Dear Mary,

x=ln gamma/beta

I don't know how to edit my original question.

I need to know whether the offset not gamma is related to the inflection point of this curve?  Offset = alpha*exp(-gamma) = g(0) in the gompertz equation g(x)=alpha*exp(-gamma*exp(-beta*x)). I think it may be the y coordinate in the (x,y) pair?

Dear Mary,

I also played or still play with the Gompertzian funcZion.

Article Kinetics of bulk polymerisation and Gompertz's law

if necessary I restricted the allowed sign of the constants by using abs(constant) and starting with the shift in the middle of the fit-range.

I use this algorithm also successfully for contact angle data. Surprisingly it works very well.(very fast conversio, nearly no deviation)

I have seen that you do not have a shift value?!

Therefore Zero is the inflexion point with f(0) as value.

Dear Dr. Schmitt,

Thank you for a most delightful reply. Your article was exactly what I needed. I did some research into differential equations.

On page 592 you state that one checks that at shift-time t=ts there is an inflection point and that ts=time of max polymerization rate. I like that. You are using the first derivative for your interpretation. What you see in the graph of the Gompertzian function in figure 4 is time vs Gompertzian function. How did you determine what is on the x axis in figure 3?

I have been given a marketing effectiveness problem involving the Gompertz function which looks the same as yours but does not include the shift time in the pdf. The development of the function from the differential equation was not done so I am not sure. I have the Gompertz model given as y=B1 exp(-B2 exp(-B3 x) + error. Equation (3) will be the same as yours with a different interpretation for M?

I am trying to use your article to understand the Gompertz function. I was introduced to it in Generalized Linear Models 2nd edition by Myers, Montgomery, Vining and Robinson on page 108.

Given a data set I want to be able to plug in a value for ts. I get that value off the graph in figure 3?

Lastly, I hate to admit that I got a different expression for equation 5 when integrating. How did you get the double log from equation 4. Is there a trick identity I don't know about?

Sincerely,

Mary A. Marion

Hi Mary

concering the axis in figure 3. I loved to express the function relative to the conversion which in an ideal case ranges from 1 = M0/M0 to zero = 0/M0. If you have a remaining double bond content (as I had within my experiments) the range in which something happens can be expressed as presented in figure 3 both counter and denominator have to be adapted.

Try the derviation from 5 to 4. PS the derviation of ln(x) is 1/x.

Prof. Schulze-Pillot as mathematician had found the Ansatz.

i have a Maple file which might also contain the derivations of this equation but I do not have Maple anymore.

Figure 3 is not relative to time. It’s equation 3. due to the relative relations this graph also to not change.

Dear Mary,

x=ln gamma/beta

In Gompertz function (ae^(-be^(-cx))) a is the higher asymptote, that the curve will reach at +infinity (a is 1 in most of the cases).

b and c are modeled in such a way that fits the data. My problem was to fit two data points (x1, y1) and (x2, y2) which will kind of bound the curve. I wanted the curve to be near 0 on 0 and 0.9 on a value p (in Z+). So I fitted the curve with two data points (0.1, 0.01) and (p, 0.9). I problem was to derive such a (b, c) combination that fits the two data points. As I cant write latex I am linking the math.stackexchange answer to complete derivation https://math.stackexchange.com/a/3270090/14832

Your function is not able to cover your desired range. the solution for x equal to zero is ae**-b and not zero. Please check the mentioned publications (we solved a similar problem by adding a shifting time).

Hi Mary A. Marion , I found youtube video on how to estimate a first value of those coefficients. Just watch the whole video, the explanation comes with the example.

https://www.youtube.com/watch?v=0ifT-7K68sk

alpha=y(max)

gamma=ln(y(max)/y(0))

beta=ln[ln(y(1)/y(max))/ln(y(max)/y(0))]