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 $x \in [0,1]$
On
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 $(0,1)$.
On
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.
On
It is clear that \begin{equation*} \sqrt{\ln\frac{1}{x}}\,=\sqrt{-\ln x}\,. \end{equation*} By virtue of the Faa di Bruno formula and several properties of the Bell polynomials of the second kind (or say, partial Bell polynomials) $\textrm{B}_{n,k}$ for $n\ge k\ge0$, we obtain \begin{align*} \bigl(\sqrt{-\ln x}\,\bigr)^{(n)}&=\sum_{k=0}^{n}\bigl(\sqrt{u}\,\bigr)^{(k)} \textrm{B}_{n,k}\Biggl(-\frac{0!}{x}, (-1)^2\frac{1!}{x^2}, \dotsc, (-1)^{n-k+1}\frac{(n-k)!}{x^{n-k+1}}\Biggr)\\ &=\sum_{k=0}^{n}\biggl\langle\frac{1}{2}\biggr\rangle_k u^{1/2-k} \frac{(-1)^n}{x^n} \textrm{B}_{n,k}(0!,1!, \dotsc, (n-k)!)\\ &=\frac{(-1)^n}{x^n} \sum_{k=0}^{n}\biggl\langle\frac{1}{2}\biggr\rangle_k (-\ln x)^{1/2-k} (-1)^{n-k}s(n,k)\\ &=-\frac{1}{x^n} \sum_{k=0}^{n}\frac{(2k-3)!!}{2^k} \frac{s(n,k)}{(-\ln x)^{k-1/2}} \end{align*} for $n\ge0$, where $u=u(x)=-\ln x$, the notation $\langle\alpha\rangle_k$ denotes the falling factorial, and the quantity $s(n,k)$ stands for the Stirling numbers of the first kind. Consequently, for $x_0\in(0,1)$, we arrive at \begin{equation*} \boxed{\sqrt{\ln\frac{1}{x}}\, =\sqrt{-\ln x}\, =-\sum_{n=0}^{\infty}\Biggl[\frac{1}{x_0^n} \sum_{k=0}^{n}\frac{(2k-3)!!}{2^k} \frac{s(n,k)}{(-\ln x_0)^{k-1/2}}\Biggr]\frac{(x-x_0)^n}{n!}, \quad |x-x_0|<\min\{x_0,1-x_0\}.} \end{equation*} In particular, since \begin{equation*} \lim_{x\to1/\textrm{e}}\bigl(\sqrt{-\ln x}\,\bigr)^{(n)} =-\textrm{e}^n \sum_{k=0}^{n}\frac{(2k-3)!!}{2^k}s(n,k) \end{equation*} and \begin{equation*} \lim_{x\to1/2}\bigl(\sqrt{-\ln x}\,\bigr)^{(n)} =-2^n \sum_{k=0}^{n}\frac{(2k-3)!!}{2^k}\frac{s(n,k)}{(\ln2)^{k-1/2}} \end{equation*} for $n\ge0$, we find \begin{equation*} \boxed{\sqrt{\ln\frac{1}{x}}\, =\sqrt{-\ln x}\, =-\sum_{n=0}^{\infty} \Biggl[\sum_{k=0}^{n}\frac{(2k-3)!!}{2^k}s(n,k)\Biggr]\frac{\textrm{e}^n}{n!}\biggl(x-\frac1{\textrm{e}}\biggr)^n, \quad \biggl|x-\frac1{\textrm{e}}\biggr|<\frac{1}{\textrm{e}}} \end{equation*} and \begin{equation*} \boxed{\sqrt{\ln\frac{1}{x}}\, =\sqrt{-\ln x}\, =-\sum_{n=0}^{\infty} \Biggl[\sum_{k=0}^{n}\frac{(2k-3)!!}{2^k} \frac{s(n,k)}{(\ln2)^{k-1/2}}\Biggr]\frac{2^n}{n!}\biggl(x-\frac12\biggr)^n, \quad \biggl|x-\frac12\biggr|<\frac{1}{2}.} \end{equation*}
All concepts, definitions, notions, properties, and notations used above can be found in the references listed below, for example.
References
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:
$$ f(x) = f({1\over e}) + f'({1\over e})(x-{1\over e}) + {f''({1\over e})(x-{1\over e})^2\over 2} + {f'''({1\over e})(x-{1\over e})^3\over 3!} + \cdots $$