I am trying to fit data to an equation of this form using non-linear regression: y = 1 − a exp ⁡ ( n ( x − x 0 ) ) I would like to select three sets of ( x , y ) data and solve the above equation for a, n and x 0 in order to improve the initial estimates of these parameters. I eliminated a using the following expression: a = ( 1 − y 1 ) exp ⁡ ( n ( x 1 − x 0 ) ) where ( x 1 , y 1 ) is one of the selected data points and then back-substituted it to give: y 2 = 1 − ( 1 − y 1 ) exp ⁡ ( n ( x 1 − x 0 ) ) exp ⁡ ( n ( x 2 − x 0 ) ) where ( x 2 , y 2 ) in another of the selected data points. This becomes y 2 = 1 − ( 1 − y 1 ) exp ⁡ ( n ( x 1 − x 2 ) ) and thus n = 1 x 1 − x 2 ln ⁡ ( 1 − y 2 1 − y 1 ) If I use point ( x 3 , y 3 ) instead I get: n = 1 x 1 − x 3 ln ⁡ ( 1 − y 3 1 − y 1 ) The problem is that I have two different expressions for n but none for x 0 because it was eliminated along the way. I get a similar result if I eliminate x 0 first.Is it possible to solve for a, n and x 0 in this manner?

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I am trying to fit data to an equation of this form using non-linear regression:  y  =  1  −      a          exp      ⁡              (                  n                      (                          x              −                                                x                  0                                                      )                          )            I would like to select three sets of   (  x  ,  y  ) data and solve the above equation for   a,   n and       x    0   in order to improve the initial estimates of these parameters. I eliminated   a using the following expression:  a  =      (          1      −                        y          1                      )    exp  ⁡      (          n              (                                            x              1                                −                                    x              0                                      )              )  where   (      x    1    ,      y    1    ) is one of the selected data points and then back-substituted it to give:            y      2        =  1  −                    (                  1          −                                    y              1                                      )            exp      ⁡              (                  n                      (                                                            x                  1                                            −                                                x                  0                                                      )                          )                    exp      ⁡              (                  n                      (                                                            x                  2                                            −                                                x                  0                                                      )                          )            where   (      x    2    ,      y    2    ) in another of the selected data points. This becomes            y      2        =  1  −      (          1      −                        y          1                      )    exp  ⁡      (          n              (                                            x              1                                −                                    x              2                                      )              )  and thus  n  =      1                            x          1                    −                        x          2                      ln  ⁡      (                            1          −                                    y              2                                                1          −                                    y              1                                            )  If I use point   (      x    3    ,      y    3    ) instead I get:  n  =      1                            x          1                    −                        x          3                      ln  ⁡      (                            1          −                                    y              3                                                1          −                                    y              1                                            )  The problem is that I have two different expressions for n but none for       x    0   because it was eliminated along the way. I get a similar result if I eliminate       x    0   first.Is it possible to solve for   a,   n and       x    0   in this manner?

Rebekah Zimmerman · Accepted Answer

y  =  1  −      a          exp      ⁡              (                  n                      (                          x              −                                                x                  0                                                      )                          )              1  −  y  =  a  exp  ⁡      (          −      n              (                  x          −                                    x              0                                      )              )    ln  ⁡  (  1  −  y  )  =  −  n  (  x  −      x    0    )  +  ln  ⁡  (  a  )This is a function on the form:  Y  =  A  x  +  B     with         {                            Y          =          ln          ⁡          (          1          −          y          )                                      A          =          −          n                                      B          =          n                                x            0                    +          ln          ⁡          (          a          )                        So, there is no need for calculus such you did. Simply, transform your data   (  x  ,  y  ) into   (  x  ,  Y  ) in computing   Y  =  ln  ⁡  (  1  −  y  ) for each point. Then, carry out a linear regression to find the approximates of   A and   B.NOTE:They are redondant parameters in your initial equation. That is the cause of the trouble that you encountered.  y  =  1  −      a          exp      ⁡              (                  n          (          x          −                      x            0                    )                )              =  1  −            a      exp      ⁡      (      n                    x        0            )              exp      ⁡              (                  n                    x                )              a  exp  ⁡  (  n      x    0    ) is a constant, thus   a  ,  n  ,      x    0   are not independent. You can arbitrary chose a value for       x    0   and then you obtain a value for   a. Or you can arbitrary chose a value for   a and you obtain a value for       x    0  . The fitting will be exactly the same what ever your choice.

I am trying to fit data to an equation of this form using non-linear regression: y = 1 &#x

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