HEJ, HU ISSN 1418-7108 Manuscript no.: ANM-991102-A

Problems with operator $\amalg$

From property (iv) of functions $\mu$ and $\pi$ we have to specify operator $\amalg$ so, that it has to satisfy the following two equalities:

$$p_1 + p_2 = \pi (\mu (p_1) \amalg \mu (p_2)) \textrm{ and}$$

$$d_1 \amalg d_2 = \mu (\pi (d_1) + \pi (d_2)).$$

Above we have applied the harmonic average to "produce" operator $\amalg$, but - as it will be presented below - we are not able to fit it exactly to these requirements. From the definitions of functions $\mu$ and $\pi$ follows

$$d_1 \amalg d_2 = -\textrm{log}_a (a^{-d_1} + a^{-d_2}),$$

so we would need

$$\frac{d_1 \cdot d_2}{d_1 + d_2} = -\textrm{log}_a (a^{-d_1} + a^{-d_2}).$$

Choosing $d_1 = d_2 \ne 0$ we get

$$\frac{d_1}{2} = -\textrm{log}_a (2a^{-d_1}),$$

$$a^{-\frac{d_1}{2}} = 2a^{-d_1},$$

$$2=a^{\frac{d_1}{2}}$$

with the obligation, that this must be held for all $d_1$. This is clearly not possible. Thus, we have ``two different'' operators $\amalg$. Using the harmonic average we get only an approach. The result is exact if one of the distances $d_1$ and $d_2$ is $\infty$, in other cases there is an error-term. However, this approach is well useable, because of its simplicity. But, by an exact transition from probability model to distance model, we have to use the logarithmic formula for operator $\amalg$.

HEJ, HU ISSN 1418-7108 Manuscript no.: ANM-991102-A