→ Honor1984 :Π [1 + b^k]是從k=1到2n 07/03 23:11
→ ballballking:感謝H大!!!! 07/04 07:33
※ 引述《ballballking (蛋蛋王)》之銘言：
: n
: Π cos [ kpi / 2n+1 ] = 1 / 2^n
: k=1
a = cos[pi/(2n+1)] + isin[pi/(2n+1)]
b = a^2
n
A = Π cos [ kpi / 2n+1 ] > 0
k=1
2n
(-1)^n A^2 = Π [a^k + (1/a)^k]/2
k=1
n
= 1/[a^((2n+1)n) 2^(2n)] Π [1 + b^k]
k=1
=1/[a^((2n+1)n) 2^(2n)]
=> A^2 = (1/2^n)^2
=> A = 1/2^n
: 請問這兩條怎樣推導出來的 感謝
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