※ 引述《arthur61106 (arthur)》之銘言：
: http://ppt.cc/TZFX
: 有想過將函數先積分在微分.... 可是後面卡住
: 不知道想法有無錯誤?
arcsin(x)
y(x) = -------------
√[1 - x^2]
- y 1
y' = ---------- + -----------
(1 - x^2) 1 - x^2
=> (1 - x^2)y' + y = 1 只要用一階微分方程式
∞
y = Σa_nx^n a_0 = 0 代入
n=0
∞ ∞
=> (1 - x^2)Σna_nx^(n-1) - xΣa_nx^n = 1
n=1 n=0
=> -Σna_nx^(n+1) + Σ(n+2)a_(n+2)x^(n+1) = 1
=> a_1 = 1
a_2 = (1/2)a_0
a_(n+2) = (n+1)/(n+2) * a_n
=> a_2k = 0 for all k = 0, 1, 2, ...
a_(2k+1) = 1 * (2/3) * (4/5) * .... [(2k)/(2k+1)]
2^(2k) (k!)^2
= -----------
(2k+1)!
arcsin(x) ∞ 2^(2n) (n!)^2
=> ----------- = Σ --------------x^(2n+1)
√[1 - x^2] n=0 (2k+1)!
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