I am trying to solve the following equation:

$$\omega^{1/3} = p_l \omega_l^{1/3} + (1-p_l)\omega_{\omega}^{1/3} \tag{1}$$

$\omega$, $\omega_l$, and $\omega_w$ are complex numbers and $p_l$ must be real (it represents a percentage).

$\omega$ is a vertical array of 200 values and $\omega_l$ is a constant single number, as well as $\omega_w$. Therefore, I should be able to obtain 200 different values for $p_l$.

I have been told that I could use the least-square fitting to solve this and obtain my 200 values for $p_l$, but I am not certain as to how to.

What I have done so far:

  • I have previously simply tried to manipulate the equation so as to isolate $p_l$ but all the results I obtained were complex numbers, which cannot happen in this case.

  • I have started by separating the real and complex parts of $(1)$, and obtained the values for $p_l$ from these two equations: $p_l = \frac{real(\omega) - real(\omega_w)}{real(\omega_l) - real(\omega_w)}$ and $p_l = \frac{imag(\omega) - imag(\omega_w)}{imag(\omega_l) - imag(\omega_w)}$, but I do not know if there is any information to obtain from this, and if there is I do not know how to obtain it.

Can I use the least-square fitting to solve this equation and obtain real values for $p_l$?

  • $\begingroup$ Couldn't you just solve for $p_l$ and substitute in the final formula 200 times? That being said, if $p_l$ is real, then $\frac{\omega^{1/3}-\omega_\omega^{1/3}}{\omega_l^{1/3}-\omega_\omega^{1/3}}$ must be real as well so not all $\omega$ values will do. Maybe some more context about what you are trying to model here? $\endgroup$ Commented Sep 28, 2023 at 16:14
  • $\begingroup$ Thanks for your help @DinosaurEgg but as I stated above, I have done what you have stated but unfortunately the equation only results in complex solutions (which makes sense mathematically), but I should also be able to obtain the real number, as it could also be fitted into that equation. I am modelling permittivity , $\omega$, over water content $\p_l$. $\endgroup$ Commented Sep 29, 2023 at 10:07
  • $\begingroup$ So there is some noise in the data that makes $p_l$ acquire an imaginary part? If so, you need to understand how sensitive the real part is to that noise somehow. $\endgroup$ Commented Sep 29, 2023 at 16:25

1 Answer 1


There is a variety of ways one can obtain an estimate for the real part of $p_l$. I will mention one here that is in the spirit of least squares and does not require any other inputs but the measurements. Assuming that the quantity $Q=||\omega^{1/3}-\omega^{1/3}_\omega+p_l(\omega^{1/3}_l-\omega^{1/3}_\omega)||$ is normally distributed with probability distribution $\mathcal{N}(0,\sigma^2)$, we can find the maximum likehlihood estimate for this quantity by minimizing the quadratic loss function

$$\mathcal{L}=\sum_{k}Q_k^2=\sum_k (X_k-a-cp_l)^2+(Y_k-b-dp_l)^2$$

where the sum is performed over the different observations of the experiment and we have denoted the observations $\omega_k^{1/3}=X_k+iY_k$ and $\omega_\omega^{1/3}=a+bi~~,~ \omega_l^{1/3}-\omega_\omega^{1/3}=c+di$. The function is quadratic and hence has a unique minimum. Finding the minimum yields the estimate

$$\frac{d\mathcal{L}}{dp_l}=0\Rightarrow\hat{p_l}=\frac{c\langle X \rangle+d \langle Y\rangle}{c^2+d^2}-\frac{ac+bd}{c^2+d^2}$$

where $\langle R\rangle=\sum_k {R_k}/n.$ The estimate is by definition real.


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