I have a non-linear term in the form of $$x(x-y)$$ I need to replace it with a linear term in the form of $$c_1x-c_2y$$ I have seen the following linear approximation in literature but I need to know the analysis that resulted in the approximation. I appreciate your help!
$$x(x-y)=(x-y)$$
Hint: x and y are always close to 1. For example: $$0.95<x<1.05$$ $$0.95<y<1.05$$
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Making the problem more general, you want to find, for the function$x(x-y)$ the best linear approximation that is to say $$\alpha + \beta x +\gamma y$$ for $a \leq x \leq b$ and $c \leq y \leq d$.
Whet you can do is to minimize with respect to $(\alpha,\beta,\gamma)$ the norm $$\Phi=\int_a^b\int_c^d \Big[x(x-y)-(\alpha + \beta x +\gamma y]\Big]^2\, dy\,dx$$
This is equivalent to a multilinear regression based on an infinite number of data points.
Setting the three partial derivatives equal to zero, you just need to solve three linear equations and the result is $$\alpha=\frac{1}{12} \left(-2 a^2+a (3 (c+d)-8 b)+b (3 (c+d)-2 b)\right)$$ $$\beta=(a+b)-\frac 12(c+d)\qquad \qquad \gamma=-\frac 12(a+b)$$ For your example, this would give $$\alpha=\frac{1}{1200}\qquad \qquad \beta=1\qquad \qquad \gamma=-1$$
One-dimension Taylor approximation of degree one of a function $f$ at $x_0$ is given by $$f(x_0)+f'(x_0)(x-x_0).$$ Two-dimensional is quite the same: $$f(x_0,y_0)+\langle\nabla f(x_0,y_0),(x-x_0,y-y_0)\rangle,$$ which gives $x-y$ in our case.