Linearization of $y(t) = \dfrac{y_M}{1+\left(\frac{y_M - y_0}{y_0}\right)\exp\left(-y_Mk(t-t_0)\right)}$.

53 Views Asked by At

I'm trying to apply Least Squares to fit to the curve $$y(t) = \dfrac{y_M}{1+\left(\frac{y_M - y_0}{y_0}\right)\exp\left(-y_Mk(t-t_0)\right)},$$ where $t_0$ and $y_0 = y(t_0)$ are constansts. Here, I'm actually looking for $y_M$ and $k$ that gives the best fit.

In my attempts, I got the following: \begin{align*}\dfrac{1}{y(t)} &= \dfrac{1}{y_M} + \left(\frac{1}{y_0}-\dfrac{1}{y_M}\right)\exp\left(-y_Mk(t-t_0)\right)\\\Rightarrow \ln\left(\dfrac{1}{y(t)} - \dfrac{1}{y_M}\right) &= \ln\left(\frac{1}{y_0}-\dfrac{1}{y_M}\right) -y_Mk(t-t_0),\\\end{align*} which bugs me, since the left hand side must be a function of $y$ and $t$, and must not include $y_M$ nor $k$, since those are the coefficients I'm looking for.

Any idea on how to progress from here? Or am I way off?

1

There are 1 best solutions below

0
On

You are on a good track writing

$$\dfrac{1}{y} = \dfrac{1}{y_M} + \left(\frac{1}{y_0}-\dfrac{1}{y_M}\right)\exp\left(-y_Mk(t-t_0)\right)$$

Give $y_M$ a value and write

$$\frac 1{y_M} \log \left(\frac{y \,(y_0-y_M)}{y_0 \,(y-y_M)}\right)=k (t-t_0)$$

Let $$Y_i=\frac 1{y_M} \log \left(\frac{y_i \,(y_0-y_M)}{y_0 \,(y_i-y_M)}\right) \qquad T_i=t_i-t_0\quad\implies\quad k=\frac {\sum_{i=1}^n Y_i\,T_i} {\sum_{i=1}^n T_i}$$

With this value of $k$, you have the $\hat Y_i$ from which you deduce $\hat y_i$ and you can compute

$$SSQ(y_M)=\sum_{i=1}^n \big[\hat y_i-y_i\big]^2$$

Vary $y_M$ until you see more or less a minimum.

At this point, you are ready for a nonlinear regression since you have good and consistent starting values.

In fact, you could even avoid the nonlinear regression and solve the problem using Newton-Raphson method using central differences.