# Why steepest descent gives a wrong direction search?

1. May 12, 2015

### ymhiq

1. The problem statement, all variables and given/known data
I have to minimize the function (x1-1)2+x23+x1x2 by the steepest descent method. The initial point is [1,1]T

2. Relevant equations

3. The attempt at a solution
The gradient of this function is ∇ƒ(x1,x2)=[2(x1-1)-x2 3x22-x1]. This gradient evaluated in the initial point is ∇ƒ(1,1)=[-1 2]. Following the steepest descent method it is mandatory to minimize the function ƒ(x0-α∇ƒ(x0)) in order to find the value of α. So ƒ(x0-α∇ƒ(x0))=-5α+15α2-8α3 and ƒ'(x0-α∇ƒ(x0))=-5+30α-24α2. This function has extreme points in α1=0.95061 and α2=5.094. In order to be a minimum of this curve ƒ''(x0-α∇ƒ(x0))=30-48α has to be positive. This is my problem ƒ''(x0-α∇ƒ(x0) evaluated at both α values is negative so they don´t minimize the direction. So what I am doing wrong?

2. May 13, 2015

### SteamKing

Staff Emeritus
Just inspecting the gradient of the original function f(x1, x2), something doesn't look right.

If you take ∂f / ∂x1, how did you obtain [2(x1-1)-x2], specifically, the ' - x2' part? I'm confused, because there were no negative signs between terms in the original definition of f(x1, x2). A similar question arises in what you show to be ∂f / ∂x2.

3. May 13, 2015

### Ray Vickson

As SteamKing has pointed out, your gradient formula is incorrect, and your initial steepest-descent direction is wrong. However, when you correct these errors, you will obtain a function $\phi(\alpha) = f(x_0 - \alpha \nabla f(x_0))$ that has no stationary points at all. What does that tell you?

4. May 13, 2015

### ymhiq

Oh! Excuse me! You are right! However I made a mistake when I wrote the original problem. Let me write it again. I have to minimize the function ƒ(x1,x2)=(x1-1)2+x23-x1x2. The initial point is [1,1]T.

5. May 13, 2015

### ymhiq

Excuse me all of you. Finally I got the mistake I made solved. It was an incorrect solutions of ƒ'(x0-α∇ƒ(x0))=-5+30α-24α2 .