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Sign confusion when taking gradient (Newton's Method)

  1. Mar 20, 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

    vanhees71

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    2016 Award

    The gradient in Cartesian components is, of course, given by
    [tex]\vec{\nabla} g=(\partial_x g,\partial_y g)=(-4(x-1)^3,-4 (y-1)^3)[/tex]
    and the Hesse Matrix by
    [tex]H_{ij}=\partial_i \partial_j g=\mathrm{diag}(-2(x-1)^2,-12(y-1)^2).[/tex]
     
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