Quod Erat Demonstrandum

2011/03/24

use series instead of lhopital

Filed under: Additional / Applied Mathematics,HKALE,Pure Mathematics — johnmayhk @ 3:10 下午

利用洛必達法則計算

\displaystyle \lim_{x\rightarrow 0}(\frac{\sin^{-1}x}{x})^{\frac{1}{x^2}}

頗煩。

或以無窮級數,粗糙地解之。

\sin^{-1}x=x+\frac{x^3}{6}+\frac{3x^5}{40}+\dots (*)

(\frac{\sin^{-1}x}{x})^{\frac{1}{x^2}}
=(1+\frac{x^2}{6}+\frac{3x^4}{40}+\dots)^{\frac{1}{x^2}}
\approx (1+\frac{x^2}{6})^{\frac{1}{x^2}}(對接近零的 x,我們忽略 x 的高次方)
=((1+\frac{x^2}{6})^{\frac{6}{x^2}})^{\frac{1}{6}}
=\sqrt[6]{e}(取 x 接近零)

(*) 源自

\sin^{-1}x
=\int_0^x \frac{d}{dt}\sin^{-1}tdt
=\int_0^x \frac{dt}{\sqrt{1-t^2}}=\int_0^t (1-t^2)^{\frac{-1}{2}}dt
=\int_0^x (1+(\frac{-1}{2})(-t^2)+\frac{(-1/2)(-3/2)}{2!}(-t^2)^2+\dots)dt
=x+\frac{x^3}{6}+\frac{3x^5}{40}+\dots

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