On the margin of continued fraction blogging I would like to present simple result as regards to continuant polynomials. Continued fraction ${[a_0,a_1,...,a_n]}$ may be expressed as quotient of two polynomials of ${(a_0,a_1,...,a_n)}$, named continuant (see continuants on Wikipedia)

${[a_0,a_1,...,a_n]} = K(a_0,a_1,...,a_n)/K(a_1,...,a_n)$

For example ${K(a_0,a_1,a_2,a_3) = a_{0} a_{1} a_{2} a_{3} + a_{0} a_{1} + a_{0} a_{3} + a_{2} a_{3} + 1}$

Continuants has many interesting recurrence relations some of which You may find in Graham, Knuth, Patashnik book „Concrete mathematics”. The most important of this recurrences is as follows:

$\displaystyle K(a_0,a_1,...,a_n,x) = xK(a_0,a_1,...,a_n)+K(a_0,a_1,...,a_n) \ \ \ \ \$

During my plays with matrix representation of continued fractions I found interesting relation:

$\displaystyle K( a_0,a_1,\ldots ,a_k,a_{(k+1)},\ldots ,a_n ) = \ \ \ \ \$

$\displaystyle K(a_0,a_1,\ldots ,a_k,1,1,a_{(k+1)}, \ldots ,a_n) - K(a_0,a_1,\ldots ,a_k,1,a_{(k+1)}, \ldots ,a_n) \ \ \ \ \$

that is – between variables ${a_k}$ and ${a_{(k+1)}}$ You put two of „1” in the first and one „1” in the second term. You may consider this as generalisation of Fibonacci recurrence – because if You put all ${a_i =1}$ You obtain Fibonacci numbers.

For example:

$\displaystyle K(a_0,a_1,a_2) = (a_0 a_ 1+1)a_2+a_0 \ \ \ \ \$

$\displaystyle K(a_0,a_1,1,1,a_2) = (2 a_0 a_1+ a_0+2)a_2+a_0 a_1+a_0+1 \ \ \ \ \$

$\displaystyle K(a_0,a_1,1,a_2) = (a_2+1)(a_0 a_1+1)+a_0 a_2 \ \ \ \ \$

And it is true that:

$\displaystyle K(a_0,a_1,a_2) = K(a_0,a_1,1,1,a_2) - K(a_0,a_1,1,a_2) \ \ \ \ \$

How to prove it? W have:

${ x = [a_{0},a_{1},a_{2},...a_{n}] =\frac{K(a_0,a_1,\ldots ,a_n)}{K(a_1,\ldots ,a_n)}= - \frac{1}{2} \frac{Tr \left( M(S- L) \right) }{Tr \left( M(S-T) \right) }}$

where ${tr}$ is trace of a matrix, and maybe You know that I define M as follows (see here ):

$\displaystyle M = S \prod_{i=0}^{n} S(ST)^{a_{i}} \ \ \ \ \$

So we have the following formulas true:

$\displaystyle K(a_{0},a_{1},a_{2} \ldots a_{n}) =\frac{1}{2} Tr \left( M(S- L) \right) \ \ \ \ \$

$\displaystyle K(a_{1},a_{2} \ldots a_{n}) = - Tr \left( M(S-T) \right) \ \ \ \ \$

Please note that upper formula has different arguments than the lower so they agree.

And there is true that ${S^2 =I}$ and ${T^2 = I+T}$ which we may use to produce „unity decomposition” as below:

$\displaystyle I = T^2 -T = S(ST)S(ST) - S(ST) \ \ \ \ \$

Then You may insert it in any place between ${S(ST)^{a_{i}}S(ST)^{a_{i+1}}}$ which gives expression above and many more if You consider ${T^p}$ for ${p \in Z}$ etc.

Remark 1: proved here formula should be possible also to prove from recursion relation directly – so if You would like to try this way – please let me know.

Remark 2: I post this on mathoverflow, and someone provide the sketch of proof by induction – but it seems to be only a sketch – and rather incomplete.

Reklamy