Proof that the best linear approximation to $f(x)$ near $a$ is given by the linear function $L_{a}(x) = f(a) + f '(a)(x-a)$

Question

The title basically says everything. The formula for linear approximation appears to be right intuitively but is there a proof for it?

Secondly, is there also a proof why to put the $\frac{1}{2}$ in front of the second derivative in the quadratic approximation? $$Q_{a}(x)=f(a)+f'(a)(x-a)+\frac{f''(a)(x-a)^2}{2}$$

Answer 1

Since by definition $$\lim_{x\to a} \frac{f(x) - f(a)}{x - a} = f'(a),$$ we have $$f(x) = f(a) + (x - a)f'(a) + o(x - a)$$ as $x \to a$. Any other approximation $f(x) = f(a) + C(x - a)$ with $C\not = f'(a)$ would behave like $(C - f'(a))(x - a)$ (which is not $o(x - a)$) near $a$. Alternatively, look at the Taylor series of $f$ at $a$, assuming $f$ has one.

For the second part: Without going through the derivation of the Taylor series, the easiest way to show that the coefficient should be $\frac{1}{2}$ and not $1$ is to take $f(x) = (x - a)^2$.

Answer 2

We can use a Taylor expansion:

If a function $f$ is infinitely differentiable at $a \in \mathbb{R}$ or $\mathbb{C}$, then $$f(x) = f(a) + f'(a)(x - a) + \frac{1}{2}f''(a)(x-a)^2 + ... + \frac{1}{n!}f^{(n)}(a)(x-a)^n + ...$$

If $O((x-a)^n)$ for some $n$ is "small enough", then it is sufficient to use the terms before that for our approximation. Hope this is helpful.