Re: [理工] [離散]生成函數
1/(1+3x)(1+2x)^2
= a/(1+3x) + b/(1+2x) + c/(1+2x)^2
= [ a ( 1+2x)^2 + b * (1+2x)(1+3x) + c(1+3x) ] / [(1+3x)(1+2x)^2]
=> a ( 1+2x)^2 + b * (1+2x)(1+3x) + c(1+3x) = 1
let x = -1/3
a ( 1 / 9) = 1 => a = 9
let x = -1/2
c ( -1/2) = 1 => c = -2
常數項 a + b + c = 1 => b = -6
有錯還請不吝指正。
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03/16 20:47, , 1F
03/16 20:47, 1F
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