6)
= [ (-1/2)x^2+ (1/21)x^4 - (1/720)x^6) ]
(1) -(1/2)[ (1/4)x^4-(1/24)x^6 ] //展开平方项，次数大于6次方的忽略，所以只计算了(-1/2)x^2的平方与 // 与(-1/2)x^2+ (1/24)x^4的乘积的2倍，下面的3次方也以同样的方式
+(1/3)[-(1/8)x^6)]
+o(x^6)

[/Quote]
注释了一下

[Crazy_hand]

Thank you very much!!

[fireseed]

很简单啊

ln cos =[(-1/2)x^2+ (1/24)x^4 - (1/720)x^6)] 
    (1)          -(1/2)[ (-1/2)x^2+ (1/24)x^4 - (1/720)x^6) ]^2 
                 +(1/3)[ (-1/2)x^2+ (1/24)x^4 - (1/720)x^6) ]^3
                   +o(x^6)

注意这里[ (-1/2)x^2+ (1/24)x^4 - (1/720)x^6) ]^2
这个二次多项展开随次数的增加收敛很快，所以只需记算x的4次项和6次项就可以了。下面那个三次多项原理一样

[Crazy_hand]

[ (-1/2)x^2+ (1/24)x^4 - (1/720)x^6) ]^2
我试着用[(x+y)+z]^2 这样算,不知道是算错了,还是怎么了.反正就是没成功...

[waizqfor]

帮顶下 吃饭去！~

[ltc_mouse]

lz是不是没有把x^8之类的项并入o(x^6)呢？