连分数
基础
欧拉连分数公式
[a_0+a_0a_1+a_0a_1a_2+cdots+a_1a_2cdots a_n=frac{a_0}{1-frac{a_1}{1+a_1-frac{a_2}{1+a_2+frac{ddots}{cdotsfrac{a_n}{1+a_{n-1}-frac{a_n}{1+a_n}}}}}}
]
这样我们得到
[egin{align}
e^x&=1+x+1cdot xcdot frac{x}{2}+1cdot xcdot frac{x}{2}cdot frac{x}{3}cdots\
&=frac{1}{1-frac{x}{1+x-frac{frac{1}{2}x}{1+frac{1}{2}x-frac{frac{1}{3}x}{1+frac{1}{3}x-ddots}}}}\
&=frac{1}{1-frac{x}{1+x-frac{x}{2+x-frac{2x}{3+x-frac{3x}{4+x-ddots}}}}}
end{align}]
特别地，我们有
[e=frac{1}{1-frac{1}{2-frac{1}{3-frac{2}{4-frac{3}{5-frac{4}{6-ddots}}}}}}
]
同理可以得到
[sin x=frac{x}{1+frac{x^2}{2cdot 3-x^2+frac{2cdot 3x^2}{4cdot 5-x^2+ddots}}}
]
[cos x=frac{1}{1+frac{x^2}{1cdot 2-x^2+frac{1cdot 2 x^2}{3cdot 4-x^2+frac{3cdot 4x^2}{5cdot 6-x^2+ddots}}}}
]
[arctan x=frac{x}{1+frac{x^2}{3-x^2+frac{(3x)^2}{5-3x^2+frac{(5x)^2}{7-5x^2+ddots}}}}
]
[ln(1+x)=frac{x}{1+frac{x}{2-x+frac{2^2x}{3-2x+frac{3^2x}{4-3x+ddots}}}}
]
[sinh x=frac{x}{1-frac{x^2}{2cdot 3+x^2-frac{2cdot 3x^2}{4cdot 5+x^2-ddots}}}
]
[cosh x=frac{1}{1-frac{x^2}{1cdot 2+x^2-frac{1cdot 2 x^2}{3cdot 4+x^2-frac{3cdot 4x^2}{5cdot 6+x^2-ddots}}}}
]
等等，具体看参考【1】
练习
（ 1 ）证明
[sqrt{frac{epi}{2}}=sum_{n=0}^infty frac{1}{(2n+1)!!}+frac{1}{1+frac{1}{1+frac{2}{1+frac{2}{1+frac{3}{1+frac{4}{1+frac{5}{1+ddots}}}}}}}
]
[color{red}{egin{align}frac{1}{sqrt{2pi}}int_z^infty e^{-frac{x^2}{2}} ext{d}x=frac{e^{-frac{z^2}{2}}}{R(z)sqrt{2pi}}end{align}}
]
其中
[R(z)=z+frac{1}{z+frac{2}{z+frac{3}{z+frac{4}{z+ddots}}}}
]
令(z=1)代入得到
[egin{align}frac{1}{R(1)}&=sqrt{e}int_1^infty e^{-frac{x^2}{2}} ext{d}x\
&=sqrt{e}cdot sqrt{frac{pi}{2}}-sqrt{e}int_0^1 e^{-frac{x^2}{2}} ext{d}x
end{align}]
其中
[egin{align}sqrt{e}int_0^1 e^{-frac{x^2}{2}} ext{d}x&=int_0^1 sum_{n=0}^infty frac{1}{n!}left(frac{1}{2}(1-x^2)
ight)^n ext{d}x\
&=sum_{n=0}^infty frac{1}{2^n n!}underbrace{int_0^1 (1-x^2)^n ext{d}x}_{x=sin heta}\
&=sum_{n=0}^infty frac{1}{2^n n!}int_0^{pi/2} cos^{2n+1} heta ext{d} heta\
&=sum_{n=0}^infty frac{1}{2^n n!}frac{(2n)!!}{(2n+1)!!}\
&=sum_{n=0}^infty frac{1}{(2n+1)!!}
end{align}]
代入即证