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 $$\log \left(\frac{1}{x}\right)=\sum_{n=0}^\infty (-1)^n \frac{e^n} n\left(x-\frac{1}{e}\right)^n$$
explanation: The document discusses the polynomial approximation of the function \(\sqrt{\log(1/x)}\) centered at \(c = 1/e\). It provides a detailed derivation of the coefficients \(a_k\) for the series expansion and compares them with the coefficients \(b_k\) of the polynomial approximation of \(\log(1/x)\). The document also includes a recursive formula for the coefficients \(a_k\) and mentions the difficulty in proving a bound on these coefficients.
  
  The document contains a question about obtaining a polynomial approximation or a bound on the coefficients of the approximation. It also includes a series of comments and a solution that provides the coefficients for the polynomial approximation of \(\sqrt{\log(1/x)}\) centered at \(x = 1/e\).
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # 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 $$\log \left(\frac{1}{x}\right)=\sum_{n=0}^\infty (-1)^n \frac{e^n} n\left(x-\frac{1}{e}\right)^n$$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document discusses the polynomial approximation of the function \(\sqrt{\log(1/x)}\) centered at \(c = 1/e\). It provides a detailed derivation of the coefficients \(a_k\) for the series expansion and compares them with the coefficients \(b_k\) of the polynomial approximation of \(\log(1/x)\). The document also includes a recursive formula for the coefficients \(a_k\) and mentions the difficulty in proving a bound on these coefficients.
  
  The document contains a question about obtaining a polynomial approximation or a bound on the coefficients of the approximation. It also includes a series of comments and a solution that provides the coefficients for the polynomial approximation of \(\sqrt{\log(1/x)}\) centered at \(x = 1/e\).
  
  Label: question-and-answer