[其他] laurents series
1/(e^z-1)
=1/(1+z+2!z^2......)-1
=(1/z-1/2+z/12-z^3/720)
想問一下
第二行如何到第三行
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推
06/04 11:10, , 1F
06/04 11:10, 1F
→
06/04 11:12, , 2F
06/04 11:12, 2F
推
06/04 12:37, , 3F
06/04 12:37, 3F
→
06/04 13:25, , 4F
06/04 13:25, 4F
→
06/04 13:27, , 5F
06/04 13:27, 5F