The first row is time and the second row is height of a plant. We need to use generalized logistic curve to model the behavior of the plant. The equation of the logistic curve is : $$N = \frac{N_*}{1+(N_*/N_0-1)e^{-a_0t}}$$ where $N_*$ is the maximum height of the plant or in other words, the supremum of the logistic function.
$a_0$ is how fast the function increases. My question is the following:
I know how to determine $N_*$, we just make up a number that is above $251$ and less than $251 + (251 - 247) $. But how to exactly determine $a_0$. I am quite lost here. How I would do it is to calculate every $a_0$ for every time and height, with equation $a_0 = \frac{1}{t}\ln(\frac{N}{N_0})$. In this equation $N$ would be the difference between heights, so let's say between $18$ and $33$ the $N$ for time equals $1$ would be $33 - 18$. Is the process here of creating a table of $\sum_{k = 1}^{10}(10 - i)$ column elements of differences, so that we take multiple $N_0$ until we run out of data from the primary table in the picture (so until $N_0 = 9$). Then we can use a formula for $a_0 = \frac{\sum a_{0l}*t_l}{\sum t_l}$ where $l$ is the number of elements.
Is my procedure for creating a logistic out of the data provided correct?
Method 1: One degree of freedom
$\newcommand{\eqd}{\triangleq}$ $\newcommand{\brp}[1]{{\left(#1\right)}}$ $\newcommand{\brs}[1]{{\left[#1\right]}}$ $\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}$ $\newcommand{\pderiv}[2]{{\frac{\partial#1}{\partial#2} }}$ Have you considered using the Method of Least Squares to find $a_0$? For that method you can define an error function $$e(t_n)\eqd N(t_n)-y_n\eqd \frac{N_{*}}{1+\brs{N_{*}/N_0-1}e^{-a_0t_n}}-y_n$$ and a cost function (the error cost) as the norm squared of $e(t_n)$ as $$cost(a_0)\eqd\norm{e}^2\eqd\sum_{n=0}^{n=10}e^2(t_n)$$ You want to find the $a_0$ that minimizes that cost; that is, you want to find where (with respect to $a_0$) the cost function "goes to the lowest point". Suppose you set $N_0\eqd N(0)=18$ and $N_*\eqd252$ $\ldots$
From the plot, it appears that $cost(a_0)$ is minimized around $a_0=0.65$.
To get a more accurate optimal $a_0$, we can differentiate $cost(a_0)$ with respect to $a_0$ and set the expression to $0$. To do that, this (lemma) may come in handy: $$\begin{align} \boxed{N'(t)} &\eqd \pderiv{}{a_0}N(t) \\&\eqd \pderiv{}{a_0}\brs{\frac{N_{*}}{1+\brp{\frac{N_*}{N_0}-1}e^{-a_0t}}} && \text{by definition of $N(t)$} \\&= \frac{0-N_*\brs{\frac{N_*}{N_0}-1}e^{-a_0t}(-t)}{\brp{1+\brs{\frac{N_*}{N_0}-1}e^{-a_0t}}^2} && \text{by Quotient Rule} \\&= \frac{N_*^2}{\brp{1+\brs{\frac{N_*}{N_0}-1}e^{-a_0t}}^2} \brs{\frac{1}{N_0}-\frac{1}{N_*}}te^{-a_0t} \\&\eqd \boxed{\brs{\frac{1}{N_0}-\frac{1}{N_*}}N^2(t)te^{-a_0t}} && \text{by definition of $N(t)$} \end{align}$$ Then$\ldots$ $$\begin{align} \boxed{0}&= \frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\cdot0 \\&=\frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\pderiv{}{a_0}\norm{e}^2 \\&\eqd \frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\pderiv{}{a_0}\sum_{n=0}^{n=10}e^2(t_n) && \text{by definition of $\norm{\cdot}$} \\&\eqd \frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\pderiv{}{a_0}\sum_{n=0}^{n=10}\brs{N(t_n)-y_n}^2 && \text{by definition of $e$} \\&= \frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\sum_{n=0}^{n=10}2\brs{N(t_n)-y_n}N'(t_n) && \text{by Chain Rule} \\&= \frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\sum_{n=0}^{n=10}2\brs{N(t_n)-y_n}\brs{\frac{1}{N_0}-\frac{1}{N_*}}N^2(t_n)te^{-a_0t_n} && \text{by (lemma)} \\&= \boxed{\sum_{n=0}^{n=10}N^2(t_n)\brs{N(t_n)-y_n}t_ne^{-a_0t_n}} \\&\eqd Dcost(a_0) && \text{(call the sum $Dcost(a_0)$)} \end{align}$$ Plotting $Dcost$ with respect to $a_0$, it appears that $Dcost(a_0)$ crosses $0$ at around $a_0=0.66$:
The uniroot function from the R stats package indicates that $Dcost(a_0)$ crosses $0$ at $a_0=0.6631183$ with estim.prec=6.103516e-05.
Using $N_0\eqd18$, $N_*\eqd252$, and $a_0\eqd0.6631183$, $N(t)$ seems to fit the 11 data points fairly well ($cost(0.6631183)=31.32307$) $\ldots$
Some R code supporting Reproducible Research: