text: # Polynomial approximation of $\sqrt{\log(1/x)}$ or bound on coefficients.
  
  I want to obtain a polynomial approximation (in a closed form) of the function $$\sqrt{\log(1/x)}$$ centered in any point $$c \in (0, 1]$$ i.e. the series $$S(x-c)= \sum_n^\infty a_k (x-c)^k$$. For convenience, we take $$c=1/e$$.
  
  If this is too difficult to obtain, I am fine with a bound on the coefficients, i.e. I want to prove (or have a counterexample) that each $$|a_k| \leq |b_k|$$, where $$b_k$$ is the coefficient of the polynomial approximation of $$\log(1/x)$$.
  
  Recall that the Taylor approximation of $$\log(1/x)$$ centered in a generic $$x_0=c$$ is: $$S^2(x-c)=\sum_{k\geq1} \frac{(-1)^k c^{-k}(x-c)^k}{k} + \log(1/c)$$ For the case of $$x_0=1/e$$, we have the more known: $$S^2(x-c)=\sum_{k\geq1} \frac{(-e)^{k}}{k} (x-\frac{1}{e})^k + 1.$$ From this, we formulate that: $$S^2(x-c) = \sum_{k=0}^\infty \left( \sum_{l=0}^k a_l a_{k-l} \right) (x-c)^k .$$
  
  We obtain the following relation of the coefficients $$a_k$$ of $$S(x-c)$$: $$\begin{cases} a_0^2 = 1, \\ \sum_{l=0}^k a_l a_{k-l} \frac{(-e)^{k}}{k} = 2a_0a_k + \sum_{l=1}^{k-1}a_la{k-l} \end{cases}$$ From the first equation we obtain that $$a_0 = 1$$. From the second equation we obtain a recursive definition of $$a_k$$, as $$a_k= \frac{1}{2a_0}\left[\frac{-c^{-k}}{k} - \sum_{l=1}^{k-1} a_l a_{k-l} \right].$$ which is equivalent to $$a_k= \frac{1}{2a_0}\left[b_k - \sum_{l=1}^{k-1} a_l a_{k-l} \right].$$
  
  From here, it is quite difficult to proceed. Experimentally, it is possible to see that the signs of the coefficients are alternating. This makes the bound more difficult to prove.
  
  Similar question:
  
  • You mean $|a_k|\le|b_k|$ Commented Oct 31, 2020 at 21:25
  • yes, thanks. i just changed it.
  – asdf
  Commented Oct 31, 2020 at 22:52
  • Hem, $\log\dfrac1x=-\log x$.
  – user65203
  Commented Nov 4, 2020 at 13:26
  • Yes, of course. so?
  – asdf
  Commented Nov 4, 2020 at 14:07
  
  Built around $$x=\frac 1e$$ (which is convenient)$$\sqrt{\log \left(\frac{1}{x}\right)}=\sum_{n=0}^\infty (-1)^n e^n a_n\left(x-\frac{1}{e}\right)^n$$ and the $$a_n$$ form the sequence $$\left\{1,\frac{1}{2},\frac{1}{8},\frac{5}{48},\frac{25}{384},\frac{209}{3840},\frac{1 961}{46080},\frac{2621}{71680},\frac{21407}{688128},\frac{1700267}{61931520},\frac {30128123}{1238630400}\right\}$$ (numerators and denominators are unknown in $$OEIS$$
  On the other side $$\log \left(\frac{1}{x}\right)=\sum_{n=0}^\infty (-1)^n \frac{e^n} n\left(x-\frac{1}{e}\right)^n$$