Changeset - ad16c1eb532f
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András Gilyén - 8 years ago 2017-09-10 16:12:37
gilyen@cwi.nl
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@@ -177,12 +177,13 @@
 
        (A simpler version: $\forall k>0: a_k^{(3)}=(k+1)(k+2)/6$)
 
		\item $\forall k\in\mathbb{N}, \forall n>m\geq 3 : a^{(n)}_k\geq a^{(m)}_k$ \label{it:geq}		
 
		\item $\forall k\in\mathbb{N}, \forall n,m > \max(k,3) : a^{(n)}_k=a^{(m)}_k$ \label{it:const}		
 
  		\item $\exists p_c=\lim\limits_{k\rightarrow\infty}1\left/\sqrt[k]{a_{k}^{(k+1)}}\right.$ \label{it:lim}			
 
	\end{enumerate}
 
	\colorbox{red}{\ref{it:pos}-\ref{it:geq} is false since $a_{1114}^{(10)}<0$ -- needs to be double checked!}
 
	I figured this out by observing that $R^{(10)}(p)$ has a pole inside the disk of radius $0.96$. This also means that $R^{(10)}(p)=\sum_{k=0}^{\infty}a_k^{(10)}p^k$ is only true in an analytic sense, since for $p>0.96$ the right hand side does not converge.
 
	
 
	We also conjecture that $p_c\approx0.61$, see Figure~\ref{fig:coeffs_conv_radius}.
 

	
 
	\begin{figure}[!htb]\centering
 
	\includegraphics[width=0.5\textwidth]{coeffs_conv_radius.pdf}
 
	%\includegraphics[width=0.5\textwidth]{log_coeffs.pdf}	
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