Suppose $f : \mathbb{R}^{n_1} \times \ldots \times \mathbb{R}^{n_k} \to \mathbb{R}^p$ is multilinear. Suppose $a = (a_1, \ldots, a_k)$ and $h = (h_1, \ldots, h_k)$ are in $\mathbb{R}^{n_1} \times \ldots \times \mathbb{R}^{n_k}$. I would like to know the best way to write out the expansion the following expression.
$$ f(a_1 + h_1, \ldots, a_k + h_k) $$
I want a good way (notation?) to express that there will be terms with a group of terms with no $h$s, a group of terms with only one $h_i$ (for some $i$), and some other terms which have at least two $h$s.
I don't know if this is relevant, but I'm asking this so that I can solve Problem 2-14 (b) from Spivak's Calculus on manifolds, where I have to show that $$Df(a_1, \ldots, a_k)(x_1, \ldots x_k) = \sum_{i = 1}^{k} f(a_1, \ldots, a_{i-1}, x_i, a_{i+1}, \ldots, a_{k}).$$
I know that on expanding $$f(a_1 + h_1, \ldots, a_k + h_k) - f(a_1, \ldots, a_k) - \sum_{i = 1}^{k} f(a_1, \ldots, a_{i-1}, h_i, a_{i+1}, \ldots, a_{k})$$ only the terms which have two or more $h$s will remain, and I want good notation so that I can express this.
Any help is appreciated.
I'm not sure whether this is what you are after, but for the exercise you should write $$f(a_1+h_1, \ldots,a_n+h_n) -f (a_1, \ldots, a_n)=\\ f(a_1+h_1, \ldots,a_n+h_n) - f (a_1+h_1, \ldots, a_{n-1} +h_{n-1},a_n) + \\f(a_1+h_1, \ldots,a_{n-1} +h_{n-1},a_n) - f(a_1+h_1, \ldots,a_{n-2} +h_{n-2},a_{n-1},a_n) + \\ \ldots \\+ f (a_1+h_1,a_2, \ldots, a_n) -f (a_1, \ldots, a_n)$$
In fact, what you have now is, by mulitilinearity, the sum of terms of the form $$f(a_1+h_1, \ldots, a_{i -1}+h_{i-1}, h_{i},a_{i+1},\ldots, a_n)$$
What you want to show is that $$f(a_1+h_1, \ldots,a_n+h_n) -f (a_1, \ldots, a_n)=\sum_i f(a_1, \ldots, a_{i-1}, h_i, a_{i+1},\ldots, a_n) + o((h_1,\ldots,h_n))$$
in other words you want to show that $$\sum_i( f(a_1+h_1, \ldots, a_{i -1}+h_{i-1}, h_{i},a_{i+1},\ldots, a_n) -f(a_1, \ldots, a_{i-1}, h_i, a_{i+1},\ldots, a_n) ) = o(h_1,\ldots,h_n)$$ For the differences in the sum you get, for fixed $i$ $$||h_i|| \left(f(a_1+h_1, \ldots, a_{i-1}+h_{i-1}, \frac{h_i}{||h_i||}, a_{i+1},\ldots, a_n) -f(a_1, \ldots, a_{i-1},\frac{h_i}{||h_i||}, a_{i+1},\ldots, a_n)\right)$$ so you need that
$$\frac{||h_i|| \left(f(a_1+h_1, \ldots, a_{i-1}+h_{i-1}, \frac{h_i}{||h_i||}, a_{i+1},\ldots, a_n) -f(a_1, \ldots, a_{i-1},\frac{h_i}{||h_i||}, a_{i+2},\ldots, a_n)\right)}{||(h_1,\ldots,h_n)||}$$
tends to $0$ if $||(h_1,\ldots,h_n)|| \rightarrow 0$ This is, however, clear since (for any choice of norm, note that I did not say which norm I'm using) $$\frac{||h_i|| }{||(h_1,\ldots,h_n)||}\le 1$$ and the remaining term in the numerator tends to $0$ by continuity, if $(h_1, \ldots, h_n)\rightarrow 0$