Waldronjw

2022-07-15

Let $y=f\left(x\right)=\left({x}_{1}^{2}+2{x}_{2},{x}_{1}{x}_{2}-3{x}_{1}\right)$
Is the linear approximation just $f\left(y\right)=f\left(x\right)+A\left(y-x\right)$ whenever y is approximately near $x$?

A differentiable function $f:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$ can be approximated by it's derivative in the sense that given a point, say, $\left(1,1\right)$, you can write:
$f\left(1+\delta x,1+\delta y\right)\approx f\left(1,1\right)+Df\left(1,1\right)\left(\delta x,\delta y\right)$
where $Df\left(1,1\right)$ is a matrix applied to the vector $\left(\delta x,\delta y\right)$. So, yes, if $y$ is near $x$ (as ($\left(1+\delta x,1+\delta y\right)$) is near ($\left(1,1\right)$)), then you can approximate the difference $f\left(y\right)-f\left(x\right)$ by the derivative evaluated at $x$, applied to the difference $y-x$. That is:
$f\left(y\right)\approx f\left(x\right)+Df\left(x\right)\left(y-x\right)$
Compare this to the one dimensional case, where the tangent line to the graph of a differentiable function locally approximates the graph, in the sense that:
$f\left(y\right)\approx f\left(x\right)+{f}^{\prime }\left(x\right)\left(y-x\right)$

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