先求出arcsin(x)在x=0的泰勒展開,為x+(1/6)*x^3+(3/40)*x^5+(5/112)*x^7+O(x^9),
通項(xiàng)為(2n-1)!/(2n)!*x^(2n+1).第n+1項(xiàng)系數(shù)為：A_(n+1)=(2n-1)!/(2n)!/(2n+1).
這個結(jié)果在很多版本的微積分、數(shù)學(xué)分析、高等數(shù)學(xué)課本上都能夠找到
然后平方,只有偶次項(xiàng),根據(jù)多項(xiàng)式乘法法則不難算出,通項(xiàng)為C_(n+1)=∑A_(k)*A_(2n+2-k)*x^(2n+2) (k=1, 2, ... , n+1),
其中,前面幾項(xiàng)為x^2+(1/3)*x^4+(8/45)*x^6+(4/35)*x^8+(128/1575)*x^10+O(x^12),