Origin Ver. and Service Release (Select Help-->About Origin): Operating System: Hi, I am trying to fit my data and found that the Boltzmann fit matches my dataset very well. I also wanted to compare the Boltzmann fit with the exponential fit: y=A*exp(-x/t). In this fit, the t-value is the value of x where the original A constant has been decreased by 1/e. I am interested in finding out if the same can be said for dx in the Boltzmann function, how I should interpret that value? According to the help function, dx is defined as: "The Y value changes drastically within a range of X variable. The width of this range is approximately dx". Does this mean that dx is the width of the steepest part of the slope?
I think it's difficult to decide what's the meaning for each parameters, because different discipline may have different terms. I used to consider dx as an "scaling factor", which makes the curve steeper or flatter.
From the sample curve, I think we can estimate the slope at (x0, (a1+a2)/2) as (a2 - a1)/(4*dx).
Meaning of dx in Boltzmann fit
Posted - 11/19/2009 : 8:49:11 PM
