※ 引述《rebe212296 (綠豆冰)》之銘言：
: 證明e^iθ=cosθ+isinθ
k k 2k k 2k+1
iθ ∞ (iθ) ∞ (-1) θ ∞ (-1) θ
e = Σ ------ = Σ --------- + i Σ ----------- = cosθ + i sinθ
k=0 k! k=0 (2k)! k=0 (2k+1)!
==========
cosθ + i sinθ
θ θ n
= lim (cos --- + i sin --- )
n->∞ n n
θ θ n 2
= lim (1 + i --- + f(---)) 其中 f(x) = cos x - 1 + i (sin x - x) = O(x )
n->∞ n n
1
------------------ (iθ + n f(θ/n))
θ θ i(θ/n) + f(θ/n)
= lim (1 + i --- + f(---))
n->∞ n n
1
----------------- lim (iθ + n f(θ/n))
θ θ i(θ/n) + f(θ/n) n->∞
= ( lim (1 + i --- + f(---)) )
n->∞ n n
iθ
= e
--
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