Please look at the attached logistic data plots. I have used excel's polynomial trend line feature to fit a curve to the logistic sigmoid curves. Is there a better way or formula to fit curves to these type of plots? eng-0.xlsx
I shall consider "column B" only.
There are of course many sigmoidal curves, the most prominent being 1/(1+e^(-t)).
Probably you want to test your data against transformations of this.
Since the values "apparently" tend to 0 for t -> -oo and to 1 for t -> +oo, we may restict our search to determine parameters a and b such that 1/(1 + e^(-a*t + b) ) is a good fit.
Note that the outer data points hardly contribute anythin to our knwledge of the curve (they've already fulfilled theiry duty by telling us about the limits as t -> +- oo). The most distinctive part of the curve is near y = 1/2.
The expression 1/(1 + e^(-a*t + b) ) becomes 1/2 exactly when t = b/a.
The derivative of 1/(1 + e^(-a*t + b) ) at the same point is a/4.
Hence I suggest you take only thos data point with value between 0.3 and 0.7, say, make a linear degression, i.e. find a best fit line y = A*t + B.
This line passes through y=1/2 at t = (1/2 - B)/A with slope A.
Thus a = 4*A and b = 2 - 4*B should give a good fit.
Thank you for your reply. I do not have enough math knowledge to understand your answer fully. Could you please use some examples to illustrate your answer. So for column B if I was supplied the X values of -4 and 8 what would I need to do to predict the Y values (somewhere between 0 and 1) ? thanks.
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There are of course many sigmoidal curves, the most prominent being 1/(1+e^(-t)).
Probably you want to test your data against transformations of this.
Since the values "apparently" tend to 0 for t -> -oo and to 1 for t -> +oo, we may restict our search to determine parameters a and b such that 1/(1 + e^(-a*t + b) ) is a good fit.
Note that the outer data points hardly contribute anythin to our knwledge of the curve (they've already fulfilled theiry duty by telling us about the limits as t -> +- oo). The most distinctive part of the curve is near y = 1/2.
The expression 1/(1 + e^(-a*t + b) ) becomes 1/2 exactly when t = b/a.
The derivative of 1/(1 + e^(-a*t + b) ) at the same point is a/4.
Hence I suggest you take only thos data point with value between 0.3 and 0.7, say, make a linear degression, i.e. find a best fit line y = A*t + B.
This line passes through y=1/2 at t = (1/2 - B)/A with slope A.
Thus a = 4*A and b = 2 - 4*B should give a good fit.