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I'm trying to formulate a regression problem such that $y=ax^b$. Previously, I formulated the $y=ax+b$ like $y=Ac+e$ where $c=
beginbmatrix
a\
b
endbmatrix$ and $A=
beginbmatrix
x_1 & 1\
x_2 & 1\
. & \
. & \
x_n & 1
endbmatrix$

Can I only change the A matrix such that $A=
beginbmatrix
x_1^b & 1\
x_2^b & 1\
. & \
. & \
x_n^b & 1
endbmatrix$ and change c as $c=
beginbmatrix
a\
0
endbmatrix$ and write the same formula $y=Ac+e$

Am I correct? Any help would be appreciated. Many thanks

No.
You need all the powers in the columns of A.

Write the problem as minimizing
$D=sum_k=1^m (y_k-sum_j=0^n a_jx_k^j)^2$.

Set $partial D/partial a_j =0$
for each $j$ and see what you get.

You have three alternatives to solve this problem.

1. Alternative: As Jens Schwaiger proposed you can rewrite the equation as $ln y_i = ln a + b ln x_i+ tildevarepsilon_i$. If we introduce the coefficient $tildea=ln a$ and the transformed outputs $tildey_i=ln y_i$ then it is possible to perform a standard linear regression. the coefficients $boldsymbolw=[tildea,b]^T=[ln a, b]^T$ can be estimated by the least squares estimate

$$hatboldsymbolw=left[boldsymbolPhi^TboldsymbolPhi right]^-1boldsymbolPhi^Ttildeboldsymboly,qquad (*)$$

in which $tildeboldsymboly=left[ln y_1, ln y_2, ldots, ln y_nright]^T$ is the transformed output vector and

$$boldsymbolPhi = beginbmatrix1 & ln x_1\1 & ln x_2 \ vdots & vdots \1 & ln x_n endbmatrix.$$

2. Alternative: We have $y_i = ax_i^b+varepsilon_i=aexp(bln x_i)+varepsilon_i$. We can now expand the exponential function by a taylor series.

$$y_i=aleft[1+bln x_i + b^2/2!left[ln x_iright]^2+b^3/3![ln x_i]^3+cdots right]+varepsilon_i $$

If we truncate the series to the $m^textth$ power then we can approximate $y_i$ as

$$y_i approx left[a+abln x_i + ab^2/2!left[ln x_iright]^2+ab^3/3![ln x_i]^3+cdots +ab^m/m!left[ln x_i right]^mright]+varepsilon_i.$$

By introducing the coefficients $w_l = ab^l/l!$ we can rewrite the previous equation as

$$y_i approx left[w_0+w_1ln x_i + w_2left[ln x_iright]^2+w_3[ln x_i]^3+cdots +w_mleft[ln x_i right]^mright]+varepsilon_i.$$

The coefficients are given by equation $(*)$ but now $tildeboldsymboly=boldsymboly=left[y_1, y_2, ldots,y_n right]^T$ and

$$boldsymbolPhi = beginbmatrix1 & ln x_1 & [ln x_1]^2 & cdots & [ln x_1]^m\1 & ln x_2 & [ln x_2]^2 & cdots & [ln x_2]^m\ vdots & vdots & vdots & ddots & vdots \1 & ln x_n & [ln x_n]^2 & cdots & [ln x_n]^m\endbmatrix.$$

After having obtained the coefficients $boldsymbolw=[a, ab, ab^2, ..., ab^m]$ you can determine $a=w_0$ and $b=w_1/w_0$ and so forth.

3. Alternative: Full nonlinear least squares (proposed by Marty Cohen), which does not have a closed form solution. Here we use the objective function $E(boldsymbolw=[a,b]^T)$ which is

$$E(boldsymbolw)=sum_i=1^n[y_i-ax_i^b]^2.$$

The partial derivatives are given by

$$dfracpartial Epartial a = sum_i=1^n2[y_i-ax_i^b](-x_i^b)$$
$$dfracpartial Epartial b = sum_i=1^n2[y_i-ax_i^b](-ax_i^bln x_i).$$

After setting these partial derivatives equal to zero you will have to solve a nonlinear equation in the coefficients $a$ and $b$. You can try to numerically solve this equation by using Newton-Raphson.