most recent 30 from http://mathoverflow.net 2013-05-18T11:53:56Z http://mathoverflow.net/feeds/question/88260 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88260/pade-approximations-of-e Yuval Filmus 2012-02-12T05:27:28Z 2012-02-12T18:25:22Z

<p>The following question came up in the analysis of some algorithm.</p> <p>Let $R_{s,t}(z)$ be the Padé approximants of $e^z$, and define $r_{s,t} = R_{s,t}(1)$. Using the explicit expression for the error term of the Padé approximant, it is easy to see that $$ (-1)^t e > r_{s,t}. $$ We focus on a secondary diagonal $s + t = n$ for $n \geq 2$. Its entries $r_{n,0},r_{n-1,1},\ldots,r_{0,n}$ are alternately lower bounds and upper bounds on $e$. The sequence of lower bounds has the property that $$ r_{n,0} &lt; r_{n,2} &lt; \cdots &lt; r_{n-2\lfloor n/4\rfloor,2\lfloor n/4\rfloor} \geq r_{n-2\lfloor n/4\rfloor-2,2\lfloor n/4\rfloor+2} > \cdots, $$ with equality when $n = 4k+3$. The sequence of upper bounds enjoys a similar property: $$ r_{n,1} > r_{n,3} > \cdots > r_{n-2\lfloor (n-2)/4 \rfloor - 1,2\lfloor (n-2)/4 \rfloor + 1} \leq r_{n - 2\lfloor (n-2)/4 \rfloor - 3,2\lfloor (n-2)/4 \rfloor + 3} &lt; \cdots, $$ with equality when $n = 4k+1$. So the tightest lower and upper bounds appear in the middle of any secondary diagonal.</p> <p>These properties of the sequence $r_{n,0},\ldots,r_{0,n}$ were observed experimentally. Why do they hold?</p> <p>Edit: Let $I_n = [r_{n-2\lfloor n/4\rfloor,2\lfloor n/4\rfloor}, r_{n-2\lfloor (n-2)/4 \rfloor - 1,2\lfloor (n-2)/4 \rfloor + 1}]$ be the interval constituting of the best bounds in the secondary diagonal $s+t=n$. Then experimentally, $$ I_2 \supset I_3 \supset I_4 \supset \cdots, $$ i.e. the bounds get tighter. Why is that?</p> <p>(Using the explicit expression for the error term, it is easy to prove that $\bigcap_{n=2}^\infty I_n = \{e\}$.)</p>