Expansión de$f(x)=\frac {\ln(1+\sin ax)-x(1+\arctan x)^{1/x}}{1-\cos x} $ para$x$ cerca de$0$

Question

Cómo encontrar la expansión finita de$$f(x)=\frac {\ln(1+\sin(ax))-x(1+\arctan x)^{1/x}}{1-\cos x} $$ to order $ 2$ and neighborhood $ 0$. My Dr. didn't expand them all to order $ 2 $, tengo problemas para entender los pedidos.

Answer 1

Usted puede escribir, para $x$ cerca de $0$, $$ 1-\cos x=\frac{x^2}{2}+\mathcal{S}(x^3) $$ $$ \sen ax=ax-\frac{a^3x^3}{6}+\mathcal{S}(x^5) $$ $$ \log(1+\sen ax)=ax-\frac{a^2x^2}{2}+\mathcal{S}(x^3) $$ $$ \arctan x=x-\frac{x^3}{3}+\mathcal{S}(x^5) $$ $$ \frac1x\log(1+\arctan x)=1-\frac{x}{2}+\frac{x^3}{12}+\mathcal{S}(x^4) $$ $$ (1+\arctan x)^{1/x}=e^{\frac1x\log(1+\arctan x)}=e^{1-\frac{x}{2}+\mathcal{S}(x^2)}=e(1-\frac{x}{2}+\mathcal{S}(x^2)) $$ y $$ f(x)=\frac {\ln(1+\sen ax)-x(1+\arctan x)^{1/x}}{1-\cos x}=\frac{ax-\frac{a^2x^2}{2}-ex(1-\frac{x}{2})+\mathcal{S}(x^3)}{x^2/2+\mathcal{S}(x^3)} $$ giving the Laurent expansion $$ \frac{2(a-e)}{x}+e-a^2+\mathcal{O}(x) $$ for $f(x)$, $x$ being near $0$.