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I have the following equation of my system:

$$\frac{\alpha}{D^2} = (\beta - \omega^2)^2 + \gamma\omega^2$$

where $\omega$ and $D$ are determined in a single experimental reading. and $\alpha\, ,\beta$ and $\gamma$ are unknowns which I would like to determine.

I try to linearise the system by performing the following substitutions:

$$\alpha \equiv x\\ \beta \equiv \sqrt{y}\\ \gamma \equiv z + 2\sqrt{y}$$

This makes my system of equations have a linearised form:

$$x - D^2y - (D\omega)^2z = (D\omega^2)^2$$

By performing $n$ experimental readings, I am able to retrieve $n$ sets of values of $D$ and $\omega$. Thus, I get a matrix equation of the form:

$$\textbf{A}_{n\times 3} \textbf{x}_{3\times 1} = \textbf{c}_{n\times1}\, \hspace{5mm},\text{with}\, \textbf{x} = \begin{bmatrix}x\\y\\z\end{bmatrix}$$

I am able to find my vector $\textbf{x}$ for the lowest squared error, by taking a pseudo inverse of the matrix:

$$\textbf{x} = (\textbf{A}^T\textbf{A})^{-1}\textbf{A}^T\textbf{c}$$

However, I have arrived at incorrect values for my unknowns $\alpha\, , \beta$, and $\gamma$ with this approach. I believe that the experiment was done correctly, and the readings are trustworthy.

Is there anything wrong in coupling the unknowns together in the definition $\gamma \equiv z + 2\sqrt{y}$? Does this coupling of unknowns violate anything in linear algebra's system of equations?

Below is an array of data set which I used

$D$ array: [5.0947e-09, 6.1859e-09, 7.3539000000000004e-09, 8.7439e-09, 1.0133e-08, 1.1583e-08, 1.3895e-08, 1.6072e-08, 1.4923e-08, 1.2402e-08, 9.847e-09, 8.451e-09, 7.188e-09, 6.2971000000000005e-09]

$\omega$ array: [219911.4857512855, 251327.41228718346, 282743.3388230814, 314159.2653589793, 345575.19189487724, 376991.1184307752, 408407.04496667313, 439822.971502571, 785398.1633974483, 816814.0899333463, 848230.0164692441, 879645.943005142, 911061.86954104, 942477.7960769379]

EDIT:

  1. Typo, $\gamma \equiv z + \sqrt{y}$ is corrected to $\gamma \equiv z + 2\sqrt{y}$. Missed out the factor of 2.
  2. Included data set used
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Expanding your original equation gives:

$\alpha = D^2\beta^2 - 2D^2\beta\omega^2 + D^2\omega^4 + D^2\gamma\omega^2$

$\Rightarrow \alpha - D^2\beta^2 - D^2\omega^2(\gamma - 2\beta) = D^2\omega^4$

To linearise this I think you need to have

$\alpha = x$

$\beta^2 = y$

$\gamma - 2\beta = z$

so $\gamma = z + 2\sqrt{y}$ instead of $\gamma = z + \sqrt{y}$

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  • $\begingroup$ Thanks for the note, yes I included that in my workings, it was a mistake that I left that factor of 2 out when I was typing out my question $\endgroup$ – Tian Jul 13 '18 at 8:43
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The model $$\frac{\alpha}{D^2} = (\beta - \omega^2)^2 + \gamma\omega^2$$ is already almost linear with respect to the parameters.

Expand the rhs and divide everything by $\alpha$ to get $$\frac{1}{D^2} =\frac{\beta^2}\alpha+\frac{\gamma-2\beta}\alpha \omega^2+\frac{1}\alpha \omega^4$$ $$\frac{1}{D^2} =a+b\,\omega^2+c\, \omega^4$$ Now, use the standard linear regression with the method of your choice. When you will have $a,b,c$, go back to $\alpha,\beta,\gamma$.

But, take care : if $D$ is the measured value, you will need, in a second step, to perform the nonlinear regression based on the model $D=f(\omega)$; but the estimates from the first step will make life quite easy.

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  • $\begingroup$ The first equation for D is already the equation of the model. I am keen in finding the unknowns in the model $\endgroup$ – Tian Jul 13 '18 at 8:47
  • $\begingroup$ @Tian. Can you tell me what is measured ? $D$, $D^2$,$\frac 1{D^2}$ ? This is crucial for a proper fit of experimental data. $\endgroup$ – Claude Leibovici Jul 13 '18 at 8:50
  • $\begingroup$ $D$ is measured. $\endgroup$ – Tian Jul 13 '18 at 8:54
  • $\begingroup$ @Tian. Then, you need the second step ! Could you post some data points ? I should show you how different could be the predicted results. $\endgroup$ – Claude Leibovici Jul 13 '18 at 9:12
  • $\begingroup$ I have included the data set in my question. Thank you for your help $\endgroup$ – Tian Jul 13 '18 at 9:19

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