A curve that gets $20\%$ closer to $y = 1$ for each $+1$ increase in $x$

Question

I'm surprised at how much I'm struggling to find an equation for the curve described below (I don't have a math degree, but it seems like it shouldn't be complicated).

The origin of the curve is $[0,0]$. The Y coordinate of this curve gets $20\%$ closer to $1$ each time the $X$ coordinate increases by $1$. Thus, the $Y$ asymptote is $1$.

Here are the coordinates for integer values of $x$ from 0 to 10:

0   0 
1   0.2 
2   0.36 
3   0.488 
4   0.5904 
5   0.67232 
6   0.737856 
7   0.7902848 
8   0.83222784 
9   0.865782272 
10  0.8926258176 

Using an online tool for least squares curve fitting, I found the formula:

$y = 1 - e ^{-0.223x}$

Which is reasonably accurate, but unintuitive. Is there a simpler formula that expresses this curve?

Answer 1

It is easier to see the pattern if you look at the distance remaining to $1$:

0   1 - 1   
1   1 - 0.8  
2   1 - 0.64   
3   1 - 0.512
etc.

These are powers of $0.8$; the formula you're looking for is $$y = 1-0.8^x.$$

You already found this answer, in fact, since $e^{-0.223\ldots} = 0.8$.

Answer 2

Denote $y$ at $x=n$ by $y_n$. Then you have$$y_{n+1}=y_n+0.2(1-y_n)=0.2+0.8y_n$$This is a linear first-order recurrence with initial condition $y_0=0$.

• The solution of the auxiliary equation is $y_1=A(0.8)^n,A\in\Bbb R$.
• The particular solution is $y_2=1$, giving the general solution as$$y_n=y_1+y_2=A(0.8)^n+1$$Using the initial condition we get $A=-1$.