# Sign confusion when taking gradient (Newton's Method)

• zmalone
In summary, the conversation is about a lecture on Newton's method with n-dimensions and the confusion about the use of a negative sign while taking the first gradient. The missing minus signs are pointed out and the need to investigate the original function is mentioned. The use of the gradient vector in determining a minimum point is also discussed.
zmalone
I'm watching a lecture on Newton's method with n-dimensions but I am kind of hung up on why the professor did not use the negative sign while taking the first gradient? Is there a rule that explains this or something that I'm forgetting? The rest makes sense but highlighted in red is the part I am confused on if anyone can clear that up I'd appreciate it, thanks!

Where g(x,y) = 1-(x-1)^4-(y-1)^4

local maximum at (1,1) ; critical point at (1,1)

F(x,y,) = [Dg(x,y,)]transpose = [4(x-1)^3 4(y-1)^3]transpose
Why not [-4(x-1)^3 -4(y-1)^3]?

DF(x,y) =
12(x-1)^2 0
0 12(y-1)^2

Screen shot which is probably easier to read:

#### Attachments

14 KB · Views: 459
zmalone said:
I'm watching a lecture on Newton's method with n-dimensions but I am kind of hung up on why the professor did not use the negative sign while taking the first gradient? Is there a rule that explains this or something that I'm forgetting? The rest makes sense but highlighted in red is the part I am confused on if anyone can clear that up I'd appreciate it, thanks!

Where g(x,y) = 1-(x-1)^4-(y-1)^4

local maximum at (1,1) ; critical point at (1,1)

F(x,y,) = [Dg(x,y,)]transpose = [4(x-1)^3 4(y-1)^3]transpose
Why not [-4(x-1)^3 -4(y-1)^3]?

DF(x,y) =
12(x-1)^2 0
0 12(y-1)^2

Screen shot which is probably easier to read:

You are right, the minus signs are missing. Anyway, D=0 at the critical point, so you have to investigate the original function.

ehild

You are using Newton's method to find what? The gradient vector points in the direction of fastest increase. If you are "following the gradient" to (numerically) find a minimum point, you want to go in the opposite direction, i.e. $-\nabla f$

## 1. What is sign confusion when using gradient in Newton's Method?

Sign confusion occurs when there is a change in the sign of the gradient during the iteration process of Newton's Method. This can lead to incorrect solutions or difficulty in converging to the actual solution.

## 2. How does sign confusion affect the accuracy of Newton's Method?

If sign confusion occurs, it can greatly affect the accuracy of Newton's Method. This is because the change in sign indicates that the algorithm is no longer moving towards the actual solution and may even diverge from it.

## 3. What causes sign confusion in Newton's Method?

Sign confusion can be caused by various factors, including choosing an initial guess that is too far from the actual solution, or when the function has multiple roots or a flat region where the gradient becomes close to zero.

## 4. How can sign confusion be avoided in Newton's Method?

To avoid sign confusion, it is important to carefully choose an initial guess that is close to the actual solution. It is also helpful to check for multiple roots or flat regions in the function and make appropriate adjustments to the algorithm.

## 5. Can sign confusion be fixed once it occurs in Newton's Method?

Yes, it is possible to fix sign confusion in Newton's Method. This can be done by restarting the algorithm with a new initial guess that is closer to the actual solution. It may also be necessary to adjust the step size or make other modifications to the algorithm to ensure convergence to the correct solution.

• Calculus and Beyond Homework Help
Replies
8
Views
756
• Calculus and Beyond Homework Help
Replies
2
Views
595
• Calculus and Beyond Homework Help
Replies
3
Views
503
• Calculus and Beyond Homework Help
Replies
1
Views
612
• Calculus and Beyond Homework Help
Replies
24
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
755
• Calculus and Beyond Homework Help
Replies
4
Views
963
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
16
Views
2K