Question

请问这个题为什么不能用泰勒公式做？

哪里出错了

Answer 1

第一次计算，展开的次数少了，分母为4阶，则上面相乘结果也要展到4阶，4阶可不止2+2，3+1也是，所以上面两个相乘都必须展到3阶

追问

没太懂，两项都都展到二阶相乘不正好是四阶吗

追答

你没理解泰勒展开，事实上泰勒展开是要展开成无穷阶，只是分母为4阶，超过4阶极限为0，直接忽略了，不是让你得到一个四阶的结果就不往后展开了，你要计算每一个四阶

Answer 2

(4)

ln(1-x^2) = -x^2 -(1/2)x^4 +o(x^4)

ln(1+x) = x -(1/2)x^2 +(1/3)x^3 +o(x^3)

ln(1-x) = -x -(1/2)x^2 -(1/3)x^3 +o(x^3)

ln(1+x).ln(1-x)

=[x -(1/2)x^2 +(1/3)x^3 +o(x^3)].[-x -(1/2)x^2 -(1/3)x^3 +o(x^3)]

=x.[-x -(1/2)x^2 -(1/3)x^3 +o(x^3)] -(1/2)x^2. [-x -(1/2)x^2 -(1/3)x^3 +o(x^3)]

+(1/3)x^3 .[-x -(1/2)x^2 -(1/3)x^3 +o(x^3)] +o(x^4)

=[-x^2 -(1/2)x^3 -(1/3)x^4 +o(x^4)] -(1/2) [-x^3 -(1/2)x^4 +o(x^4)]

+(1/3)[-x^4 +o(x^4)] +o(x^4)

=-x^2 +(-1/2 + 1/2)x^3 +( -1/3 +1/4 -1/3)x^4 +o(x^4)

=-x^2 - (5/12)x^4 +o(x^4) 

ln(1+x)ln(1-x) -ln(1-x^2)

=[-x^2 - (5/12)x^4 +o(x^4) ]-[-x^2 -(1/2)x^4 +o(x^4)]

=(-5/12 + 1/2)x^4 +o(x^4)

=(1/12)x^4 +o(x^4)

lim(x->0) [ln(1+x)ln(1-x) -ln(1-x^2) ]/x^4

=lim(x->0) (1/12)x^4/x^4

=1/12

追问

请问ln(1+x)这两项为什么要展开到三阶，展开到二阶相乘不正好是四次方吗

追答

答案在我的展开之中

ln(1+x).ln(1-x)

=[x -(1/2)x^2 +(1/3)x^3 +o(x^3)].[-x -(1/2)x^2 -(1/3)x^3 +o(x^3)]

=x.[-x -(1/2)x^2 -(1/3)x^3 +o(x^3)] 

不展开到-(1/3)x^3 ， 少了一项 -(1/3)x^4 

-(1/2)x^2.[-x -(1/2)x^2 -(1/3)x^3 +o(x^3)]

+(1/3)x^3. [-x -(1/2)x^2 -(1/3)x^3 +o(x^3)]

不展开到 (1/3x^3 ， 少了一项 -(1/3)x^4 

这导致泰勒展开的不完整！

追问

好滴，谢谢！