- Thread starter
- #1

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?

- Thread starter ianchenmu
- Start date

- Thread starter
- #1

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?

- Feb 13, 2012

- 1,704

In case of higher dimension You cal substitute the derivative $\displaystyle f^{\ '}(x)= \frac{d f(x)}{d x}$ with the

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?

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

Kind regards

$\chi$ $\sigma$

Last edited:

- Thread starter
- #3

Thanks. But how about $f:\mathbb{R}^n\rightarrow \mathbb{R}$In case of hogher dimension You cal substitute the derivative $\displaystyle f^{\ '}(x)= \frac{d f(x)}{d x}$ with thegradient, 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

$\chi$ $\sigma$

- Thread starter
- #4

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?In case of higher dimension You cal substitute the derivative $\displaystyle f^{\ '}(x)= \frac{d f(x)}{d x}$ with thegradient, 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

$\chi$ $\sigma$

(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?)

Last edited:

- Feb 13, 2012

- 1,704

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...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?

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

Kind regards

$\chi$ $\sigma$

- Thread starter
- #6

But this is what I need to prove. To clarify, I need to prove this: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

$\chi$ $\sigma$

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 $a$.

- Moderator
- #7

- Feb 7, 2012

- 2,793

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 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 $a$.

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

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.