how can we proof this :

direction of steepest descent of a differentiable function F of M variables is the vector

($\displaystyle -cdF/dw_1,...,-cdF/dw_M$)

where c is a positive constant of proportionality.

thanks

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- Aug 8th 2006, 05:40 AMAMRsteepest descent
how can we proof this :

direction of steepest descent of a differentiable function F of M variables is the vector

($\displaystyle -cdF/dw_1,...,-cdF/dw_M$)

where c is a positive constant of proportionality.

thanks - Aug 8th 2006, 07:41 AMThePerfectHackerQuote:

Originally Posted by**AMR**

*gradient*of a diffrenciable function.

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I can prove it for $\displaystyle f(x,y)$.

Let $\displaystyle \nabla f_0\not = 0$ at a point.

Let, $\displaystyle \bold{u}$ be any unit vector.

Then, directional derivative of,

$\displaystyle D_u f=\nabla f_0 \cdot \bold{u} = ||\nabla f_0||\cdot ||\bold{u}||\cos \theta$

The largest possible value of,

$\displaystyle \cos \theta =1$

Thus, maximum is,

$\displaystyle ||\nabla f_0||(1)$

Thus, it is in the in the direction of the gradient. Thus, it is a multiple of the vector, hence the constant of proportionality.

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I am thinking about several variables.

I guess the proof is based on finding the critical points. Since the critical points are determined by partial derivatives they appear over here. I will think more about it. - Aug 8th 2006, 10:15 AMThePerfectHacker
You have a function,

$\displaystyle f:\mathbb{R}^n\to \mathbb{R}$.

At point, $\displaystyle P(x_1,x_2,...,x_n)$

You have, $\displaystyle \nabla f\not = \bold{0}$.

Consider, a unit vector,

$\displaystyle \bold{u}=<u_1,...,u_n>$ and, $\displaystyle \sum_{k=1}^n u_k^2 = 1$

You need to find, what the components of $\displaystyle \bold{u}$ such as, $\displaystyle D_u f$ (directional derivative along this vector) is extremized.

But,

$\displaystyle D_u f =\nabla f \cdot \bold{u}$

Thus, you want to maximize,

$\displaystyle g(u_1,...,u_n)=\nabla f\cdot \bold{u}$

Thus,

$\displaystyle g(u_1,...,u_n)=\sum_{k=1}^n \frac{\partial f}{\partial x_k}u_k$

With constraint curve,

$\displaystyle c(u_1,...u_n)=\sum_{k=1}^n u_k^2=1$

Thus, Lagrange Multipliers,

$\displaystyle \nabla g=\kappa \nabla c$

Thus,

$\displaystyle \left< \frac{\partial f}{\partial x_1},...,\frac{\partial f}{\partial x_n} \right>=\left<2\kappa u_1,...,2\kappa u_n\right>$

Since,

$\displaystyle \nabla f\not = \bold{0}$ thus, $\displaystyle \kappa \not = 0$

Thus,

$\displaystyle \bold{u}=\frac{1}{2\kappa}\left< \frac{\partial f}{\partial x_1},...,\frac{\partial f}{\partial x_n} \right>=\frac{1}{2\kappa} \nabla f$

Thus, we see that every extremized directional derivative is a multiple of the gradient of the function.

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Note, I used Lagrange Multipliers. The problem might be that the prove of Lagrange Multipliers for multi-variables more than 2 is based on the fact of the max/min property of the gradient. Thus, my proof was not a proof because it uses a fact that depends on itself. - Aug 8th 2006, 11:34 AMAMR
thanks alot :D