Exponential Regression

What's that?

Regression is finding an approximation to predict future and current data based on a set of given input data. Usually the approximation is put in a linear way, the so-called Linear Regression.

However, sometimes a simple line does not approximate the data's interaction and correlations properly. In such cases one has to switch to a polynomial or exponential regression. The decision is based upon the growth and growth expectance of the data. If this growth is in a similar extreme way as an exponential function (first a huge amount of almost no growth, then explosive growth) exponential regression should be the way.

How to install

$ go get github.com/nillga/exponential-regression

Example Usage

Caution: This example absolutely ignores error handling!

package main
import (
    "fmt"

    "github.com/nillga/exponential-regression"
)

func main() {
    r := exponential-regression.Regression{}
    r.Init([]struct{
        x float64
        y float64
    }{
        {1.0, 2.2},{1.1, 2.2},{1.2, 2.22},{1.3, 2.23},
        {1.4, 2.26},{1.5, 2.28},{1.6, 2.3},{1.7, 2.35},
        {1.8, 2.43},{1.9, 2.6},{2.0, 3.2},{2.1, 4.8},
    })
    r.Convert()
    r.Run()
    a,b,_ := r.Result()

    fmt.Printf("Regression formula:\ny=%.4f*%.4f^x\n",a,b)
}

The output of this example will look like this:

$ Regression formula:
  y=1.2036*1.6098^x
$

Realisation

This package uses the base logic of this package: github.com/sajari/regression

It utilizes a linear regression only, but it is possible to calculate an exponential regression with the toolbox of a linear regression. For doing this, a simple trick needs to be used.

Linear Regression

Linear regression will output an equation of the scheme y = c + m * x, that is able to predict values for x. As it is easily visible, this scheme is the scheme of a regular linear function.

Exponential Regression

Therefore, for exponential regression a formula in an exponential scheme (y = a * b^x) is expected. But how can we transform this into a linear equation? This is where the trick comes into play:

1.) We take the log of the equation: ln(y)=ln(a*b^x)

2.) By logarithm laws, we can drag this apart: ln(y)=ln(a)+ln(b^x)

3.) We can drag it further apart: ln(y)=ln(a)+x*ln(b)

ln(a) and ln(b) are constants, so we do end up with a linear equation.

4.) We perform a linear regression with the given x-values and the logarithm of our y-values

5.) LReg(x,ln(y)) ==> ln(y) = c + m*x

6.) Now we need to bring this back into shape: y = e^(c+m*x)

7.) This can be re-dragged again: y = e^c * e^(m*x)

8.) And finally in the shape of y = a * b^x: y = e^c * (e^m)^x

As easy as this, we have our exponential regression done.