unique root. The special case $\symCountVariate_{\symRegRange+1} = \symNumReg$, for which all registers have reached the maximum possible value, the estimate would be positive infinite.

Using $\symFunc(\symPoissonRateEstimate)=0$ and $1-\frac{\symX}{2}\leq \frac{\symX}{e^\symX-1}\leq 1$ we obtain following bounds for $\symPoissonRateEstimate$ $$\label{equ:inequality}
\frac{\symNumReg(\symNumReg-\symCountVariate_0)}
{\symCountVariate_0+\frac{3}{2}\sum_{\symRegVal=1}^{\symRegRange}\frac{\symCountVariate_\symRegVal}{2^\symRegVal} + \frac{\symCountVariate_{\symRegRange+1}}{2^{\symRegRange+1}}}
\leq
\symPoissonRateEstimate
\leq
\frac{\symNumReg (\symNumReg-\symCountVariate_0)}
{\sum_{\symRegVal=0}^{\symRegRange}
\frac{\symCountVariate_\symRegVal}{2^\symRegVal}}.$$ If the cardinality is in the intermediate range, where $\symCountVariate_0=\symCountVariate_{\symRegRange+1}=0$ the lower and the upper bound differ only by a constant factor and both are proportional to the harmonic mean of $\lbrace \symNumReg 2^{\symRegValVariate_1},\ldots,\symNumReg 2^{\symRegValVariate_\symNumReg} \rbrace$. Hence, consequent application of the method would have directly suggested to use a cardinality estimator that is proportional to the harmonic mean without knowing the raw estimator in advance. The history of the algorithm shows that the raw estimator was first found after several attempts using the geometric mean [@Flajolet2007; @Durand2003].

For $\symRegRange=0$, which corresponds to the linear counting case, the estimator can be found by analytical means. The result is exactly the linear counting estimator which makes us optimistic that the method in combination with the Poisson model also works well for the more general case $\symRegRange>0$. Since $f$ is convex, Newton-Raphson iteration and the secant method [@Press2007] will both converge to the root, provided 