What is the vector form of Taylor's Theorem?

Question

I checked most of the posts about Taylor expansion with scalar functions. Could anyone tell me what is the multivariate version of Taylor's Theorem, and how I can use it?

Answer 1

It is not entirely clear which of the following you are asking for: (a) Taylor expansion for a vector-valued function or (b) Taylor expansion for a function defined on $\mathbb{R}^N$ with $N > 1$. In the former case, you just use the usual Taylor's theorem component by component. In the latter, look at this for the statement and proof.

Answer 2

For a scalar-valued function $f$ of $n$ variables $x_k$ the multivariate form of Taylor's theorem can be brought into the following form: $$f(p+X)=\sum_{r=0}^n {1\over r!} d^r f(p)(X)+ R_n.$$ Here $d^r f(p)(X)$ is a homogeneous polynomial of degree $r$ in the coordinates $X_1$, $\ldots$, $X_n$ of the increment vector $X$, namely $$d^r f(p)(X)=\sum_{k_1,\ldots,k_r}f_{.k_1\ldots k_r}(p)X_{k_1}\ldots X_{k_r},$$ where the indices $k_i$ run from $1$ to $n$; so formally there are $n^r$ summands in total. The remainder term $R_n$ can assume various forms, e.g. $R_n(X)= o(|X|^n) \ (X\to 0)$.

Among the $n^r$ terms of $d^r f(p)(X)$ many are equal. If we have just two variables $x$,$y\ $ then the $2^r$ terms of $d^r f(p)(X,Y)$ (now the increment vector is $(X,Y)$ with scalar $X$, $Y$) can be arranged into $$d^r f(p)(X,Y) =\sum_{k=0}^r {r \choose k}\ {\partial^r f\over \partial x^k\partial y^{r-k}}(p)\ X^k Y^{r-k}.$$ In the last formula the expressions ${\partial^r f\over \partial x^k\partial y^{r-k}}(p)$ are ${\it constants}$. In particular one has $$df(p)(X,Y)=f_x(p)X+f_y(p)Y, \qquad d^2f(p)(X,Y)=f_{xx}(p)X^2+2 f_{xy}(p) XY +f_{yy}(p)Y^2 .$$