Re: [math-fun] rwg's matrix product for Clausen's Lambert identity
Your perseverance rewards us all. Glitch: $K(k,\infty)$ for $\lim_{n\to\infty}{K(k,n)}$ would be clearer as $N(k,\infty)$ for $\lim_{n\to\infty}{N(k,n)}$ --rwg Joerg>Finally. Typset document at http://www.jjj.de/lambert-paper/ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Gosper's matrix form of Clausen's identity} % \jjfile{gosper-path-invariant.gp} % Gosper defines \cite{Gosper-mathfun} matrices $K(k,n)$ and $N(k,n)$ as % \[ %% rwg: % {k, {{q^(1 + 2*k + n), -((q*(-1 + q^(2*k + n)))/((-1 + q^k)*(-1 + q^(k +n))))}, {0, 1}}}, %% jj: % Km(k,n)={ [ q^(n+2*k+1), q*(1-q^(2*k+n))/( (1-q^(k)) * (1-q^(k+n)) ) ; 0, 1 ]; } K(k,n)= \begin{bmatrix} q^{n+2k+1} & {q\,(1-q^{2k+n})}/{\left((1-q^k) \, (1-q^{k+n})\right)} \\ 0 & 1 \end{bmatrix} \] % \[ %% rwg: % {n, {{q^k, q/(1 - q^(k + n))}, {0, 1}}} %% jj: % Nm(k,n)={ [ q^(k), q/(1-q^(k+n)); 0, 1 ]; } N(k,m)= \begin{bmatrix} q^{k} & {q}/{(1-q^{k+n})} \\ 0 & 1 \end{bmatrix} \] % These matrices satisfy \[ %% rwg: % handy invariance checker % Nm(k,n) Km(k,n+1) = Km(k,n) Nm(k+1,n), %% jj: % Nm(k,n)*Km(k,n+1) - Km(k,n)*Nm(k+1,n) \\ == zero, OK N(k,n) \cdot K(k,n+1) = K(k,n) \cdot N(k+1,n) \] % Writing $K(\infty,n)$ for $\lim_{k\to\infty}{K(k,n)}$ and $K(k,\infty)$ for $\lim_{n\to\infty}{K(k,n)}$ we obtain \eqref{rel:clausen-lambert} as upper right entries on both sides of \[ %% rwg: % MProd[[N] /. k -> 1, {n, 0, oo}] . MProd[Limit[[K], n -> oo], {k, 1, oo}] %% jj: % Ln = prod(n=0,N, Nm(1,n) ) \\ == [0, Lam; 0, 1] % Lk = prod(k=1,N, Km(k,oo) ) \\ == [0, 1; 0, 1]] % L = Ln * Lk \\ == [0, Lam; 0, 1] \\ OK \prod_{n\geq{}0}{N(1,n)} \cdot \prod_{k\geq{}1}{K(k,\infty)} = % == %% rwg: % MProd[[K] /. n -> 0, {k, 1, oo}] . MProd[Limit[[N], k -> oo], {n, 0, oo}]] %% jj: % Rk = prod(k=1,N, Km(k,0) ) \\ == [0, Lam; 0, 1] % Rn = prod(n=0,N, Nm(oo,n) ) \\ == [0, q; 0, 1] % R = Rk * Rn \\ [0, Lam; 0, 1] \\ OK \prod_{k\geq{}1}{K(k,0)} \cdot \prod_{n\geq{}0}{N(\infty,n)} % % d = L - R \\ == zero \] %%%%%%%%%%%%%%%%%%%%%
* Bill Gosper <billgosper@gmail.com> [Feb 28. 2012 08:14]:
Your perseverance rewards us all. Glitch:
$K(k,\infty)$ for $\lim_{n\to\infty}{K(k,n)}$
would be clearer as
$N(k,\infty)$ for $\lim_{n\to\infty}{N(k,n)}$
--rwg
Thanks! Fixed it as follows Writing $K(k,\infty)$ for $\lim_{n\to\infty}{K(k,n)}$ and $N(\infty,n)$ for $\lim_{k\to\infty}{N(k,n)}$ we obtain \eqref{rel:clausen-lambert} as upper right entries on both sides of \[ \prod_{n\geq{}0}{N(1,n)} \cdot \prod_{k\geq{}1}{K(k,\infty)} = \prod_{k\geq{}1}{K(k,0)} \cdot \prod_{n\geq{}0}{N(\infty,n)} \] And updated
Typset document at http://www.jjj.de/lambert-paper/
Added the following (easily obtained from Osler/Hassen): t * sum(n>=1, x^n*q^n/(1-t*q^n)) == x * sum(n>=1, t^n*q^n/(1-x*q^n)) == sum(n>=1, x^n*t^n*q^(n^2)/(1-t*q^n) + x^(n+1)*t^n*q^(n*(n+1))/(1-x*q^n) ) == sum(n>=1, x^n*t^n*q^(n^2)/(1-x*q^n) + x^n*t^(n+1)*q^(n*(n+1))/(1-t*q^n) ) == sum(n>=1, (1-x*t*q^(2*n)) * x^n*t^n*q^(n^2) / ( (1-x*q^n) * (1-t*q^n) ) ) Also direct derivation of the (less-) generalized Lambert from (Wrench/Knuth)'s beautiful identity: \sum_{n\geq{}1}{ \frac{a_n\,q^n}{ 1 - q^n } } = \sum_{n\geq{}1}{ \left[ a_n + \sum_{k\geq{}1}{ (a_n+a_{n+k})\,q^{k\,n} } \right]\,q^{n^2} } (this is how I originally found it). I'll submit tomorrow unless someone shouts "Stop the presses!"
participants (2)
-
Bill Gosper -
Joerg Arndt