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Power series of a function of 2 variables

  1. Oct 12, 2007 #1
    I have learned that if a function of one real variable can be defined as a power series, then this one is its Taylor series.

    Does the same occur with functions of 2 real variables? I mean, if a function f(x, y) can be defined as a power series, does this series is the Taylor series of f(x, y)?

    Thanks for help.
     
  2. jcsd
  3. Oct 13, 2007 #2

    CompuChip

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    I thought an expansion like this was possible...
    [tex]
    f(x, y) = f(x_0, y_0)
    + \left. \frac{\partial f(x, y_0)}{\partial x} \right|_{x = x_0} (x - x_0)
    + \left. \frac{\partial f(x_0, y)}{\partial x} \right|_{y = y_0} (y - y_0)
    + \frac12 \left. \frac{\partial^2 f(x, y_0)}{\partial x^2} \right|_{x = x_0} (x - x_0)^2
    [/tex][tex]
    + \frac12 \left. \frac{\partial^2 f(x_0, y)}{\partial y^2} \right|_{y = y_0} (y - y_0)^2
    + \frac12 \left. \frac{\partial^2 f(x, y)}{\partial x \partial y}(x-x_0)(y-y_0) \cdots
    + \mathcal{O}(x, y)^3
    [/tex]
     
    Last edited by a moderator: Oct 13, 2007
  4. Oct 13, 2007 #3

    HallsofIvy

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    That should be
    [tex]
    f(x, y) = f(x_0, y_0)
    + \left. \frac{\partial f(x, y_0)}{\partial x} \right|_{x = x_0} (x - x_0)
    + \left. \frac{\partial f(x_0, y)}{\partial x} \right|_{y = y_0} (y - y_0)
    + \frac12 \left. \frac{\partial^2 f(x, y_0)}{\partial x^2} \right|_{x = x_0} (x - x_0)^2
    [/tex][tex]
    + \frac12 \left. \frac{\partial^2 f(x_0, y)}{\partial y^2} \right|_{y = y_0} (y - y_0)^2
    + \frac12 \left. \frac{\partial^2 f(x, y)}{\partial x \partial y}(x-x_0)(y-y_0)\cdots
    + \mathcal{O}(x, y)^3
    [/tex]
    where I have added [itex](x-x_0)(y-y_0)[/itex] after the mixed second derivative.
     
  5. Oct 14, 2007 #4

    CompuChip

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    I believe that's what I meant by the [itex]\cdots[/itex], sorry for being unclear.
     
  6. Oct 14, 2007 #5

    arildno

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    This is incorrect, Halls!
    You have a 1/2 in front of the mixed second partial; it should be a 1 instead.

    For OP:
    Here's how we can DEDUCE the look of the Taylor polynomial for a 2-variable function.
    Now, given a function f(x,y); we may as a first step regard this as a single variable function G(x;y)=f(x,y); where "y" in G is just some fixed parameter.
    G can be expanded in a 1-variable Taylor series in x about the point (x0,y), so switching to f-notation, we have:
    [tex]f(x,y)=\sum_{n=0}^{\infty}\frac{1}{n!}\frac{\partial^{n}f}{\partial{x}^{n}}(x_{0},y)(x-x_{0})^{n}[/tex]
    where the 0'th derivative of a function means the function itself.

    Now, each of these derivatives is a function of y, with a fixed parameter x0. Thus, they can be expanded as Taylor series, and we get:
    [tex]f(x,y)=\sum_{n=0}^{\infty}\frac{1}{n!}\sum_{m=0}^{\infty}\frac{1}{m!}\frac{\partial^{(n+m)}f}{\partial{x}^{n}\partial{y}^{m}}(x_{0},y_{0})(x-x_{0})^{n}(y-y_{0})^{m}[/tex]
    Regrouping our double series in term of the total derivative index s=n+m, we readily get:
    [tex]f(x,y)=\sum_{s=0}^{\infty}\frac{1}{s!}\sum_{n=0}^{s}\binom{s}{n}\frac{\partial^{s}f}{\partial{x}^{n}\partial{y}^{(s-n)}}(x_{0},y_{0})(x-x_{0})^{n}(y-y_{0})^{(s-n)}[/tex]
    where I have utilized [tex]\frac{1}{s!}\binom{s}{n}=\frac{1}{n!}\frac{1}{(s-n)!}[/tex]


    This form is readily extendable to functions with more than two variables as well.
     
    Last edited: Oct 14, 2007
  7. Oct 14, 2007 #6
    One easy way to remember the Taylor series in higher dimensions is to write it like

    [tex]
    f(x+u) = f(x)\; +\; u\cdot\nabla f(x)\; +\; \frac{1}{2!}(u\cdot\nabla)^2 f(x)\; +\; \frac{1}{3!}(u\cdot\nabla)^3 f(x)\; + \cdots
    [/tex]

    You can get those coefficients for the partial derivatives by computing [tex](u\cdot\nabla)^n[/tex] open. For example

    [tex]
    (u_1\partial_1 + u_2\partial_2)^2 = u_1^2\partial_1^2 + 2u_1u_2\partial_1\partial_2 + u_2^2\partial_2^2
    [/tex]
     
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