Editing Executive governance (section)

==== Constant fees ====
For constant fees [[Validation Pool#Vote|<math>f'(t):=f_0'</math>]] we get[[Validation Pool#Vote|<math display="block">R(t)=R_0+\int_{0}^t m*f' (s)ds=R_0+mf_0't </math>]]<nowiki/>and so Equation 11 gives[[Validation Pool#Vote|<math display="block">P_0^1(t)  =exp\Biggl( p_3m\int _0^t \frac{f_0'}{R_0+mf_0's}\Biggr)ds=exp\Biggl( p_3 \biggl(ln(R_0+mf_0's)-ln(R_0)\biggr)\Biggr) =\biggl(1+\frac{mf_0'}{R_0}t\biggr)^{p_3}.</math>]]If (as we are assuming) all members police all the time, then[[Validation Pool#Vote|<math display="block">P_0^{R_0}(t)=R_0 \biggl(1+\frac{mf_0'}{R_0}t\biggr)^{p_3}.</math>]]<nowiki/>gives the formula for the collection of all policing tokens [[Validation Pool#Vote|<math display="inline">P_0^{R_0}</math>]] generated from the original tokens [[Validation Pool#Vote|<math display="inline">R_0</math>]] from time [[Validation Pool#Vote|<math display="inline">t=0</math>]]. The policing tokens from a single token  grow without bound, however the relative power that all original tokens [[Validation Pool#Vote|<math display="inline">R_0</math>]] maintain under automated policing is[[Validation Pool#Vote|<math display="block">\frac{P_0^{R_0}(t)}{R(t)}=\frac{R_0 \biggl(1+\frac{mf_0'}{R_0}t\biggr)^{p_3}}{R_0+mf_0t}=\biggl(1+\frac{mf_0'}{R_0}t\biggr)^{p_3-1}\rightarrow 0</math>]]<nowiki/>as <math>t\rightarrow \infty</math> anytime [[Validation Pool#Vote|<math>p_3<1</math>]]. This proves policing alone cannot maintain your relative power in the DAO, nor the power of the original cohort . The inflationary minting mechanism will dilute your power if you don’t continue to contribute to the DAO with further work. So the entire starting DAO members will eventually be usurped if they don’t add more original contributions beyond automated policing.

We also can calculate the income stream [[Validation Pool#Vote|<math display="inline">f_0^{1,P}(t)</math>]] of 1 token with policing [[Validation Pool#Vote|<math display="inline">P_0^1(t)</math>]] and its present value under the assumption of constant fees:[[Validation Pool#Vote|<math display="block">f_0^{1,P}(t)=\int_0^t f'(s)\frac{P_0^1(s)}{R(s)}ds=\int_0^t f_0'\frac{\biggl(1+\frac{mf_0'}{R_0}s\biggr)^{p_3}}{R_0+mf_0's}ds=\frac{f_0'}{R_0}\int_0^t \biggl(1+\frac{mf_0'}{R_0}s\biggr)^{p_3-1}ds=\frac{1}{mp_3}\Biggl(\biggl(1+\frac{mf_0'}{R_0}t\biggr)^{p_3}-1\Biggr).</math>]]<nowiki/>and<math display="block">PVf_0^{1,P}=\int_0^\infty e^{-rt} \frac{d}{dt} f_0^{1,P} (t)dt=\frac{f_0'}{R_0}\int_0^\infty e^{-rt}\biggl(1+\frac{mf_0'}{R_0}t\biggr)^{p_3-1}dt</math>This last expression is finite, but not expressible using elementary functions. This is common for any expression that derives from a complicated continuous process—most elementary functions lack elementary antiderivatives. Technically it is an incomplete gamma function. When necessary, we can make tables of values for any relevant parameters we happen to be using in order to gain intuition for how it behaves.