# Does this Linear Approximation Contain an Error?

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?

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