I have a bunch of curves that look roughly like the example below. Each curve has two 'asymptotes' a constant value for $x\rightarrow0$ and a linear curve for $x\rightarrow\infty$ (although, as in the example, infinity is not that far away, around $x=6 \text{ or } 7$).

Assuming that I know the slope of the linear function and the asymptotic values for $x\rightarrow0$ for each curve, I would like to scale/transform $y$ and $x$ such that all curves fall on top of each other. My question is: how do I do this scaling/transforming of $y$ and $x$ to achieve a collapse of all data?

I can see that simply shifting $y$ with $y(0)$ is not going to work out, because that also shifts the linearly sloped parts and purely stretching results in different linear slopes so I guess I need some smart combination but I really don't know where to start.

enter image description here


There are infinitely many such transformations, however, if you seek for a linear transformation, any curve with data $(x,y)$ can be rescaled to a new curve $(x',y')$ using

$$\begin{bmatrix}x' \\ y' \end{bmatrix}= \begin{bmatrix}x \\ y \end{bmatrix}+t\cdot\begin{bmatrix}x \\ y-y_0 \end{bmatrix}$$ and some $t\in\mathbb{R}$, where $y_0$ is the height, at which the sloping tangent meets $x=0$ (i.e. $y_0= 1.7$ in this example).

In other words, any curve with $y(0)=k$ can be mapped onto another curve with $y(0)=k'$ by choosing $t=(k'-k)/(y_0-k')$.

Geometrically, this just expresses a rescaling relatively to the point $(0,y_0)$.

  • $\begingroup$ I was indeed looking for a linear transformation. Sorry, forgot to mention that. Thanks for the answer! - Just for future reference: the choice of $t$ to map everything on the red curve in my case would be $t=(1-y_{as})/(y_0-y_{as})$ where $y_{as}$ is the value of the horizontal asymptote of the given function $\endgroup$ – Michiel Mar 12 '14 at 11:48
  • $\begingroup$ @Michiel yes thanks, just added this to the answer. $\endgroup$ – flonk Mar 12 '14 at 14:36
  • $\begingroup$ Just a quick follow-up, if the lefthand asymptote is also a linear function instead of a constant, could I still apply a similar transformation?! $\endgroup$ – Michiel Mar 13 '14 at 12:42
  • $\begingroup$ @Michiel would the lefthand asymptote still be fixed for all curves while the other one varies? $\endgroup$ – flonk Mar 13 '14 at 12:51
  • $\begingroup$ The slope would be fixed if that's what you mean. Essentially it will be the same as the graph in my question but with all curves having (all the same) slope $dy/dx\neq0$ at $x=0$ instead of the current case with $dy/dx=0$ at $x=0$ $\endgroup$ – Michiel Mar 13 '14 at 14:22

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