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.
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