$\newcommand{\s}{\text{superpercent}}$My friend made up the following problem and asked me the problem(he knows the answer, but wants me to solve it). Here's the exact wording:
Let's say that there's a function named $\s$ that takes in two arguments, $a$ and $b$, such that $a,b\in\mathbb R^+$. Define the value of $\s(a,b)$ as the quantity when you decrease $a$ by $b$ percent, then increase the resulting value by $b$ percent(note that this won't give you back $a$). For example, to calculate $\s(100,50)$, we first decrease $100$ by $50\%$ to get $50$, then increase $50$ by $50\%$ to get $75$. So $\s(100,50)=75$. Let's now define a function $\s_n(a,b)$ for $n\in\mathbb Z_0^+$ and $a,b\in\mathbb R^+$ as follows:$$\s_n(a,b)=\begin{cases}\s_{n-1}(\s(a,b),b)&\text{if }n>1\\\s(a,b)&\text{if }n=1\\a&\text{if }n=0\end{cases}$$ Find and simplify a formula for $\s_n(a,b)$ without using the $\s$ or $\s_n$ function(i.e. in terms of $a$, $b$, and $n$).
I have the formula for $\s(a,b)$, but I'm stuck on finding a formula for $\s_n(a,b)$. I got that $\s(a,b)=a*\frac{100-b}{100}*\frac{100+b}{100}$, but the piecewise function(I think it's also a recursion?) that defines $\s_n(a,b)$ confuses me. I know what it's saying, but I just don't know how to approach a problem like this. Any tips? Thanks for the help.
$\newcommand{\s}{\text{superpercent}}$ Note that $\s(a,b)=a*(1-\beta^2)$ where $\beta=\beta(b)=\frac{b}{100}$. So $$\s_2(a,b)=\s(a*(1-\beta^2),b)=a*(1-\beta^2)*(1-\beta^2)=a*(1-\beta^2)^2.$$ Does this help to calculate $\s_3(a,b), \ldots$?