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Sign confusion when taking gradient (Newton's Method)

  1. Mar 21, 2014 #1
    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:
     

    Attached Files:

  2. jcsd
  3. Mar 21, 2014 #2

    ehild

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


    ehild
     
  4. Mar 21, 2014 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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. [itex]-\nabla f[/itex]
     
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