# proof of calculus theorem used in the Lagrange method

Let $f(\mathbf{x})$ and $g_{i}(\mathbf{x}),i=0,{\ldots},m$ be differentiable   scalar functions; $\mathbf{x}\in R^{n}$.

We will find local extremes of the function $f(\mathbf{x})$ where $\nabla f=0$. This can be proved by contradiction   :

 $\nabla f\neq 0$
 $\Leftrightarrow\exists\epsilon_{0}>0,\forall\epsilon;0<\epsilon<\epsilon_{0}:f% (\mathbf{x}-\epsilon\nabla f)

but then $f(\mathbf{x})$ is not a local extreme.

Now we put up some conditions, such that we should find the $\mathbf{x}\in S\subset R^{n}$ that gives a local extreme of $f$. Let $S=\bigcap_{i=1}^{m}S_{i}$, and let $S_{i}$ be defined so that $g_{i}(\mathbf{x})=0\forall\mathbf{x}\in S_{i}$.

Any vector $\mathbf{x}\in R^{n}$ can have one component   perpendicular   to the subset $S_{i}$ (for visualization, think $n=3$ and let $S_{i}$ be a flat surface). $\nabla g_{i}$ will be perpendicular to $S_{i}$, because:

 $\exists\epsilon_{0}>0,\forall\epsilon;0<\epsilon<\epsilon_{0}:g_{i}(\mathbf{x}% -\epsilon\nabla g_{i})

But $g_{i}(\mathbf{x})=0$, so any vector $\mathbf{x}+\epsilon\nabla g_{i}$ must be outside $S_{i}$, and also outside $S$. (todo: I have proved that there might exist a component perpendicular to each subset $S_{i}$, but not that there exists only one; this should be done)

By the argument  above, $\nabla f$ must be zero - but now we can ignore all components of $\nabla f$ perpendicular to $S$. (todo: this should be expressed more formally and proved)

So we will have a local extreme within $S_{i}$ if there exists a $\lambda_{i}$ such that

 $\nabla f=\lambda_{i}\nabla g_{i}$

We will have local extreme(s) within $S$ where there exists a set $\lambda_{i},i=1,{\ldots},m$ such that

 $\nabla f=\sum\lambda_{i}\nabla g_{i}$
Title proof of calculus theorem used in the Lagrange method ProofOfCalculusTheoremUsedInTheLagrangeMethod 2013-03-22 13:29:51 2013-03-22 13:29:51 mathcam (2727) mathcam (2727) 6 mathcam (2727) Proof msc 15A18 msc 15A42