Strictly speaking it is still non-linear because the errors are centred on x not e^-x. Probably won't make a big difference though
"Justin Davis" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > > "Horst Kraemer" <[EMAIL PROTECTED]> wrote in message > news:[EMAIL PROTECTED] > > On Sat, 24 Apr 2004 08:02:25 +0200, "Konrad Den Ende" > > <[EMAIL PROTECTED]> wrote: > > > > > Suppose you know that a process follows a function > > > y(t) = a + b e^-x, t >= 0. > > > > y(t) = a + b e^-t, t >= 0. > > > > > ALso, suppose you have following data. > > > t: { 0, 1, 2, 3 } > > > y: { 2.2, 1.4, 0.87, 0.44 } > > > > > > How does one estimate the values of a and b? > > > > Isn't this a simple least square problem which reduces to a linear > > regression y = a + b*x for > > > > x: { 1 , e^-1, e^-2 ,e^-3 } > > y: { 2.2, 1.4, 0.87 ,0.44 } > > > > > > -- > > Horst > > > > Horst, > > You're right. I read over and saw a functional form so similar to one I've > worked with for a while (a + b * exp (c x)) that I immediately put a > constant in the exponent. Sorry for the earlier wrong reply. > > Justin Davis > > . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
