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Taylor expansion-multivariable calculus(basic question)

  1. Oct 15, 2009 #1
    What's the Taylor expansion of F(x,y,z) in the neighborhood of (a,b,c)?

    Thank you
     
  2. jcsd
  3. Oct 15, 2009 #2

    HallsofIvy

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    Just extend the Taylor's expansion for functions of one variable.

    For any n, let i, j, k be any non negative integers such that i+ j+ k= n.

    One term in the Taylor's expansion, about [itex](x_0,y_0,z_0)[/itex], is
    [tex]\frac{1}{n!}\frac{\partial f}{\partial x^i\partial y^j\partial z^k}[/itex][itex](x- x_0)^i(y- y_0)^j(z- z_0)^k[/itex].
    The entire Taylor's expansion is the sum for n= 0 to infinity, of the sum of all such terms for all "partitions" of n.

    Specifically, the "0" term has only i= j= k= 0 and is [itex]f(x_0,y_0,z_0)[/itex]. For n= 1, we have i=1, j= k= 0 so [itex]\partial f/\partial x(x_0,y_0,z_0)(x- x_0)+ \partial f/\partial y (x- y_0)[/itex][itex]+ \partial f/\partial z(x_0,y_0,z_0)(z- z_0)[/itex]. The second order terms are [itex](1/2)\partial^2f/\partial x^2 (x_0,y_0,z_0)(x- x_0)^2[/itex][itex]+ (1/2)\partial^2 f/\partial y^2(x_0,y_0,z_0)(y- y_0)^2[/itex][itex]+ (1/2)\partial^2 f/\partial z (x_0,y_0,z_0)(z-z_0)^2[/itex][itex]+ (1/2)\partial f^2/\partial x\partial y(x_0,y_0,z_0)(x-x_0)(y-y_0)+[/itex][itex] (1/2)\partial f^2/\partial x_0\partial z_0(x_0,y_0,z_0)(x-x_0)(z-z_0)[/itex][itex]+ (1/2)\partial f^2/\partial y_0\partial z_0(x_0,y_0,z_0)(y-y_0)(z-z_0)[/itex].

    Do you get the idea? "3" can be partitioned as 3+ 0+ 0, 2+ 1+ 0, 2+ 0+ 1, 1+ 2+ 0, 1+ 0+ 2, 1+ 1+ 1, 0+ 3+ 0, 0+ 0+ 3, 0+ 1+ 2, and 0+ 2+ 1 so there are (if I counted correctly) 10 third degree terms
     
    Last edited: Oct 15, 2009
  4. Oct 16, 2009 #3
    Thanks a lot HallsofIvy for this clear explanation!
    I got the idea.
     
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