# Padé approximations of $e$

The following question came up in the analysis of some algorithm.

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 $$r_{n,0} < r_{n,2} < \cdots < r_{n-2\lfloor n/4\rfloor,2\lfloor n/4\rfloor} \geq r_{n-2\lfloor n/4\rfloor-2,2\lfloor n/4\rfloor+2} > \cdots,$$ with equality when $n = 4k+3$. The sequence of upper bounds enjoys a similar property: $$r_{n,1} > r_{n,3} > \cdots > r_{n-2\lfloor (n-2)/4 \rfloor - 1,2\lfloor (n-2)/4 \rfloor + 1} \leq r_{n - 2\lfloor (n-2)/4 \rfloor - 3,2\lfloor (n-2)/4 \rfloor + 3} < \cdots,$$ with equality when $n = 4k+1$. So the tightest lower and upper bounds appear in the middle of any secondary diagonal.

These properties of the sequence $r_{n,0},\ldots,r_{0,n}$ were observed experimentally. Why do they hold?

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?

(Using the explicit expression for the error term, it is easy to prove that $\bigcap_{n=2}^\infty I_n = \{e\}$.)

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It smells like there could be a connection between the Padé approximations of $e^z$ and the continued fraction expansion of $e$. I don't understand these things nearly well enough myself, but I'd also be curious to see the answer! – Igor Khavkine Feb 12 '12 at 13:45
The continued fraction expansion apparently involves only diagonal or near-diagonal elements. But formulas for the Padé approximants do appear in Perron's text on continued fractions. – Yuval Filmus Feb 12 '12 at 18:16
By the way, it sounds from your last comment that you already know how this works, but arxiv.org/abs/math.NT/0601660 gives an exposition of the continued fraction connection. – Henry Cohn Feb 12 '12 at 18:51
For the record: I have worked this out already in 2012. If anyone is interested, let me know. – Yuval Filmus Nov 16 '15 at 6:15