Proof of the Formula for Linear Approximation With $n$ Variables?

Question

The linear approximation of $f(\vec{x})$ where $\vec{x}$ has $n$ elements is given by $$f(\vec{x}) \approx f(\vec{a}) + \sum_{j=1}^n \frac{\partial f}{\partial x_j} (\vec{a})(x_j-a_j)$$ for $\vec{a}$ near $\vec{x}$. I understand this formula, but don't understand the proof. Can someone please explain this to me?

Answer 1

We say that a linear transformation $T: \mathbb{R}^n\to\mathbb{R}$ is the derivative of $f$ at point $a$ if $$ f(x) = f(a) + T(x-a) + r(x-a) $$ where $\lim_{h\to0} \frac{r(h)}{||h||} = 0$. And we denote $f'(a) := T$. By the definition of limit, for any $\epsilon > 0$ we can get $\delta > 0$ such that if $||h||<\delta$ then $|r(h)| \leq \epsilon ||h||$. Therefore, taking $h = x-a$:

$$f(a) + f'(a)(x-a) - \epsilon ||x-a||\leq f(x) \leq f(a) + f'(a)(x-a) + \epsilon ||x-a|| $$

Since $\epsilon||x-a||$ is small when $||x-a||$ is small we say it is an approximation. In fact, since the error is of order $||x-a||$ we say that it is a first order approximation.

Now just notice that $f'(a)(x-a) = \sum_{j=1}^n \frac{\partial f}{\partial x_j} (a) (x_j-a_j)$.