how can we proof this :

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

( )

where c is a positive constant of proportionality.

thanks

Printable View

- Aug 8th 2006, 06:40 AMAMRsteepest descent
how can we proof this :

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

( )

where c is a positive constant of proportionality.

thanks - Aug 8th 2006, 08:41 AMThePerfectHackerQuote:

Originally Posted by**AMR**

*gradient*of a diffrenciable function.

---

I can prove it for .

Let at a point.

Let, be any unit vector.

Then, directional derivative of,

The largest possible value of,

Thus, maximum is,

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

---

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, 11:15 AMThePerfectHacker
You have a function,

.

At point,

You have, .

Consider, a unit vector,

and,

You need to find, what the components of such as, (directional derivative along this vector) is extremized.

But,

Thus, you want to maximize,

Thus,

With constraint curve,

Thus, Lagrange Multipliers,

Thus,

Since,

thus,

Thus,

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

---

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, 12:34 PMAMR
thanks alot :D