Expression of difference in *Calculus on Manifolds* I am looking at Spivak's Calculus on Manifolds, and on p. 35 I read
$$f^i(y) - f^i(x) = \sum_{j=1}^n [ f^i(y^1, \dotsc, y^j, x^{j+1}, \dotsc, x^n) - f(y^1, \dotsc, y^{j-1}, x^j, \dotsc, x^n) ].$$
Can someone explain this assertion to me? I've never seen this mixing of the components of $x$ and $y$.
Thanks
 A: For simplicity, let's suppose the domain of $f$ is a cube $A \subset \mathbb{R}^3$ with $y=(y_1,y_2,y_3)$ and $x=(x_1,x_2,x_3)$ both in $A$. Denote by $(f^1,f^2,f^3)$ the components of $f$. Then we can write
\begin{align}
f^i(y)-f^i(x)
 &= f^i(y_1,y_2,y_3)-f^i(x_1,x_2,x_3) \\
 &= [f^i(y_1,y_2,y_3)-f^i(y_1,y_2,x_3)]+[f^i(y_1,y_2,x_3)-f^i(y_1,x_2,x_3)]+[f^i(y_1,x_2,x_3)-f^i(x_1,x_2,x_3)].
\end{align}
Think of the points $x$ and $y$ as describing a cube aligned with the coordinate axes, then we look at the variation of $f^i$ along each leg of the cube. When we add these up, the intermediate terms cancel. Note that you have to creatively interpret the subscript $y_0$ as meaning no $y_i$'s appear.
Along each leg of the cube, as we're only varying $f^i$ in one dimension we can apply the usual mean value theorem. Summing over $n=3$ legs of the cube (to get a path from $x$ to $y$) and then over all $n=3$ component functions yields the mean value inequality
$$
|f(y)-f(x)| \leq n^2M\cdot|x-y|,
$$
where $M$ is a uniform bound on all the partial derivatives in the cube $A$.
