# Achieving the “mirror” of exponential decay

I'm working on a product that has a visual transition. I've found that applying a simple filter that results in an exponential decay (starting fast, then tapering off) is pleasing in one direction. The problem is that going the other direction with the same filter looks bad. The filter I'm using is of the form:

$$Y_\text{next} = Y_\text{current} + \alpha\left(\text{target}-Y_\text{current}\right)$$

What I basically need is the reverse behavior, where the transition starts out slow and then speeds up. Imagine playing the exponential decay I currently see "in reverse". I'm stumped as to the formulation, however. Note that I needn't have one formula to cover all cases, as I can switch the behavior depending on direction and/or current value.

This feels like a recurrence-type problem, but my familiarity with that topic is woefully out of date.

How about using $\alpha^{-1}$, for exponential growth? –  Neal Mar 11 '12 at 1:18
α is a constant, and $\left(target - Y_{current}\right)$ is still decreasing as target is approached thus that term is getting smaller. –  LVB Mar 11 '12 at 1:23
I don't quite understand the question. What exactly do you mean by "the other direction"? Are $Y$ and $target$ functions of time? If so, are you allowed to "look into the future" in your formulation? –  Rahul Mar 11 '12 at 1:42
Okay, I thought about this a little more and I'm not going to be stupid this time. You might try fitting to an exponential growth that terminates at $t = T$ with value $target$: $Y(t) = target \exp{\alpha(t - T)}$. In these terms, your current model is something like $Y(t) = target( 1 - \exp{\alpha t} ).$ –  Neal Mar 11 '12 at 2:10