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Newton's method to approximate critical point

  1. Oct 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Let ##F(x,y)=4sin(xy)+x^3+y^3## Use Newton's method to approximate the critical point that lies near ##(x,y)=(-1,-1)##

    2. Relevant equations


    3. The attempt at a solution
    I have a problem here because the derivative is not a square matrix. Hence, I can't find the inverse needed for the form $$x_{k}=x_{k-1}-[Df(x_{k-1})]^{-1}f(x_{k-1})$$
    In this case, the derivative is zero, not the function, so I think the form above must be modified?
    If derivative is zero, I get a form which doesn't make sense, which is ##z-f(x_o,y_o)=0##
    Any hints?
     
  2. jcsd
  3. Oct 13, 2016 #2

    mfb

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    You can use symmetry to reduce the problem to a one-dimensional problem.

    The more textbook-like solution: calculate the first and second derivatives, then find the simulateous minimum of the first derivatives.
     
  4. Oct 13, 2016 #3

    andrewkirk

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    You could take it in turns to move in directions parallel to the x and y axes.

    step 1 take a Newton step parallel to the y axis towards the point near (-1,-1) on the line ##x=x_0=-1## where ##F_y(x_0,y)=0##. Label the landing point of that step ##(x_{1},y_{1})##
    step 2 take a Newton step parallel to the x axis towards the nearby point on the line ##y=y_{1}## where ##F_x(x,y_1)=0##. Label the landing point of that step ##(x_{2},y_{2})##
    step 3 take a Newton step parallel to the y axis............
     
    Last edited: Oct 13, 2016
  5. Oct 13, 2016 #4

    Ray Vickson

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    You CAN use Newton's method to solve the two equations ##f_x(x,y) = 0## and ##f_y(x,y) = 0##. You end up with the problem of inverting a 2x2 matrix at each step. See, examples in
    https://www.math.ohiou.edu/courses/math3600/lecture13.pdf or
    http://www.seas.ucla.edu/~vandenbe/103/lectures/newton.pdf [Broken] .

    Just Google "Newton's method for system of equations".
     
    Last edited by a moderator: May 8, 2017
  6. Oct 13, 2016 #5
    But I just don't get it. Newton's method is used to get the root of an equation, whereas this is to get a critical point where the derivative is zero. Can the above form apply here?
     
  7. Oct 13, 2016 #6

    Ray Vickson

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    Yes, it applies to the univariate problem of finding the root of the derivative.

    However, in this case you have TWO equations in the TWO unknowns x and y, because you need to set TWO partial derivatives to zero in order to find the stationary point. So, you need to use the multivariate version of Newton's method that I gave you links for in my previous response. Have you read them, or parts of them at least?
     
  8. Oct 13, 2016 #7
    Yes I've read them but I still have no idea on how the links you gave me can be applied to this problem. Sorry.

    But I will show you what I have tried so far. I made the partial derivatives of ##x## and ##y## as the new functions I need to evaluate using Newton's method. I reason that by getting the roots for these functions, I would be getting the value for the critical point.

    So for ##f_x##, $$z-z_0=f_x(x-x_0)$$
    By setting ##z=0##, $$x-x_0=\frac{-z_0}{f_x}$$
    Hence, $$x_{k-1}=x_{k}-\frac{z_0}{f_x}$$

    Is this a correct approach?
     
  9. Oct 13, 2016 #8

    Ray Vickson

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    NO. As I stated very explicitly in post #4, and stated again in post #6, you have two equations ##F_x(x,y) = 0## and ##F_y(x,y) = 0##. Here, ##F_x = \partial F/\partial x## and ##F_y = \partial F/ \partial y##. The question is asking you for stationary (or critical) points of ##F##, and those are the points where both partial derivatives vanish.

    So, to repeat: you have two nonlinear equations in two unknowns and cannot, therefore, just use the simple 1-variable version of Newton's method; you need the multivariate version as in the links.
     
  10. Oct 13, 2016 #9
    Ah, I think I've got it. Here are my first steps:

    Let ##g(x,y)=(4y \cos(xy)+3x^2, 4x\cos(xy)+3y^2)##. The 2 functions there are the partial derivatives ##F_x## and ##F_y## respectively. By taking derivative of the function, I can get an invertible 2x2 matrix. And from here I can proceed to use Newton's method, where setting ##g(x_0)+Dg(x_0)(x-x_0)=0## makes sense.

    I'm quite sure I'm right this time.:smile:
     
  11. Oct 14, 2016 #10

    mfb

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    I think technically it is not necessary. You want to find the point where both partial derivatives are zero - you can find the point where their product is zero, which leads to the classical problem of finding a root of a R^2 -> R function.
     
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