Taylor expansion-multivariable calculus(basic question)

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What's the Taylor expansion of F(x,y,z) in the neighborhood of (a,b,c)?

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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 (x_0,y_0,z_0), is
\frac{1}{n!}\frac{\partial f}{\partial x^i\partial y^j\partial z^k}[/itex](x- x_0)^i(y- y_0)^j(z- z_0)^k.<br /> The entire Taylor&#039;s expansion is the sum for n= 0 to infinity, of the sum of all such terms for all &quot;partitions&quot; of n.<br /> <br /> Specifically, the &quot;0&quot; term has only i= j= k= 0 and is f(x_0,y_0,z_0). For n= 1, we have i=1, j= k= 0 so \partial f/\partial x(x_0,y_0,z_0)(x- x_0)+ \partial f/\partial y (x- y_0)+ \partial f/\partial z(x_0,y_0,z_0)(z- z_0). The second order terms are (1/2)\partial^2f/\partial x^2 (x_0,y_0,z_0)(x- x_0)^2+ (1/2)\partial^2 f/\partial y^2(x_0,y_0,z_0)(y- y_0)^2+ (1/2)\partial^2 f/\partial z (x_0,y_0,z_0)(z-z_0)^2+ (1/2)\partial f^2/\partial x\partial y(x_0,y_0,z_0)(x-x_0)(y-y_0)+(1/2)\partial f^2/\partial x_0\partial z_0(x_0,y_0,z_0)(x-x_0)(z-z_0)+ (1/2)\partial f^2/\partial y_0\partial z_0(x_0,y_0,z_0)(y-y_0)(z-z_0).<br /> <br /> Do you get the idea? &quot;3&quot; 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
 
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Thanks a lot HallsofIvy for this clear explanation!
I got the idea.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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