What is the equation I can use to create the inverse expoential graph?

Question

I am seeking an equation I can apply to provide a score based on a measurable ability rating. It would be great to understand how the equation works so I can alter the curvature of this flexibly based on different scoring matrix.

Answer 1

Perhaps it's

$$ f(x) = 100 \cdot \big(1 - e^{-\alpha x}\big) $$

or perhaps it's

$$ f(x) = 100 \cdot \big(1 - \tfrac{1}{1+\beta x}\big) $$

or perhaps it's

$$ f(x) = 100 \cdot \big(\tfrac{2}{\pi}\arctan(\gamma x)\big) $$

Answer 2

Maybe you can try the rational polynomial functions as:

$f(x)=\dfrac{a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0}{b_mx^m+b_{m-1}x^{m-1}+...+b_{1}x+b_0}$

You can control the limit in $\infty$ with $n, m, a_n$ and $b_m$. For this case $n=m$ and $\dfrac{a_n}{b_m}=100$. Also, you could control the point around $0$ with $a_0$ and $b_0$ and curvature with the rest of the coefficients.

Answer 3

You're searching for an asymptotic function. The simplest asymptotic function I can think of is $y=\frac{1}{x}$:

This already looks a bit like what we want. However, we can tweak it a bit. First of all, we want $y$ to increase with increasing $x$, so let's negate one of the two sides:

$y=-\frac{1}{x}$

Now, when $x$ approaches infinity, both sides will go towards 0. We want $y$ to approach $100$ in that situation, so let's fix that. We now have:

$y-100=-\frac{1}{x}$.

We also want the curve to pass through the origin, where $x=y=0$. We can achieve this by changing the right-hand side slightly:

$y-100=-\frac{100}{x+1}$

Now, when $x$ is small, the right hand side goes towards $-100$, whereas for large $x$ the denominator is dominant and it behaves quite like the $-1/x$ we had before.

One last tweak we can do is put a constant (call it $a$) in front of $x$ to control how fast the function approaches $100$ with increasing $x$. While we're at it, let's also add 100 to both sides. This results in our final function:

$y=-\frac{100}{ax+1}+100$

This is the curve for $a=0.2$:

Answer 4

Sigmoid

$$f(x) = \dfrac{M}{1+e^{-kx}} $$

where:

$M$ - the asymptotic value

$k$ - controls how steep is the curve

Power and exponential function

$$f(x)=M\left(1-e^{-kx^p}\right)$$

where:

$M$ - the asymptotic value

$k$ - controls how steep is the curve

$p$ - controls the bend of the knee

For example $p=1$:

$p=2$