I am working to derive a system of linear equations for use in a matrix. One of these equations is nonlinear, but can be linearly approximated to arrive at a solution. I have been provided with a linear approximation, however I suspect an error might have been made.

The nonlinear equation in question is

$$ \frac{\mathbf{h_0}}{1-\chi \mathbf{\epsilon_0}}=\frac{r_1\mathbf{h_3}}{1-\chi \mathbf{\epsilon_3}} $$

where $\chi$ is a constant, $\mathbf{h_0}$, $\mathbf{\epsilon_0}$, $\mathbf{h_3}$, $\mathbf{\epsilon_3}$, and $r_1$ are variables to be solved for.

The given linear approximation is $$ \mathbf{\delta h_0}-\chi \mathbf{\epsilon_3}=\mathbf{\delta h_3}-\chi \mathbf{\epsilon_3}+\delta r_1 $$

where $\mathbf{h_0}=\mathbf{\hat{h_0}}(1+\mathbf{\delta h_0})$, $\mathbf{h_3}=\mathbf{\hat{h_3}}(1+\mathbf{\delta h_3})$, and $\mathbf{r_1}=\mathbf{\hat{r_1}}(1+\mathbf{\delta r_1})$.

By my calculations using the binomial approximation, I obtain a different solution (signs in front of $\chi$). $$ \mathbf{\delta h_0}+\chi \mathbf{\epsilon_3}=\mathbf{\delta h_3}+\chi \mathbf{\epsilon_3}+\delta r_1 $$

This is because the original nonlinear equation has a binomial approximation of $$ \mathbf{h_0}(1+\chi \mathbf{\epsilon_0})=r_1\mathbf{h_3}(1+\chi \mathbf{\epsilon_3}) $$ Which by assuming nominal values and solving for delta terms produces $$ \mathbf{\hat{h_0}}(1+\delta\mathbf{h_0})(1+\chi \mathbf{\epsilon_0})= \mathbf{\hat{r_1}}(1+\delta\mathbf{r_1})\mathbf{\hat{h_3}}(1+\delta\mathbf{h_3})(1+\chi \mathbf{\epsilon_3}) $$ The given term $\mathbf{\hat{r_1}}=\frac{\mathbf{\hat{h_0}}}{\mathbf{\hat{h_3}}}$ cancelling the leading constants and the rest of the terms expand. Because the $\delta$ and $\epsilon$ terms are small in magnitude, their square and cube components can be ignored, leaving my stated approximation.

Why do these solutions differ? Is this considered an error, or is there a mistake in my derivation?

  • $\begingroup$ if $\epsilon_0$ is a vector then any expression of the form $a/\epsilon_0$ doesn't make sense because vectors cannot be divided $\endgroup$
    – Masacroso
    Apr 23 '20 at 15:09
  • $\begingroup$ The division is element-wise, but I will edit the representation for purposes of this question. @Masacroso $\endgroup$
    – Groger
    Apr 23 '20 at 15:12

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