作者mathmanliu (烦恼即是菩提)
看板trans_math
标题Re: POWER SERIES
时间Mon Mar 9 00:49:12 2009
※ 引述《ADAH33 (逐渐消失的生命)》之铭言：
: 请问要怎样求 ln(1-x^2) 的 power series
1
── = 1- x + x^2 - x^3 + .... |x|<1
1+ x
两边定积分0~ x, 得
ln|1+x| = x - x^2/2 + x^3/3 - x^4/ 4 + ..... -1<x<= 1 ----(A)
=>ln|1-x| = -x - x^2/2 - x^3/3 - .... -1<= x < 1 ----(B)
(A)+(B) => ln|1-x^2| = -2 (x^2/2 + x^4/4 + x^6/6 + ....) -1<x<1
另法:
由(A), 将 x 改为 -t^2, 得
ln|1-t^2| = -t^2 - t^4/2 - t^6/ 3 - .... - t^2n/n -... -1<t<1 ---(C)
再将(C)之t换为 x即可!
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请参观yahoo游艺数学
http://tw.group.knowledge.yahoo.com/math-etm
不虚此行喔!
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