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Define $$ \Lambda_{i}(u) = \sum_{r=0}^{\infty} \Lambda_{i,r}u^{r}, \Psi_{i}(u) = \sum_{m=0}^{\infty} \psi_{i,m}u^{m}=k_i\frac{\Lambda_{i}(uq_i^{-1})}{\Lambda_{i}(uq_i)}. $$ How can we show that $$ \Lambda_{i,r}=\frac{1}{k_i(q_i^r-q_i^{-r})}\sum_{t=1}^{r}q_i^{r-t}\psi_{i,t}\Lambda_{i,r-t} $$ and $$ \psi_{i,r}=-k_i(q_i^r-q_i^{-r})\Lambda_{i,r}-\sum_{t=1}^{r-1}q_i^{r-t}\psi_{i,t}\Lambda_{i,r-t} ?$$ Many thanks.

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It seems the index $i$ doesn't figure in the question? It would make the formulas a lot easier to read if you omit it. – joriki Jul 20 '11 at 3:28
up vote 2 down vote accepted

I'll supress the index $i$ as suggested in my comment.

Multiplying through by $\Lambda(uq)$ yields


or, expressed in terms of the series,

$$\left(\sum_{m=0}^\infty \psi_mu^m\right)\left(\sum_{r=0}^\infty\Lambda_r(uq)^r\right)=k\sum_{r=0}^\infty\Lambda_r(uq^{-1})^r\;.$$

The multiplication on the left leads to a convolution of the coefficient sequences, and comparing coefficients of the individual powers of $u$ on both sides gives


Noting that $\psi_0=\Psi(0)=k\Lambda(0)/\Lambda(0)=k$, we can separate out the $t=0$ term and take it to the other side:


Now solving for $\Lambda_r$ yields the first of your equations, up to a sign that one of us must have gotten wrong. For the second equation, additionally separate out the $t=r$ term and solve for $\psi_r$; however, you need to divide through by $\Lambda_0$ in the process, which isn't in your equation; are you assuming $\Lambda_0=1$?

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thank you very much. Yes, $\Lambda_0=1$. – LJR Jul 20 '11 at 14:16

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