<s>[INST] Predicting increasing income within a given time I have this game where player earns "gold" and "reputation". With higher reputation, the player nets more income (gold).


*

*Each reputation points yields 1% bonus in income (gold)

*For every 20 gold received, player receives 1 reputation points

*Player starts with receiving 10 gold every second. This increases as reputation point increases.


How much gold and reputation will the player receive after 1000 seconds?
So I'm asking for an equation/formula and how you came up with it. I tried simplifying the example so I can explain it easily. I think I need a quadratic equation of some sort, but i'm just speculating.
I'm not really that good with providing my own formula :(
EDIT:
Here's an excel simulation of what I described above: https://www.dropbox.com/s/vz5wizqg4x43599/offline%20progression%20simulation.xlsx?dl=0
The first row holds the initial value (Time = 0)
 [/INST] Let $G(n)$ and $R(n)$ denote the amounts of gold and reputation after $n$ seconds, respectively. Then $G(0)=R(0)=0$ and $R(n)=\lfloor\tfrac{G(n)}{20}\rfloor$ and 
$$G(n+1)=G(n)+10(1+0.1R(n))=G(n)+10(1+0.01\lfloor\tfrac{G(n)}{20}\rfloor).$$
If we ignore the floor function, this becomes $G(n+1)=1.005G(n)+10$, which can be written as
$$\begin{pmatrix}
G(n+1)\\1
\end{pmatrix}
=
\begin{pmatrix}
1.005&10\\
0&1
\end{pmatrix}
\begin{pmatrix}
G(n)\\1
\end{pmatrix}.$$
This gives us an explicit expression for $G(n+1)$ in terms of $n$;
$$\begin{pmatrix}
G(n+1)\\1
\end{pmatrix}
=
\begin{pmatrix}
1.005&10\\
0&1
\end{pmatrix}^{n+1}
\begin{pmatrix}
0\\1
\end{pmatrix}$$
Luckily, powers of this matrix are easy to compute because $\tbinom{0\ 10}{0\ \hphantom{1}0}^2=0$, so by the binomial theorem
\begin{eqnarray*}
\begin{pmatrix}
1.005&10\\
0&1
\end{pmatrix}^{n+1}
&=&\left(
\begin{pmatrix}
1.005&0\\
0&1
\end{pmatrix}
+\begin{pmatrix}
0&10\\
0&0
\end{pmatrix}\right)^{n+1}\\
&=&\begin{pmatrix}
1.005&0\\
0&1
\end{pmatrix}^{n+1}
+(n+1)\begin{pmatrix}
1.005&0\\
0&1
\end{pmatrix}^n
\begin{pmatrix}
0&10\\
0&0
\end{pmatrix}\\
&=&
\begin{pmatrix}
1.005^{n+1}&10(n+1)1.005^n\\
0&1
\end{pmatrix}
\end{eqnarray*}
This shows that
$$\begin{pmatrix}
G(n+1)\\1
\end{pmatrix}
=
\begin{pmatrix}
1.005^{n+1}&10(n+1)1.005^n\\
0&1
\end{pmatrix}
\begin{pmatrix}
0\\1
\end{pmatrix},$$
and hence that
$$G(n+1)=10(n+1)1.005^n.$$
In particular, after $n=1000$ seconds this gives us
$$G(1000)\approx1458464.$$
Keep in mind that we have ignored the floor function at the start; in this model you do not gain $1$ reputation for every $20$ gold in increments, but you simply gain $1/20$-th of your gold in reputation.
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