Sign confusion when taking gradient (Newton's Method)

[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)

Gradient of g(x,y):

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]?

Gradient of F(x,y):

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

Screen shot which is probably easier to read:

 

[ehild]

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)

Gradient of g(x,y):

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]?

Gradient of F(x,y):

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

 

[HallsofIvy]

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. $-\nabla f$