Full and truncated counting differ for a power map

The function $f(z)=z^d$ has exactly one zero, at $z=0$, of order $d$. Hence for every $t\ge 0$: $n(t,0;f)=d$, $\bar{n}(t,0;f)=1$ and $n_1(t,0;f)=d-1$.

Centre regularisation of the first count: $n(0,0;f)=d$, so $N(r,0;f)=d\log r+{\int }_0^r\frac{d-d}{t}dt=d\log r$ for every $r>0$.

For the truncated counts the centre values are $1$ and $d-1$ respectively, so $\bar{N}(r,0;f)=1\cdot \log r+{\int }_0^r\frac{1-1}{t}dt=\log r$ and $N_1(r,0;f)=(d-1)\log r$.

Consistency: $d\log r=\log r+(d-1)\log r$, in agreement with $N=\bar{N}+N_1$; and $N(r,0;f)=d\log r>\log r=\bar{N}(r,0;f)$ because $d\ge 2$ and $\log r>0$.

The rational degree of $z^d$ is $d$, so [F3] gives $T(r,f)=d\log r+O(1)$; thus $T=N(r,0;f)+O(1)$ while the truncated count is smaller by $(d-1)\log r$. The truncated count omits the weight $(d-1)\log r$ of the multiple zero. Replacing the full count by the truncated count in the First Main Theorem therefore changes its equality by this unbounded term. In a Second Main Theorem inequality with truncated counts on the right, replacing them by the larger full counts preserves the inequality but gives a weaker bound.