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Let $$y(x)=a\cdot\frac{x^c}{b^c}$$

Assume that we have a number of data samples $(x_i,y_i)$.

The objective is to find $a, b,$ and $c$ that minimizes the square error function, i.e.,

$$J(a,b,c) = \sum_{\forall i} (y(x_i)-y_i)^2$$

As always, to minimize, we set the derivatives to 0, i.e.,

$$\frac{\partial J}{\partial a}=0 \iff 2\sum_{\forall i} (a\cdot\frac{x_i^c}{b^c}-y_i)\cdot \frac{x_i^c}{b^c}=0 \iff \sum_{\forall i} (a\cdot\frac{x_i^c}{b^c}-y_i)\cdot x_i^c=0$$

$$\frac{\partial J}{\partial b}=0 \iff 2\sum_{\forall i} (a\cdot\frac{x_i^c}{b^c}-y_i)\cdot c \cdot a \cdot x_i^c \cdot b^{-c-1}=0 \iff \sum_{\forall i} (a\cdot\frac{x_i^c}{b^c}-y_i)\cdot x_i^c=0$$

$$\frac{\partial J}{\partial c}=0 \iff 2\sum_{\forall i} (a\cdot\frac{x_i^c}{b^c}-y_i)\cdot a\cdot\frac{x_i^c}{b^c} \cdot \log\frac{x_i}{b}=0 \iff \sum_{\forall i} (a\cdot\frac{x_i^c}{b^c}-y_i)\cdot x_i^c \cdot \log\frac{x_i}{b}=0$$

As you may have noticed, I'm supposed to get a 3-by-3 system of equations, but due to cancellation, I got 2 equations and 3 unknowns (the 1st and 2nd equations are identical after simplification).

Does this mean the existence of many solutions? Or did I do some mistake in deriving the equations? Please note that all the parameters are assumed to be positive reals and greater than zero.

Please help.

user8420488483439
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2 Answers2

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The problem is that there are infinitely many least-square solutions. This can for example be seen by noticing that $J(k^ca,kb,c) = J(a,b,c)$ for all $k\not= 0$. We can only constraint the parameter combinations $(a/b^c,c)$ so your fitting-function has one parameter that cannot be constrained. You can get around this problem by instead trying to fit to the functional form $f(x) = a x^c$ instead. With this form you will get a well defined equation-system.

You can also, if wanted, reduce the problem to a standard linear least-square problem by taking $\log$'s which allows you to use the standard well-known formulas for this case directly. If we define $\tilde{y}_i = \log(y_i)$ and $\tilde{x}_i = \log(x_i)$ (assuming for simplicity that $y_i > 0$; otherwise we need to work with the absolute value) and try to fit these points to $y(\tilde{x}) = \alpha \tilde{x}+\beta$ then the least-square solution for $(\alpha,\beta)$ are related to the least-square solution for $(a,c)$ via $c = \alpha$ and $a = e^{\beta}$.

Winther
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  1. The problem statement has redundant parameters; eliminate one parameter. Consider the form $$ y(x) = \alpha x^{c} $$ which has two fit parameters, $\alpha$ and $c$. The constant term is $$ \alpha = a b^{-c}. $$

  2. Use least squares to constrain the constant parameter which appears in a linear form. Using the new trial function from part one, and $m$ measurements $\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}$, the least squares solution is defined as $$ \left[ \begin{array}{c} \alpha \\ c \end{array} \right]_{LS} = \left\{ \left[ \begin{array}{c} \alpha \\ c \end{array} \right] \in \mathbb{R}^{2} \colon r^{2} (\alpha,c) = \sum_{k=1}^{m} \left( y_{k} - \alpha x_{k}^{c} \right)^{2} \text{ is minimized} \right\} $$ The minimization criterion $$ \frac{\partial}{\partial \alpha} r^{2} = -2 \sum_{k=1}^{m} \left( y_{k} - a x_{k}^{c} \right) x_{k}^{c} = 0 $$ provides the constraint $$ \hat{\alpha}(c) = \frac{\sum_{k=1}^{m} y_{k} x_{k}^{c}} {\sum_{k=1}^{m} x_{k}^{2c}}. $$ The two parameter nonlinear least squares problem is now reduced to a one dimensional minimization problem $$ r^{2}(\hat{\alpha}(c),c) = r^{2}(c) = \sum_{k=1}^{m} \left( y_{k} - \frac{\sum_{k=1}^{m} y_{k} x_{k}^{c}} {\sum_{k=1}^{m} x_{k}^{2c}} x_{k}^{c} \right)^{2}. $$ This solution will be unique.

dantopa
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