Kiara Haas

2022-03-23

Maclaurin series for a function
Provided I have the function
$f\left(x\right)={\left(1+x\right)}^{\frac{1}{x}},$
and I want to calculate a 3rd order Maclaurin series, how can that be done without taking direct derivatives (as this seems hard..). I know that
${\left(1+x\right)}^{\frac{1}{x}}={e}^{\mathrm{ln}\frac{\left(1+x\right)}{x}},$
and the Maclaurin series for ${e}^{x}$ is easy to prove, so I think it's a good direction..

aznluck4u72x4

$\frac{\mathrm{ln}\left(1+x\right)}{x}=1-\frac{x}{2}+\frac{{x}^{2}}{3}-\frac{{x}^{3}}{4}+o\left({x}^{3}\right)$
Substitute -$\frac{x}{2}+\frac{{x}^{2}}{3}-\frac{{x}^{3}}{4}$ to u in the development of eu, first computing the succesive powers of u:
${u}^{2}=\frac{{x}^{2}}{4}-\frac{{x}^{3}}{3}+o\left({x}^{3}\right)$
${u}^{3}-\frac{{x}^{3}}{8}+o\left({x}^{3}\right),$
so that
${\left(1+x\right)}^{\frac{1}{x}}=e\left(1+u+\frac{{u}^{2}}{2}+\frac{{u}^{3}}{6}+o\left(u\right)\right)$
$=e\left(1-\frac{x}{2}+\frac{11}{48}{x}^{2}-\frac{7}{16}{x}^{3}+o\left({x}^{3}\right)\right).$

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