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Taylor expansions in two variables

  1. Mar 17, 2012 #1
    1. Problem: if f(1,3)=7, use Taylor expansion to describe f(1.2,3.1) and f(.9,2.8) if the partials of f are give by

    df/dx=.2
    d^2f/dx^2=.6
    df/dy=.4
    d^2f/dy^2=.9

    (you do not need to go beyond the second derivative for this problem)

    2. I know from class how to do this if one variable changes. I thought I knew how to do it with 2, but I am getting the wrong answer.

    if I wanted to do f(1.2,3) I would do f(1,3)+df/dx*(.2)+d2f/dx2*(.2)^2=7+(.2)*(.2)+(.2)^2/2*(.6).

    so I thought if I wanted to do both, I would expand in both variables:

    f(1.2,3.1) ~ f(1,3)+df/dx*(.2)+d2f/dx2*(.2)^2+df/dy*(.1)+d2f/dy2*(.1)^2

    Why is this coming out to be the wrong answer?
     
  2. jcsd
  3. Mar 17, 2012 #2

    Dick

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    Science Advisor
    Homework Helper

    Perhaps you are missing some factorials?
     
  4. Mar 17, 2012 #3

    HallsofIvy

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    Science Advisor

    If f(x, y) is a function of the two variables, x and y, then its Taylor series, about x= a, y= b, is
    [tex]f(a,b)+ f_x(a,b)(x- a)+ f_y(a,b)(y- b)+ (f_{xx}(a,b)/2!)(x- a)^2+ (f_{xy}(a,b)/2!)(x- a)(y- b)+ (f_{yy}(a,b)/2!)(y- b)^2+ (f_{xxx}(a,b)/3!)(x- a)^3+ (f_{xxy}(a,b)/3!)(x- a)^2(y- b)+ (f_{xyy}(a,b)/3!)(x-a)(y-b)^2+ (f_{yyy}(a,b)/3!)(y- b)^2+ \cdot\cdot\cdot[/tex]
    Get the idea?
     
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