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The following question comes up during analysis of Padé approximants to $e^x$ (see my related question in MathOverflow for more background). Recall that the continued fraction expansion of $e$ is $ e = [2,1,2,1,1,4,1,1,6,1,1,8,\ldots]. $ Consider $ \begin{align*} R_{k,k} &= [2,1,2,1,1,4,\ldots,1,1,2(k-1),1,1] \text{ and } \\ R_{k+1,k-1} &= [2,1,2,1,1,4,\ldots,1,1,2(k-1),1,1,k]. \end{align*} $ Empirically, we always have $ |R_{k+1,k-1} - e| < |R_{k,k} - e|. $ Can anyone explain why?

Edit: Now I can explain why. We have $ \frac{R_{k,k} + R_{k+1,k-1}}{2} = [2,1,2,1,1,4,\ldots,1,1,2k,1,1,2(k-1),\ldots,1,1,4,1,1,3]. $ When $k$ is even (odd), this will differ from the continued fraction in an even (odd) position, by being too small. This means that the value is larger (smaller) than $e$. Since $R_{k,k}$ is an upper (lower) bound on $e$, and $R_{k+1,k-1}$ is a lower (upper) bound on $e$, we get what we want.

Now it remains to prove this continued fraction expansion of $(R_{k,k} + R_{k+1,k-1})/2$...

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    @YuvalFilmus: If I am right the approximants $A_{n}/B_{n}$ of the given c.f. for $ e=2+K_{j=1}^{\infty }\left( 1/b_{j}\right) $ satisfy $ A_{3k+1}/B_{3k+1}=R_{k,k}$ while the modified approximants $C_{n}/D_{n}$ satisfy $C_{3k+1}/D_{3k+1}=\left( A_{3k+1}+A_{3k}/k\right) /\left( B_{3k+1}+B_{3k}/k\right) =R_{k+1,k-1}$. Since b_{k}>0, $R_{k+1,k-1}$ converges to $e$ provided that $\lambda _{k}:=1-w_{k-1}(b_{k}+w_{k})\neq 0$ for $k=1,2,\ldots $. The sequence $\left( w_{k}\right) _{k\geq 0}$ would be $% w_{3k}=1/k$, $w_{3k\pm 1}=0$.2012-02-19

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