Fit data to function $g(t) = \frac{100}{1+\alpha e^{-\beta t}}$ by using least squares method (projection/orthogonal families of polynomials)

Question

t --- 0 1 2 3 4 5 6

F(t) 10 15 23 33 45 58 69

Adjust $F$ by a fnction of the type $$g(t) = \frac{100}{1+\alpha e^{-\beta t}}$$ by the discrete least squares method

I'm studying orhotogonal families of polynomials and projection onto subspaces in the context of least squares method.

I think need to see this problem as a projection onto some subspace and use some inner product but I'm lost.

UPDATE:

Shouldn't the function $g(t)$ be a member of a vector space? I tried applying $\ln$ to see if I'd get something from a vector space but it also won't work

Answer 1

Forgetting (projection/orthogonal families of polynomials), the problem is quite easy to solve using standard nonlinear regression.

As usual, we need good or at least consistent estimates of parameters $(\alpha, \beta)$ and these can be obtained by a linearization of the model. $$g = \frac{100}{1+\alpha e^{-\beta t}} \implies \color{red}{y}=\log \left(\frac{100}{g}-1\right)=\log(\alpha)-\beta\,t=\color{red}{a+b t}$$

Consider the data to be $$\left( \begin{array}{ccc} t & g & y=\log \left(\frac{100}{g}-1\right) \\ 0 & 10 & +2.197225 \\ 1 & 15 & +1.734601 \\ 2 & 23 & +1.208311 \\ 3 & 33 & +0.708185 \\ 4 & 45 & +0.200671 \\ 5 & 58 & -0.322773 \\ 6 & 69 & -0.800119 \end{array} \right)$$ A preliminary linear regression leads to $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & +2.21599 & 0.01226 & \{+2.18195,+2.25003\} \\ b & -0.50409 & 0.00340 & \{-0.51353,-0.49465\} \\ \end{array}$$ corresponding to $R^2=0.999878$ which is already very good.

This gives as estimates $\alpha=e^a=9.17046$ and $\beta=-b=0.50409$.

Now, we can start the nonlinear regression and obtain $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ \alpha & 9.22336 & 0.13438 & \{8.85027,9.59645\} \\ \beta & 0.50576 & 0.00350 & \{0.49603,0.51549\} \\ \end{array}$$ corresponding to $R^2=0.999972$ which is very good. PLease, notice how good are the initial estimates.

Below are reproduced the data as well as the predicted values $$\left( \begin{array}{ccc} t & g & g_{pred} \\ 0 & 10 & 9.782 \\ 1 & 15 & 15.24 \\ 2 & 23 & 22.97 \\ 3 & 33 & 33.08 \\ 4 & 45 & 45.05 \\ 5 & 58 & 57.62 \\ 6 & 69 & 69.27 \end{array} \right)$$

If we had in advance known that the model was good (based on physics for example) and the data in small errors (because of accurate measurements), we could have skipped th first step and used the first and last data points to estimate $(\alpha, \beta)$

$$10=\frac {100}{1+\alpha} \implies \alpha=9$$ $$69=\frac {100}{1+9 e^{-6\beta}}\implies \beta=\frac{1}{6} \log \left(\frac{621}{31}\right)=0.499557$$

Answer 2

Make the transformations: $$g(t) = \frac{100}{1+\alpha e^{-\beta t}} \iff \alpha e^{-\beta t}=\frac{100}{g(t)}-1 \iff \underbrace{\ln \left(\frac{100}{g(t)}-1\right)}_{y(x)}=\underbrace{-\beta t}_{ax}+\underbrace{\ln \alpha}_{b}$$ Hence: $$\begin{array}{c|r|r} &x&y(x)&xy&x^2\\ \hline &0&2.20&0.00&0\\ &1&1.73&1.73&1\\ &2&1.21&2.42&4\\ &3&0.71&2.13&9\\ &4&0.20&0.80&16\\ &5&-0.32&-1.60&25\\ &6&-0.80&-4.80&36\\ \hline \text{Total}&21&4.93&0.68&91\\ \end{array}\\ \begin{align}a&=\frac{\sum xy-\frac{\sum x \sum y}{n}}{\sum x^2-\frac{(\sum x)^2}{n}}=\frac{0.68-\frac{21\cdot 4.93}{7}}{91-\frac{21^2}{7}}=-0.5\\ b&=\bar{y}-a\bar{x}=\frac{4.93}{7}-(-0.5)\frac{21}{7}=2.2\\ \ln \alpha&=b=2.2 \Rightarrow \alpha =9.03\\ \beta &=-a=0.5\end{align}$$ So, the final answer: $$g^*(t) = \frac{100}{1+9.03 e^{-0.5t}}\\ \begin{array}{c|c|c} t&g(t)&g^*(t)\\ \hline 0&10&9.97\\ 1&15&15.44\\ 2&23&23.14\\ 3&33&33.17\\ 4&45&45.00\\ 5&58&57.43\\ 6&69&68.99 \end{array}$$