(Possibly light question) Data linearization in curve fitting

Question

I was studying curve fitting in numerical methods and met an approximation curve of the form $y=\frac{x}{A+Bx}$. We were supposed to linearize this curve before applying the least square method, but two linearizations:

1. $Y=1/y$ and $X=1/x$, and
2. $Y=x/y$ and $X=x$

gave two different results for $A$ and $B$ respectively. I kept making sure that I put the right values in the normal equations, but is it just possible that two different linearization methods give different constants? I forget about it, but I remember an equation that I linearized in two different ways produced the same constants. Or is it just me?

Note: No.1 is obtained by taking reciprocal of both sides. No.2 is obtained by switching y and the denominator on the right hand side.

Answer 1

What AlexanderJ93 answered is perfectly correct if there is no error in the data.

If there are, just as you observed, two diffferent linearizations will provide different answers and this is normal since the considered sum of squares are diiferent $$SSQ_1=\sum_{i=1}^n \left(B+\frac A {x_i}-\frac 1{y_i} \right)^2$$ $$SSQ_2=\sum_{i=1}^n \left(A+B {x_i}-\frac {x_i}{y_i} \right)^2$$

Let us take as an example the following data which contain very small errors $$\left( \begin{array}{cc} x & y \\ 1 & 28 \\ 2 & 34 \\ 3 & 36 \\ 4 & 38 \\ 5 & 39 \\ 6 & 39 \end{array} \right)$$

The first regression would give $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ A & 0.012299 & 0.000374 & \{0.011109,0.013490\} \\ B & 0.023395 & 0.000186 & \{0.022801,0.023988\} \\ \end{array}$$ while the second would give $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ A & 0.012124 & 0.001026 & \{0.008858,0.015390\} \\ B & 0.023450 & 0.000266 & \{0.022611,0.024288\} \\ \end{array}$$

The corefficients are similar but not identical.

In any manner, after linearization, you need to go to nonlinear regression since what is measured is $y$ and not $\frac 1 y$ or $\frac x y$.

Using any of the above estimates, using nonlinear regression, you would get $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ A & 0.012281 & 0.000563 & \{0.010490,0.014071\} \\ B & 0.023400 & 0.000209 & \{0.022736,0.024064\} \\ \end{array}$$

Answer 2

These two linearizations work for any $A,B$. The difference between them is that they swap the roles of $A$ and $B$. The first one evaluates to $Y = AX+B$, while the second evaluates to $Y=BX+A$. Whatever the values for $A$ and $B$ are, they are the same between the equations, but in the first one $A$ is the slope and $B$ is the intercept, while in the second one $B$ is the slope and $A$ is the intercept.