更新1:
thanks... i thought that it's about taylor series tim -_-""
problem about taylor series
2010-01-04 8:07 pm
expand [ 1 / sqrt ( 1 - x ) ] in ascending powers of x as far as the term x to the power 3. please give a brief explanation. thanks so much.
回答 (2)
2010-01-04 9:56 pm
✔ 最佳答案
Using Newton's generalized binomial theorem (1-x)^(-1/2)
=1+(-1/2)(-x)+[(-1/2)(-1/2-1)/2](-x)^2+[(-1/2)(-1/2-1)(-1/2-2)/3!](-x^3)
=1+(1/2)x+(3/8)x^2+(5/16)x^3
2010-01-04 11:20 pm
By general binomial series formula:
1/(1-x)^(1/2)
=(1-x)^(-1/2)
=1+(-1/2)x+(-1/2)(-1/2-1)x^2/2!+(-1/2)(-1/2-1)(-1/2-2)x^3/3!+...
=1-x/2+3x^2/8-5x^3/48+...
1/(1-x)^(1/2)
=(1-x)^(-1/2)
=1+(-1/2)x+(-1/2)(-1/2-1)x^2/2!+(-1/2)(-1/2-1)(-1/2-2)x^3/3!+...
=1-x/2+3x^2/8-5x^3/48+...
收錄日期: 2021-04-26 14:00:46
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100104000051KK00478