Exercise

Given the values

$$ \begin{array}{|c|ccccc|} \hline t & 1 & 2 & 3 & 4 & 5\\ \hline Q & 0.47 & 0.27 & 0.20 & 0.15 & 0.12 \\ \hline \end{array} $$

fit, using the least squares criterion, the curve

$$Q\left(t\right)=\frac{Q_{0}}{1+Q_{0}Kt}$$

---

We linearize the function to approximate. If we take into account that

$$ Q=\frac{Q_{0}}{1+Q_{0}K\,t} \quad \Longrightarrow \quad \frac{1}{Q}=\frac{1+Q_{0}K\,t}{Q_{0}} \quad \Longrightarrow \quad \frac{1}{Q}=\frac{1}{Q_0}+K\,t $$

And if we call

$$ y_k = \frac{1}{Q_k}, \quad x_k =t_k, \quad a_0 = \frac{1}{Q_0}, \quad a_1=K $$

we have

$$ \frac{1}{Q_k}=\frac{1}{Q_0}+K\,t_k \quad \Longrightarrow \quad y_k = a_0 + a_1\, x_k $$

The problem is now to fit a least squares regression line

$$ P_1(x) = a_0 + a_1\, x $$

to the transformed data $(x_k,y_k),\quad k = 1,\ldots,5$ with the system

$$ \left(\begin{array}{cc} \sum_{k=1}^{5}1 & \sum_{k=1}^{5}x_{k}\\ \sum_{k=1}^{5}x_{k} & \sum_{k=1}^{5}x_{k}^{2} \end{array}\right)\left(\begin{array}{c} a_{0}\\ a_{1} \end{array}\right)=\left(\begin{array}{c} \sum_{k=1}^{5}y_{k}\\ \sum_{k=1}^{5}x_{k}y_{k} \end{array}\right) $$

We compute the elements of these matrices

$$ \begin{array}{c|c|c|c|c|c|c|} \hline & x_k = t_k & Q_k & y_{k}=1/Q_k & x_{k}^{2} & x_{k}\,y_k\\ \hline & 1 & 0.47 & 2.128 & 1 & 2.127 \\ & 2 & 0.27 & 3.704 & 4 & 7.407 \\ & 3 & 0.20 & 5.000 & 9 & 15.00 \\ & 4 & 0.15 & 6.667 & 16 & 26.67\\ & 5 & 0.12 & 8.333 & 25 & 41.67\\ \hline \sum & 15 & & 25.831 & 55 & 92.87\\ \hline \end{array} $$

And

$$ \begin{array}{rcrcc} 5\,a_{0} & + & 15\,a_{1} & = & 25.83\\ 15\,a_{0} & + & 55\,a_{1} & = & 92.87 \end{array} $$

And the solution is $a_0 = 0.5540 \quad a_1=1.5474$. As

$$ a_0 =\frac{1}{Q_0}, \quad a_1=K \quad \Longrightarrow Q_0 =\frac{1}{a_0}\approx 1.8 \quad K = a_1\approx 1.5 $$

The approximate curve is

$$Q\left(t\right)=\frac{1.8}{1+1.8\times 1.5\,t},$$