This type of up­grade has an identical equa­tion to the equa­tion men­tioned above, just with dif­fer­ent con­stants in the places of

$a$

$b$

$Value = cd^{Level}\qquad (1-2)$

In which

$c>0$

$d>1$

▶︎ Lemma 1.1 Permalink: lemma-11#

$(L\in\mathbb{N})$

$u$

$\rho = Cost(u,L)$

$u$

$\rho$

$u = K\rho^{r}$

$K\in\mathbb{R}^+$

$r\in\mathbb{R}^+$

$r$

▶︎ Proof Permalink: proof#

$\rho = ab^{L}$

$a \neq 0$

$\frac{\rho}{a} = b^{L}$

$b > 0$

$b$

$\log_b\left[\frac{\rho}{a}\right] = L$

$\log_b(\rho) - \log_b(a) = L\qquad (1-3)$

$u = cd^{\log_b(\rho) - \log_b(a)}$

$= \frac{c}{d^{\log_b(a)}}d^{\log_b(\rho)}$

$= \frac{c}{d^{\frac{\log_d(a)}{\log_d(b)}}}d^{\frac{\log_d(\rho)}{\log_d(b)}}$

$= \frac{c}{\left[d^{\log_d(a)}\right]^{\frac{1}{\log_d(b)}}}\left[d^{\log_d(\rho)}\right]^{\frac{1}{\log_d(b)}}$

$= \frac{c}{a^{\log_b(d)}}\rho^{\log_b(d)}$

$r = \log_b(d)$

$b>1$

$d>1$

$\log_b(d)>0$

$a>0$

$c>0$

$\frac{c}{a^{\log_b(d)}}>0$

$K$

$r$

▶︎ Lemma 1.2 Permalink: lemma-12#

$u$

$r \in \mathbb{R}^+$

$u$

$\rho$

$u(\rho) \in {\Theta(\rho^r)}$

▶︎ Proof Permalink: proof-1#

$(L>0)$

$u$

$(L+1)$

$ab^{L} \le \rho < ab^{L+1}$

$u’$

$a’ = ab\quad c’= c$

$b’ = b \quad d’= d$

$u$

$u’$

$r$

$K$

$K$

$K’$

$b>1$

$K’< K$

$K’\rho^r \le u \le K\rho^r$

$u(\rho) \in \Theta(\rho^r)$

This type of vari­able fol­lows a rather un­ortho­dox path of pro­gres­sion, an ex­plan­a­tion of its mech­anic is ex­plained here. Let’s sup­pose there ex­ist a Step­wise vari­able

$v$

1. A cycle of

   $v$

   spans

   $m>0$

   levels.
2. On the first cycle of

   $v$

   , it in­creases by

   $p>0$

   per level.
3. On each cycle, the amount of in­crease per level gets mul­ti­plied by

   $q>1$

   .

▶︎ Lemma 1.3 Permalink: lemma-13#

$v$

$(c_1, c_2) \in \mathbb{R}^+$

$d \in \mathbb{R}^+$

$(L \in \mathbb{N} \ge m)$

$c_1d^L \le v_L \le c_2d^L$

▶︎ Proof Permalink: proof-2#

$(k+1 \ge 2)$

$v$

$v$

$(0 < l \in \mathbb{N}\ \le m$

$v_{mk+l}$

$v_{mk+l} = \frac{1}{q-1}p\left\{[m+l(q-1)]q^k-m\right\}\qquad (1-4)$

$m$

$\begin{eqnarray}

\frac{1}{q-1}\left\{[m+l(q-1)]q^k-m\right\}\qquad (1-5) \

< \frac{1}{q-1}p[m+l(q-1)]q^k=\frac{1}{q-1}mpq^k\left[1+\frac{l}{m}(q-1)\right]

\end{eqnarray}$

$0 \le \frac{l}{m} \le 1$

$q>1$

$\begin{flalign*}

&\frac{1}{q-1}mpq^k\left[1+\frac{l}{m}(q-1)\right] < \frac{1}{q-1}mpq^k(1+(q-1)) &&\

&= \frac{1}{q-1}mpq^{(k+1)} < \frac{1}{q-1}mpqq^{\frac{mk+l}{m}} &&\

&= \frac{1}{q-1}mpq(q^\frac{1}{m})^{mk+l}

\end{flalign*}$

$v_{mk+l} < mp\frac{q}{q-1}\left(q^\frac{1}{m}\right)^{mk+l}\qquad (1-6)$

$k>1$

$q^{k-1}>1$

$\begin{eqnarray}

\frac{1}{q-1}p\left\{[m+l(q-1)]q^k-m\right\} \


\frac{1}{q-1}p\left\{[m+l(q-1)]q^k-q^{k-1}m\right\}

\end{eqnarray}$

$= \frac{1}{q-1}p\left[mq^k-mq^{k-1}+l(q-1)q^k\right]$

$= \frac{1}{q-1}p\left[m(q-1)q^{k-1}+l(q-1)q^k\right]$

$= mpq^{k-1}\left(1+\frac{l}{m}q\right)\qquad (1-7)$

$0 \le \frac{l}{m} \le 1$

$q>-1$

$1+\frac{l}{m}q \ge (1+q)^{\frac{l}{m}}$

$mpq^{k-1}\left(1+\frac{l}{m}q\right) \ge mpq^{k-1}(1+q)^{\frac{l}{m}}>mpq^{k-1}q^{\frac{l}{m}}$

$= \frac{mp}{q}\left(q^{\frac{1}{m}}\right)^{mk+l}$

$v_{mk+l} > \frac{mp}{q}\left(q^{\frac{1}{m}}\right)^{mk+l}\qquad (1-8)$

$\frac{mp}{q}\left(q^{\frac{1}{m}}\right)^{mk+l} < v_{mk+l} < mp\frac{q}{q-1}\left(q^\frac{1}{m}\right)^{mk+l}$

$mk+l = L$

$\frac{mp}{q}\left(q^{\frac{1}{m}}\right)^L < v_L < mp\frac{q}{q-1}\left(q^\frac{1}{m}\right)^L$

$\left(\frac{mp}{q},\frac{mpq}{q-1}\right) \in \mathbb{R}^+$

$q^\frac{1}{m} \in \mathbb{R}^+$

$(c_1,c_2)$

$d$

▶︎ Lemma 1.4 Permalink: lemma-14#

$v$

$r \in \mathbb{R}^+$

$v$

$\rho$

$v(\rho) \in {\Theta(\rho^r)}$

▶︎ Proof Permalink: proof-3#

$c_1d^L \le v_L \le c_2d^L$

$v_1$

$v_2$

$v_1 \le v \le v_2\qquad (1-9)$

$v_1$

$v_2$

$v_1$

$v_2$

$v_1 \in \Theta(\rho^r)$

$v_2 \in \Theta(\rho^r)\qquad 

(1-10)$

$r \in \mathbb{R}^+$

$v(\rho) \in \Theta(\rho^r)$

For any pur­chas­able vari­able

$v$

$r \in \mathbb{R}^+$

$v$

$\rho$

$v(\rho) \in {\Theta(\rho^r)}$

▶︎ Lemma 2.1 Permalink: lemma-21#

$a$

$t$

$(x,y) \in \mathbb{R}^+$

$a(\rho,t) \in \Theta(\rho^xt^y)$

▶︎ Proof Permalink: proof-4#

$t$

$\frac{∂a}{∂t}$

$\mathbb{R}^+$

$x \in \mathbb{R}^+$

$\frac{∂ a(\rho,t)}{∂ t} \in \Theta({\rho^x}t^y)$

$a = \Theta(\rho^xt^y)t+C$

$a \in \Theta(\rho^xt^{y+1})$

$y \in \mathbb{R}^+$

$(y+1) \in \mathbb{R}^+$

We can de­term­ine the asymp­totic nota­tion of a vari­able of this kind by solv­ing a first or­der ODE, be­low is an ex­ample with the

$q$

$\frac{dq}{dt} = \frac{8q_1q_2}{q+1}$

This is a clas­sic ex­ample of a self-in­flu­enced vari­able, for­tu­nately, the ODE is sep­ar­able:

$(q+1)dq = 8q_1q_2dt$

In­teg­rate both sides,

$\frac{q^2}{2}+q = 8q_1q_2t + C$

As

$\rho$

$q$

$\sqrt{16q_1q_2t}$

In more com­plic­ated cases such as the sys­tems that con­tain mul­tiple powers of

$t$

1. De­term­ine the strongest term by ex­per­i­ment and use the pre­vi­ous two meth­ods (which is how we ana­lyze T6, which has an in­tric­ate main equa­tion).
2. Use Nu­mer­ical Meth­ods (which the The­ory Sim­u­lator in­cor­por­ates).
3. Re­verse en­gin­eer the ex­po­nent of

   $\rho$

   of a spe­cific vari­able from the fi­nal De­cay Factor ex­per­i­ment­ally.

If we man­age to get everything sor­ted, we will get a power­ful con­clu­sion:

For any the­ory sys­tem in which Lemma 2.1 can ap­ply fully on or can be ana­lyzed asymp­tot­ic­ally by other means, there ex­ists

$(x,y) \in \mathbb{R}^+$

$\frac{d\rho}{dt} \in \Theta(\rho^xt^y)$

For any the­ory sys­tem

$T$

$\frac{d\rho}{dt}=K\rho^{x}t^{y}\rho^{z}_{pub}$

the long-term mean Pub­lic­a­tion Mul­ti­plier of

$T$

$\rho \rightarrow \infty$

$e^{\frac{z(y+1)}{1-x}}$

$\frac{d\rho}{dt}=K\rho^{x}t^{y}\rho^{z}_{pub}$

This is a sep­ar­able dif­fer­en­tial equa­tion, after some re­arrange­ment we have:

$\rho^{-x}d\rho = Kt^{y}\rho^{z}_{pub}dt$

In­teg­rate both sides,

$\int\rho^{-x}d\rho = \int Kt^{y}\rho^{z}_{pub}dt$

Since

$K\rho^{z}_{pub}$

$\int\rho^{-x}d\rho = K\rho^{z}_{pub}\int t^{y}dt$

Eval­u­ate both In­teg­rals,

$\frac{\rho^{1-x}}{1-x} = \frac{K\rho^{z}_{pub}}{y+1}t^{y+1} + C$

In an En­dgame situ­ation,

$t \rightarrow \infty$

$C$

$\frac{\rho^{1-x}}{1-x} = K’\rho^{z}_{pub}t^{y+1}$

We solve the equa­tion to get:

$\rho = \left[K’(1-x)\rho^{z}_{pub}t^{y+1}\right]^{\frac{1}{1-x}}\qquad (3-3)$

The goal of ours is to find the Pub­lic­a­tion Mul­ti­plier that gives the max­imum speed meas­ured in a \log­ar­ithmic scale, which can be ex­pressed with the equa­tion be­low:

$Speed ∝ \frac{\ln(\rho) - \ln(\rho_{pub})}{t}\qquad (3-4)$

We sub­sti­tute (3-3) into (3-4) and get:

$\frac{\ln(\rho) - \ln(\rho_{pub})}{t} = \frac{\ln\left[K’(1-x)\rho^{z}{pub}t^{y+1}\right]^{\frac{1}{1-x}} - \ln(\rho)}{t}$

$ = \frac{\frac{\ln\left[K’(1-x)\right]+\ln\left(\rho^{z}{pub}\right)+\ln\left(t^{y+1}\right)}{1-x} - \ln(\rho)}{t}\qquad (3-5)$

In the En­dgame,

$\ln\left[K’(1-x)\right]$

$\frac{\frac{\ln\left[K’(1-x)\right]+\ln\left(\rho^{z}{pub}\right)+\ln\left(t^{y+1}\right)}{1-x} - \ln(\rho)}{t} \simeq \frac{\frac{\ln\left(\rho^{z}{pub}\right)+\ln\left(t^{y+1}\right)}{1-x} - \ln(\rho)}{t}$

$ = \frac{\left(\frac{z}{1-x} - 1\right)\ln\rho_{pub} + \frac{y+1}{1-x}\ln( t )}{t}$

$ = \frac{x+z-1}{1-x}\ln\rho_{pub}\cdot\frac{1}{t} + \frac{y+1}{1-x}\cdot\frac{\ln( t )}{t}\qquad (3-6)$

We use dif­fer­en­ti­ation to find the max­imum of the ex­pres­sion above:

$\frac{d}{dt}\left(\frac{x+z-1}{1-x}\ln\rho_{pub}\cdot\frac{1}{t} + \frac{y+1}{1-x}\cdot\frac{\ln( t )}{t}\right) = 0$

$\frac{x+z-1}{1-x}\ln\rho_{pub}\cdot\frac{d}{dt}\frac{1}{t} + \frac{y+1}{1-x}\cdot\frac{d}{dt}\frac{\ln( t )}{t} = 0$

$\frac{x+z-1}{1-x}\ln\rho_{pub}\cdot\left(-\frac{1}{t^2}\right) + \frac{y+1}{1-x}\cdot\frac{d}{dt}\frac{\frac{1}{t}\cdot t-\ln( t )}{t^2} = 0$

$\frac{y+1}{1-x}\cdot\frac{1-\ln( t )}{t^2} = \frac{x+z-1}{1-x}\ln\rho_{pub}\frac{1}{t^2}$

Since

$t>0, x \neq 1$

$(y+1)[1-\ln(t)]  = (x+z-1)\ln\rho_{pub}$

Thus,

$\ln( t ) = 1-\frac{x+z-1}{y+1}\ln\rho_{pub}\qquad (3-7)$

The pub­lic­a­tion mul­ti­plier of the the­ory is:

$r = \left(\frac{\rho}{\rho_{pub}}\right)^z$

Since

$r>0$

$\ln( r ) = z\left[\ln(\rho)-\ln(\rho_{pub})\right]$

$\ln( r ) = zt\left(\frac{x+z-1}{1-x}\ln\rho_{pub}\cdot\frac{1}{t} + \frac{y+1}{1-x}\cdot\frac{\ln( t )}{t}\right)$

$= z\left(\frac{x+z-1}{1-x}\ln\rho_{pub} + \frac{y+1}{1-x}\ln(t)\right)\qquad (3-8)$

Sub­sti­tute (3-7) into (3-8),

$\ln®=$

$z\left[\frac{x+z-1}{1-x}\ln\rho_{pub}+\frac{y+1}{1-x}\left(1-\frac{x+z-1}{y+1}\ln\rho_{pub}\right)\right]$

$ = z\left[\frac{(x+z-1)\ln\rho_{pub} + (y+1)(1-\frac{x+z-1}{y+1}\ln\rho_{pub})}{1-x}\right]$

$ = z\left[\frac{(x+z-1)\ln\rho_{pub} + (y+1) - (x+z-1)\ln\rho_{pub}}{1-x}\right]$

$ = \frac{z(y+1)}{1-x}\qquad (3-9)$

From (3-9), we get:

$r = e^{\frac{z(y+1)}{1-x}}$

Q.E.D.

The main equa­tion in the En­dgame is:

$\frac{d\rho}{dt} = q_1q_2\left(c_1^{1.15}c_2 + c_3\rho^{0.2} + c_4\rho^{0.3}\right)$

Us­ing Lemma 1.5, we can get:

$q_1, c_1 \in \Theta\left(\rho^{0.1}\right)$

$q_2, c_2 \in \Theta\left(\rho^{\log2}\right)$

$c_3 \in \Theta\left(\rho^{\frac{2}{9}}\right)$

$c_4 \in \Theta\left(\rho^{\frac{1}{8}}\right)$

Us­ing Lemma 2.2, we can see that the dom­in­ant term in the par­en­theses on the right is

$c_4\rho^{0.3}$

$K\rho^{0.164}_{pub}$

$\frac{d\rho}{dt} \in \Theta\left(\rho^{0.3+0.125+0.1+\log2}t^0\rho^{0.164}_{pub}\right)$

There­fore,

$x = 0.525+\log2, y = 0, z = 0.164$

Thus, from The­orem 1 we get:

$r = e^{\frac{0.164}{0.475-\log2}} \simeq 2.5669$

This is very close to the geo­met­ric mean of the res­ults of The­ory Sim­u­lator from e1000 to e1500.

The main equa­tion is:

$\frac{d\rho}{dt} = (q_1r_1)^{1.15}$

This the­ory fea­tures four lay­ers of non-main equa­tions, all in the form of:

$q_{n-1} = c_nq_n$

or

$r_{n-1} = d_nr_n$

In which

$c_n$

$d_n$

$c_1, d_1 \in \Theta\left(\rho^{0.1}\right)$

$c_2, d_2 \in \Theta\left(\rho^{0.1}\right)$

$c_3, d_3 \in \Theta\left(\rho^{\frac{\log2}{10\log3}}\right)$

$c_4, d_4 \in \Theta\left(\rho^{0.05}\right)$

Us­ing Lemma 2.2, we can ana­lyze all the lay­ers and factor in the pub­lic­a­tion mul­ti­plier equa­tion of

$K\rho^{0.198}$

$\frac{d\rho}{dt} \in \Theta\left(\rho^{1.152(0.1+0.1+\frac{\log2}{10\log3}+0.05)}t^{9.2}\rho^{0.198}_{pub}\right)$

There­fore,

$x \simeq 0.72011 , y = 9.2, z = 0.198$

Thus, from The­orem 1 we get:

$r \simeq e^{\frac{0.198*10.2}{0.27989}} \simeq 1360.61$

Which is close to the res­ults of The­ory Sim­u­lator without T2MC from e1000 to e1500. This be­ha­vior is ex­pec­ted since Coast­ing – The main concept of T2MC is not con­sidered at all in this ana­lysis.

1. Ghost (Playsprout)
2. Spiral (a_spir­al­ist)
3. Megam­inx

1. Mr. Op­timal (NGZ)
2. Speedrun­ner (Jooo)
3. scide (scbose)
4. Ghost (Playsprout)
5. Helper (Axiss)
6. nubest

Thanks the en­tire com­munity for sup­port­ing my on the way of re­search­ing and cre­at­ing such a won­der­ful en­vir­on­ment!