Why steepest descent gives a wrong direction search?

[ymhiq]

1. Homework Statement
I have to minimize the function (x₁-1)²+x₂³+x₁x₂ by the steepest descent method. The initial point is [1,1]^T

Homework Equations

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?

 

[SteamKing]

1. Homework Statement
I have to minimize the function (x₁-1)²+x₂³+x₁x₂ by the steepest descent method. The initial point is [1,1]^T

Homework Equations

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?

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.

 

[Ray Vickson]

1. Homework Statement
I have to minimize the function (x₁-1)²+x₂³+x₁x₂ by the steepest descent method. The initial point is [1,1]^T

Homework Equations

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?

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?

 

[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.

 

[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 .