What is the Maclaurin series for (1+x)^(1/x)?

Question

abb0t · Answer

A maclaurin series is a Taylor series expansion of a function about 0.
So this will be long. But try and look for a pattern.
Start by taking a few derivatives.

ghazi · Answer

series expansion at x=0, 1 or what?

ghazi · Answer

it is expansion about zero $$e-\frac{ ex }{ 2 }+\frac{ 11ex^2 }{ 24 }-\frac{ 7ex^3 }{ 16 }+ ......$$

abb0t · Answer

$$f(x) = f(0) + f'(0)x + \frac{ f''(0) }{ 2! }x^2+\frac{ f^{(3)}(0) }{ 3! }x^3 $$

abb0t · Answer

and so forth.

zehanz · Answer

Maybe the problem is finding the derivatives?
Start by writing $$f(x)=(1+x)^{\frac{ 1 }{ x }}$$as$$f(x) = e ^{\ln(1+x)^{\frac{ 1 }{ x }}}=e ^{\frac{ 1 }{ x }\ln(1+x)}$$Now the first derivative is:$$f'(x)=e ^{\frac{ 1 }{ x }\ln(1+x)}(-\frac{ 1 }{ x^2 }\ln(1+x)+\frac{ 1 }{ x }\frac{ 1 }{ 1+x })$$$$=\frac{ 1 }{ x }\left( \frac{ 1 }{ 1+x }-\frac{ 1 }{ x }\ln(1+x) \right)e ^{\frac{ 1 }{ x }\ln(1+x)}$$Hmm, finding f'(0) is not *that* straightforward...