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Fermat's theorem (stationary points) of higher dimensions

ianchenmu

Member
Feb 3, 2013
74

chisigma

Well-known member
Feb 13, 2012
1,704
Look at this page and the Proof part,

Fermat's theorem (stationary points) - Wikipedia, the free encyclopedia

How to change the proof 2 into a proof of higher dimensions or can you give a proof of Fermat's theorem of higher dimensions?
In case of higher dimension You cal substitute the derivative $\displaystyle f^{\ '}(x)= \frac{d f(x)}{d x}$ with the gradient, defined as...


$\displaystyle \nabla f(x_{1},\ x_{2}, ...,\ x_{n}) = \frac{ \partial f}{\partial x_{1}}\ \overrightarrow {e}_{1} + \frac{ \partial f}{\partial x_{2}}\ \overrightarrow {e}_{2} + ... + \frac{ \partial f}{\partial x_{n}}\ \overrightarrow {e}_{n}$ (1)

Kind regards

$\chi$ $\sigma$
 
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ianchenmu

Member
Feb 3, 2013
74
In case of hogher dimension You cal substitute the derivative $\displaystyle f^{\ '}(x)= \frac{d f(x)}{d x}$ with the gradient, defined as...


$\displaystyle \nabla f(x_{1},\ x_{2}, ...,\ x_{n}) = \frac{ \partial f}{\partial x_{1}}\ \overrightarrow {e}_{1} + \frac{ \partial f}{\partial x_{2}}\ \overrightarrow {e}_{2} + ... + \frac{ \partial f}{\partial x_{n}}\ \overrightarrow {e}_{n}$ (1)

Kind regards

$\chi$ $\sigma$
Thanks. But how about $f:\mathbb{R}^n\rightarrow \mathbb{R}$
 

ianchenmu

Member
Feb 3, 2013
74
In case of higher dimension You cal substitute the derivative $\displaystyle f^{\ '}(x)= \frac{d f(x)}{d x}$ with the gradient, defined as...


$\displaystyle \nabla f(x_{1},\ x_{2}, ...,\ x_{n}) = \frac{ \partial f}{\partial x_{1}}\ \overrightarrow {e}_{1} + \frac{ \partial f}{\partial x_{2}}\ \overrightarrow {e}_{2} + ... + \frac{ \partial f}{\partial x_{n}}\ \overrightarrow {e}_{n}$ (1)

Kind regards

$\chi$ $\sigma$
But what's then? what $\frac{ \partial f}{\partial x_{1}},\frac{ \partial f}{\partial x_{2}},...,\frac{ \partial f}{\partial x_{n}}$ equal to?

(I mean, is that $\frac{ \partial f}{\partial x_{1}}=\frac{ \partial f}{\partial a_{1}}=0$,$\frac{ \partial f}{\partial x_{n}}=\frac{ \partial f}{\partial a_{2}}=0$,...,$\frac{ \partial f}{\partial x_{n}}=\frac{ \partial f}{\partial a_{n}}=0$, (where $a$ is a local maximum) ,why?)
 
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chisigma

Well-known member
Feb 13, 2012
1,704
But what's then? what $\frac{ \partial f}{\partial x_{1}},\frac{ \partial f}{\partial x_{2}},...,\frac{ \partial f}{\partial x_{n}}$ equal to?
If You write as $\displaystyle \overrightarrow {x}$ a generic vector of dimension n and as $\displaystyle \overrightarrow {0}$ the nul vector of dimension n, then $\displaystyle \overrightarrow {x}_{0}$ is a relative maximum or minimum only if is...

$\displaystyle \nabla f (\overrightarrow {x}_{0}) = \overrightarrow {0}$ (1)

Kind regards

$\chi$ $\sigma$
 

ianchenmu

Member
Feb 3, 2013
74
If You write as $\displaystyle \overrightarrow {x}$ a generic vector of dimension n and as $\displaystyle \overrightarrow {0}$ the nul vector of dimension n, then $\displaystyle \overrightarrow {x}_{0}$ is a relative maximum or minimum only if is...

$\displaystyle \nabla f (\overrightarrow {x}_{0}) = \overrightarrow {0}$ (1)

Kind regards

$\chi$ $\sigma$
But this is what I need to prove. To clarify, I need to prove this:
Let $E\subset \mathbb{R}^n$ and $f:E\rightarrow\mathbb{R}$ be a continuous function. Prove that if $a$ is a local maximum point for $f$, then either $f$ is differentiable at $x=a$ with $Df(a)=0$ or $f$ is not differentiable at $a$.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
But this is what I need to prove. To clarify, I need to prove this:
Let $E\subset \mathbb{R}^n$ and $f:E\rightarrow\mathbb{R}$ be a continuous function. Prove that if $a$ is a local maximum point for $f$, then either $f$ is differentiable at $x=a$ with $Df(a)=0$ or $f$ is not differentiable at $a$.
When I saw this problem, I thought that it would be easy to tackle it by reducing it to the one-dimensional case. In fact, let $b\in\mathbb{R}^n$. If $f$ has a local maximum at $a$, then the function $g:\mathbb{R}\to\mathbb{R}$ defined by $g(t) = f(a+tb)$ must have a maximum at $t=0$. What we need to do here is to choose the vector $b$ suitably. Then, provided that $f$ is differentiable at $a$, we can use the fact that $g'(0) = 0$ to deduce that $Df(a) = 0.$

But that turns out to be a bit tricky. The reason is that if $a\in\mathbb{R}^n$ then the derivative $Df(a)$ belongs to the dual space $\mathbb{R}^n$. In other words, if you think of $a$ as a column vector, then $Df(a)$ will be a row vector. So suppose we take $b=(Df(a))^{\text{T}}$, the transpose of $Df(a)$. According to the higher-dimensional chain rule, $g'(0) = Df(a)\circ b = Df(a)\circ (Df(a))^{\text{T}} = \bigl|Df(a)\bigr|^2.$ But since $0$ is a local maximum for $g$ it follows that $g'(0) = 0$ and hence $Df(a) = 0.$

If you really want to get to grips with duality, and the reasons for distinguishing between row vectors and column vectors, then you will have to come to terms with covariance and contravariance.