Taylor series of $\sqrt{\ln\left(\frac{1}{x}\right)}$ I am trying to compute the Taylor series of:
$\sqrt{\ln\left(\frac{1}{x}\right)}$
I have computed the derivatives and evaluated them in $x=1/e$ but I cannot find the formula for the sequence of the coefficients.
Any other ideas?
Thanks!
PS: The approximation should be valid over the interval $x \in [0,1]$
 A: This is a note on how you might compute the derivatives using implicit differentiation. It isn't really an answer, but was too long for a comment, and someone may see how to use this in an answer.
$$y=\left(-\ln(x)\right)^{\frac 12}$$
$$y'=\frac 12\left(-\ln(x)\right)^{-\frac 12}\cdot-\frac 1x=-\frac 1{2xy}$$
Whence $$2xyy'=-1$$ and $$2yy'+2xy'^2+2xyy''=0$$ From which you can cancel a factor two and keep going to compute higher derivatives.
Using $2xyy'=-1$ you can multiply through by $y'$ and obtain:
$$2yy'^2+2xy'^3-y''=0$$ or $$y''=2xy'^3+2yy'^2$$ Which you can differentiate as often as you like.
A: Let's try to find the expansion about $x=1/2$.  Also, note that $\ln(1/x) = -\ln(x)$.
$$
f(x) = \sqrt{-\ln(x)};\quad f(1/2) = \sqrt{\ln(2)}\\
f'(x)= \frac{1}{2}(-\ln(x))^{-1/2}\cdot(-1/x); \quad f'(1/2) = -1/(\sqrt{\ln(2)})
$$
Whatever this Taylor Series is, it's only going to get worse from here.  To get something workable, I think you'll have to settle for something that converges only over a part of $(0,1)$.
A: Hint: the general Taylor series expansion is
$$
\begin{align}
f(x) &= f(x_0) + {f'(x-x_0)(x-x_0)} + {f''(x_0) (x-x_0)^2\over 2!} + \cdots\\
&= \sum_{n=0}^\infty {{d^n\over dx^n}f(x)|_{x=x_0} \times (x-x_0)^n \over n!}
\end{align}
$$
This works wherever all the derivatives exist at $x_0$ and the function is continuous between $x$ and $x_0$.
For this specific problem:
$$
f(x) = f({1\over e}) + f'({1\over e})(x-{1\over e}) + {f''({1\over e})(x-{1\over e})^2\over 2} + {f'''({1\over e})(x-{1\over e})^3\over 3!} + \cdots
$$
