# How do I prove the partial denominators formula of the Bauer-Muir transformation of a generalized continued fraction?

Notation: $b_{0}+\underset{n=1}{\overset{\infty }{\mathbb{K}}}\left( a_{n}/b_{n}\right)$ is the Gauss Notation for generalized continued fractions.

Description of the Bauer-Muir transformation

(Based on pp. 76-77 of Lisa Lorentzen and Haakon Waadeland's book (A), chapter II, theorems 11, and Theorem 7; also section 5 of J. Mc Laughlin, and Nancy J. Wyshinski's online paper (C), and section 5 of Bruce C. Berndt, Sen-Shan Huang, Jaebum Sohn, and Seung Hwan Son's online paper (B) ).

Given a convergent c.f. $\underset{n=1}{\overset{\infty }{\mathbb{K}}}(a_{n}/b_{n})=\lim_{n\rightarrow \infty }A_{n}/B_{n}$ and a sequence ${w_{n}}$ we can construct a new c.f $\underset{n=1}{\overset{\infty }{\mathbb{K}}}({c_{n}/d_{n}})=\lim_{n\rightarrow \infty }C_{n}/D_{n}$ (if convergent) which is its Bauer-Muir transform with respect to ${w_{n}}$.

By theorem 11, chapter II, of Lisa Lorentzen and Haakon Waadeland's book (A), pp.76-77, the relations between $A_{n},B_{n}$ and $C_{n},D_{n}$ are given by:

$C_{n}=A_{n}+A_{n-1}w_{n}$, $D_{n}=B_{n}+B_{n-1}w_{n}$, with the initial conditions $C_{-1}=1,D_{-1}=0.$

If for $n\geq 1$, $\lambda_{n}=a_{n}-w_{n-1}(b_{n}+w_{n})\neq 0$, then

$$c_{n}=a_{n-1}\lambda_{n}/\lambda_{n-1},$$

$$d_{n}=b_{n}+w_{n}-w_{n-2}\lambda_{n}/\lambda _{n-1},$$

for $n\geq 2,$ and

$$\underset{n=1}{\overset{\infty }{\mathbb{K}}}(a_{n}/b_{n})=w_{0}+\dfrac{\lambda_{1}}{b_{1}+w_{1}+\underset{n=1}{\overset{\infty }{\mathbb{K}}}(c_{n}/d_{n})}.$$

The elements of $\underset{n=1}{\overset{\infty }{\mathbb{K}}}(a_{n}/b_{n})$ are computed by an application of Theorem 7, chapter II, of Lisa Lorentzen and Haakon Waadeland's book (A), which transforms a sequence into a continued fraction. I was able to derive $c_{n}$ but not $d_{n}$.

Example: Application to the $\log (1+t)$ expansion. By choosing $w_{0}=w_{1}=0,w_{n}=(n-1)t$, for $n\geq 2$, one can derive

$$\underset{n=1}{\overset{\infty }{\mathbb{K}}}\left( n^{2}t/\left( \left( n+1\right) -kt\right) \right) =\dfrac{t}{2+\underset{n=3}{\overset{\infty }{% \mathbb{K}}}\left( \left( n-2\right) ^{2}t/\left( n-\left( n-3\right) t\right) \right) }.$$

Hence (see Wikipedia) the expansion

$$\log (1+t)=\displaystyle\sum_{n=1}^{\infty }\dfrac{(-1)^{n-1}t^{n}}{n}=\dfrac{t}{1+% \underset{n=1}{\overset{\infty }{\mathbb{K}}}\left( n^{2}t/\left( \left( n+1\right) -nt\right) \right) },$$

can be improved with respect to the convergence speed to this one

$$\log (1+t)=\dfrac{t}{1+\dfrac{t}{2+\underset{n=3}{\overset{\infty }{\mathbb{% K}}}\left( \left( n-2\right) ^{2}t/\left( n-\left( n-3\right) t\right) \right) }}.$$

Derivation of $c_n$

Here is how I got $c_n$. In page 77 of reference A it is proved that

$$C_{n-1}D_n-D_{n-1}C_n=(A_{n-1}B_{n-2}-A_{n-2}B_{n-1})\lambda_n.$$

Hence

$$C_{n-2}D_{n-1}-D_{n-2}C_{n-1}=(A_{n-2}B_{n-3}-A_{n-3}B_{n-2})\lambda_{n-1}.$$

For $n\ge 2$ Theorem 7 of reference A (derived from the fundamental c.f. recurrence) gives

$$c_n=\dfrac{C_{n-1}D_n-D_{n-1}C_n}{-(C_{n-2}D_{n-1}-D_{n-2}C_{n-1})}=\dfrac{(A_{n-1}B_{n-2}-A_{n-2}B_{n-1})\lambda_n}{-(A_{n-2}B_{n-3}-A_{n-3}B_{n-2})\lambda_{n-1}}.$$

By the determinant formula we have

$$A_{n-1}B_{n-2}-A_{n-2}B_{n-1}=-a_{n-1}(A_{n-2}B_{n-3}-A_{n-3}B_{n-2}).$$

Thus

$$c_n=a_{n-1}\dfrac{\lambda_n}{\lambda_{n-1}}.$$

QUESTION 1: How does one prove

$$d_{n}=b_{n}+w_{n}-w_{n-2}\lambda_{n}/\lambda_{n-1}$$

from

$$d_{n}=\dfrac{C_{n}D_{n-2}-D_{n}C_{n-2}}{C_{n}D_{n-1}-D_{n}C_{n-1}}\qquad\text{for}\quad n\ge 2$$

and

$\lambda_{n}=a_{n}-w_{n-1}(b_{n}+w_{n})$, $C_{n}=A_{n}+A_{n-1}w_{n}$, $D_{n}=B_{n}+B_{n-1}w_{n}$ ?

References

A - Lisa Lorentzen and Haakon Waadeland, Continued Fractions and Applications, North-Holland, Amsterdam, 1992. (pdf file of pp. 76-77)

B - Bruce C. Berndt, Sen-Shan Huang, Jaebum Sohn, and Seung Hwan Son, A Transformation Formula in Rogers--Ramanujan Continued Fraction. (section 5)

C - J. Mc Laughlin, and Nancy J. Wyshinski, Real numbers with polynomial continued fraction expansions, arXiv, 2004.

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well thats over my head – WalterJ89 Sep 20 '10 at 11:20
I think you could shunt this over to mathoverflow – Sam Lichtenstein Sep 21 '10 at 15:07
This sort of question looks like nobody will answer it unless they're right in the trenches studying this theory as they read it.. – anon Sep 23 '10 at 17:38

Answer to Question 1:

Using the following relations (in Perron, Die Lehre von den Kettenbrüchen, Band II, 1957 and Sergey Khrushchev, Orthogonal Polynomials and Continued Fractions: From Euler's Point of View, 2008)

$$$$A_{n}B_{n-1}-A_{n-1}B_{n}=\left( -1\right) ^{n-1}a_{1}a_{2}\cdots a_{n} \nonumber$$$$

$$$$A_{n}B_{n-2}-A_{n-2}B_{n}=\left( -1\right) ^{n}b_{n}a_{1}a_{2}\cdots a_{n-1}$$$$

$$$$A_{n}B_{n-3}-A_{n-3}B_{n }=(-1)^{n-1}a_{1}a_{2}\cdots a_{n -2}\left( b_{n}b_{n-1}+a_{n}\right)$$$$

we can derive

$$\begin{eqnarray} &&\left( A_{n}+w_{n}A_{n-1}\right) \left( B_{n-1}+w_{n-1}B_{n-2}\right) \\ &&-\left( A_{n-1}+w_{n-1}A_{n-2}\right) \left( B_{n}+w_{n}B_{n-1}\right) \\ &=&-\left( -1\right) ^{n }a_{1}a_{2}\cdots a_{n-1}\left( a_{v}-w_{n-1}\left( b_{n}+w_{n}\right) \right) \end{eqnarray}$$

and

$$\begin{eqnarray} &&\left( A_{n -1}+w_{n-1}A_{n-2}\right) \left( B_{n-2}+w_{n-2}B_{n-3}\right) \nonumber \\ &&-\left( A_{n-3}+w_{n-3}A_{\nu -3}\right) \left( B_{n-1}+w_{n-1}B_{n -2}\right) \\ &=&-\left( -1\right) ^{n -1}a_{1}a_{2}\cdots a_{n-2}\left( a_{n-1}-w_{n-2}\left( b_{n-1}+w_{n-1}\right) \right) \end{eqnarray}$$

as well as

$$\begin{eqnarray} &&\left( A_{n}+w_{n}A_{n-1}\right) (B_{n-2}+w_{n-2}B_{n-3}) \\ &&-(A_{n-2}+w_{n-2}A_{n-3})\left( B_{n}+w_{n}B_{n -1}\right) \\ &=&\left( -1\right) ^{n}b_{n }a_{1}a_{2}\cdots a_{n -1}+w_{n -2}(-1)^{n -1}a_{1}a_{2}\cdots a_{n -2}\left( b_{n}b_{n-1}+a_{n}\right) \\ &&+w_{n}\left( -1\right) ^{n -2}a_{1}a_{2}\cdots a_{n-1}+w_{n-2}\left( -1\right) ^{n-1}b_{n-1}a_{1}a_{2}\cdots a_{n-2} \end{eqnarray}$$

Therefore

$$\begin{eqnarray*} d_{n} &=&\frac{\left( A_{n}+w_{n}A_{n-1}\right) (B_{n-2}+w_{n-2}B_{n-3})-(A_{n-2}+w_{n-2}A_{n-3})\left( B_{n}+w_{n}B_{n-1}\right) }{% (A_{n-1}+w_{n-1}A_{n-2})(B_{n-2}+w_{n-2}B_{n-3})-(A_{n-2}+w_{n-2}A_{n-3})(B_{n-1}+w_{n-1}B_{n-2})% } \\ &=&\frac{\left( -1\right) ^{n}b_{n}a_{1}a_{2}\cdots a_{n-1}+w_{n-2}(-1)^{n-1}a_{1}a_{2}\cdots a_{n-2}\left( b_{n}b_{n-1}+a_{n}\right) }{-\left( -1\right) ^{n -1}a_{1}a_{2}\cdots a_{n-2}\left( a_{n-1}-w_{n-2}\left( b_{n -1}+w_{n-1}\right) \right) } \\ &&+\frac{w_{n }\left( -1\right) ^{n-2}a_{1}a_{2}\cdots a_{n-1}+w_{n-2}\left( -1\right) ^{n-1}b_{n -1}a_{1}a_{2}\cdots a_{n-2}}{% -\left( -1\right) ^{n-1}a_{1}a_{2}\cdots a_{n-2}\left( a_{v-1}-w_{n-2}\left( b_{n-1}+r_{n-1}\right) \right) } \\ &=&\dfrac{\left( b_{n}+w_{n}\right) \left( a_{n-1}-w_{n-2}(b_{n-1}+w_{n-1})\right) -w_{n-2}\left( a_{n}-w_{n-1}(b_{n}+w_{n})\right) }{a_{n-1}-w_{n-2}(b_{n-1}+w_{n-1})} \\ &=&b_{n }+w_{n}-w_{n-2}\dfrac{a_{n }-w_{n-1}(b_{n }+w_{n})}{a_{n -1}-w_{n-2}(b_{n-1}+w_{n-1})} \end{eqnarray*}$$

This transformation appears also in "Die Transformation von Bauer und Muir", §7 of Oskar Perron, Die Lehre von den Kettenbrüchen, Band II, 1957.

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