From: Walter Roberson on 7 May 2010 09:49 John D'Errico wrote: > "C. " <vogel.jan(a)googlemail.com> wrote in message > <hs0j69$pc5$1(a)fred.mathworks.com>... >> c is 1 for the first plot, 2 for the second and so on... There are >> multiple equations i'd like to fit to multiple data sets, but i have >> to get one setof parameters a, b. >> >> >> Walter Roberson <roberson(a)hushmail.com> wrote in message >> <LaQEn.418$V%2.304(a)newsfe08.iad>... >> > C. wrote: >> > > Hello, >> > > I'm trying to fit one equation y=a*(b^c)*t to multiple plots. > NO. You cannot do this fit. It is not possible to > estimate both a and b here. Not even if c is fixed > at some SINGLE value! > > Whoever has told you that it is possible is wrong. The fact that you have different c's is what allows it to be solved, by establishing ratios amongst the different y's. The a's fall away in the ratio, allowing one to fit amongst the various b estimates that one gets.
From: Walter Roberson on 7 May 2010 11:44
Walter Roberson wrote: > Now take a weighted average of the bh. On a finite precision machine, > the first of them should be the most accurate if b>1 and the last of > them should be the most accurate if b<1. I'm no longer sure that that is the case; now I'm thinking that perhaps the first one will always be the most accurate. > divide the second by the first > bh{2} = (y{2} ./ t) / (y{1} ./ t) Note that the / there is intended to be a least-squares fit operation. / will do a LSF but I'm weak on exactly what the inputs need to be. |