Linear or non-linear regression Given an equation, say, $y^{1/n} = x^{1/n} + z^{1/n}$ and a bunch of 3-dimensional sample points, what is the best way to find the optimal value for $n$ that best fits the sample points? I suppose least-squares can be a metric, but is the regression non-linear? Can the equation somehow be made linear?
 A: You have $k$ data points $(x_i,z_i,y_i)$ and the model you want to fit is in fact
$$y=\left(x^{\frac{1}{n}}+z^{\frac{1}{n}}\right)^n$$ since what is measured is $y$ and not any of its possible transform.
So, what you want to minimize is
$$SSQ(n)=\sum_{i=1}^k \Big[\left(x_i^{\frac{1}{n}}+z_i^{\frac{1}{n}}\right)^n-y_i \Big]^2$$ which is extremely nonlinear with respect to the parameter. This means that you need at least a reasonable guess.
What I should do is to try a few values of $n$ until you see more or less a minimum. When this is done, you are ready for the nonlinear regression.
Suppose that we have the following data set
$$\left(
\begin{array}{ccc}
x & z & y\\
 12 & 11 & 100 \\
 13 & 13 & 110 \\
 14 & 15 & 130 \\
 15 & 17 & 140 \\
 16 & 19 & 150 \\
 17 & 21 & 170 \\
 18 & 23 & 180 \\
 19 & 25 & 190 \\
 20 & 27 & 210
\end{array}
\right)$$
Trying with a fixed step size (we could do better), you will have
$$\left(
\begin{array}{cc}
n & SSQ(n) \\
 1.0 & 132705 \\
 1.5 & 102443 \\
 2.0 & 66281 \\
 2.5 & 28528 \\
 3.0 & 1956 \\
 3.5 & 18027 \\
 4.0 & 148044
\end{array}
\right)$$ So, $n=3$ seems to be quite good.
Now, the nonlinear regression would give $(R^2=0.99964)$
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a & 3.1391 & 0.0104 & \{3.1146,3.1636\} \\
\end{array}$$ corresponding to $SSQ=80.2602$.
The predicted values would be
$$\{101.2,114.5,127.7,140.8,153.8,166.7,179.7,192.6,205.5\}$$
If you do not access a nonlinear regression software, you could continue zooming more and more around the minimum. There is also an algebraic way to do the job; if you are concerned, let me know.
