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Recently I encounter an identity $$z_{0}^{-1}\delta(\frac{z_{1}-z_{2}}{z_{0}})=z_{1}^{-1}\delta(\frac{z_{0}+z_{2}}{z_{1}})$$ where $\delta(x)=\Sigma_{n\in \mathbb{N}}x^{n}$.

I tried to expand both sides. I get $$ LHS=z_{0}^{-1}\Sigma_{p\in \mathbb{Z}}\Sigma_{q \in \mathbb{N}}(\frac{z_{1}}{z_{0}})^{p-q}(\frac{z_{2}}{z_{0}})^{q}(-1)^{q}$$

$$RHS=z_{1}^{-1}\Sigma_{p\in \mathbb{Z}}\Sigma_{q \in \mathbb{N}}(\frac{z_{0}}{z_{1}})^{p-q}(\frac{z_{2}}{z_{1}})^{q}$$. When I compare the coefficients of both sides, left hand side has negative term. Is this identity wrong or did I miss something?

Jack
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1 Answers1

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I get $\, \delta(x) = 1/(1-x)\,$ if $\,|x|<1 \,$ and an identity $\, z_{0}^{-1}\delta(\frac{z_{1}-z_{2}}{z_{0}}) = \frac1{z_0-z_1+z_2} = - z_{1}^{-1}\delta(\frac{z_{0}+z_{2}}{z_{1}}). \,$ For convergence, we need that $\, \max(z_1-|z_0|, -|z_1|-z_0) < z_2 < \min(z_1+|z_0|, |z_1|-z_0). \,$

Somos
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