I am new to Taylor expansions and I would like to calculate the Taylor polynomial of the function $x^{1/x}=e^{(1/x)\log x}$. Since the function is not defined at $x=0$, how should I choose the point which I will calculate the expansion at? I would like to understand the following result

$$x^{1/x}=1-\frac1x \log\frac1x+O(x^{-2})\qquad (1)$$

The first derivative of my function is $$f'(x)=x^{1/x-2}\cdot(1-\log x)$$ Given that the Taylor polynomial is $f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots$, I can't get how this first derivative is used in $(1)$. For which $a$ is $(1)$ calculated? Moreover, why I cannot obtain the same first two terms using the derivative?

Should I use the Taylor expansion of $e^u$ substituting $u$ with my $u(x)$? Is this possible? Anyway, I should obtain the same result in every situation.

Could you please guide me to understand these points? What's my problem?

Thank you for your help.

  • $\begingroup$ Your formula makes no sense. What is $n$? $\endgroup$ Commented Sep 11, 2014 at 8:53
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    $\begingroup$ Actually, Taylor expansions can be useful here: write $$x^{1/x}=e^{(1/x)\log x}$$ first. $\endgroup$ Commented Sep 11, 2014 at 8:57
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    $\begingroup$ I assume you know the Taylor expansion of $e^u$ as a function of $u$. Just replace $u$ everywhere with $(1/x)\log x$. $\endgroup$ Commented Sep 11, 2014 at 9:02
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    $\begingroup$ @Gerry: but that would give error term $O(\log(x) / x^2)$. But I actually suspect the claimed equation is not true. $\endgroup$
    – user14972
    Commented Sep 11, 2014 at 9:59
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    $\begingroup$ You have to decide once and for all the point around which you seek an expansion of the function. When $x\to\infty$, your (1) "almost" holds since $\log x/x\to0$ and $e^u=1+u+\Theta(u^2)$ hence $x^{1/x}=1+\log x/x+\Theta((\log x)^2/x^2)$. Thus, in (1), the main term $1+\log x/x$ is correct but the remainder term $O(1/x^2)$ is wrong. The expansion when $x\to0$ (which you mysteriously invoke in the question) is quite unrelated to (1) (and to the other question you link to, I must add). $\endgroup$
    – Did
    Commented Sep 11, 2014 at 11:14


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