How to prove Taylor's theorem?

Question

Here is a proof of mean value theorem:

Consider a line passing through the points $(a, f(a))$ and $(b, f(b))$. The equation of the line is

$y-f(a) = \displaystyle\frac{f(b)-f(a)}{(b-a)} (x-a)$

or $y = f(a)+ \displaystyle\frac{f(b)-f(a)}{(b-a)} (x - a)$

Let h be a function that defines the difference between any function f and the above line.

$h(x) = f(x) – f(a) – \displaystyle\frac{f(b)-f(a)}{(b-a)} (x - a)$

Using “Rolle’s theorem”, we have

$h'(x) = f'(x) – \displaystyle\frac{f(b)-f(a)}{(b-a)}$

Or $f(b) - f(a) = f'(x) (b - a)$. Hence, proved.

The source of this proof is this. It is clear why they used $h(x) = f(x) – f(a) – \displaystyle\frac{f(b)-f(a)}{(b-a)} (x - a)$ and how it arrives.

Now here is a proof of Taylor's theorem:

Consider the function $p(x)$ defined in $(a, b)$ by $p(x) = q(x) - \displaystyle\frac{(b-a)^n}{(b-x)^n} q(a)$ where $q(x) = f(b) - f(x) - (b - x)f'(x) - \displaystyle\frac{(b-x)}{2!}f''(x) -... \displaystyle\frac{(b-x)^{n-1}}{(n-1)!}) f^{n-1}(x)$

Then they show that, $p(a) = p(b)=0$ and apply roll's theorem and proves Taylor's theorem.

I don't understand why they use, $p(x) = q(x) - \displaystyle\frac{(b-a)^n}{(b-x)^n} q(a)$. They apply Rolls theorem on $h(x) = f(x) – f(a) – \displaystyle\frac{f(b)-f(a)}{(b-a)} (x - a)$ because it has something to do with a line passing through the points $(a, f(a))$ and $(b, f(b))$. What is the relation of this function, $p(x) = q(x) - \displaystyle\frac{(b-a)^n}{(b-x)^n} q(a)$, with the points? How do we come with this function? Can you show any proof as it was shown how we come with the function, $h(x) = f(x) – f(a) – \displaystyle\frac{f(b)-f(a)}{(b-a)} (x - a)$ from the difference between any function f and the above line? The book I used was as if it took for granted that $p(x) = q(x) - \displaystyle\frac{(b-a)^n}{(b-x)^n} q(a)$

Hope I've clarified my problem.

Answer 1

I'm a math newbie, so sorry if it's not very rigorous.

The key resides in considering the function $g(x) = R_{n,f,x}(b)$. That is, the error of the aproximation of $f(b)$ given by the degree n Taylor polynomial centered at $x$. We will apply the mean value theorem to this function to prove the residue formula. Notice that here the variable is the center of the polynomial, not the value interpolated.

Also, we have the results: $$g(a) = R_{n,f,a}(b)\quad; \quad g(b) = 0$$ $$g(x) = f(b)-\sum_{k=0}^{n}{\frac{f^{(k)}(x)}{k!}(b-x)^k}$$ Since each of the terms of $g(x)$ is derivable (We supose $f\in C^{n+1}$), we have $$\frac {d}{dx}\left({\frac{f^{(k)}(x)}{k!}(b-x)^k}\right) = {\frac{f^{(k+1)}(x)}{k!}(b-x)^k} - {\frac{f^{(k)}(x)}{(k-1)!}(b-x)^{k-1}} $$ This provides a nice telescopic series, so that it's not difficult to see that $$g'(x)= -\frac{f^{(n+1)}(x)}{n!}(b-x)^{n}$$ The proof of the theorem resides on this property of function $g$. If we only apply the mean value theorem to $g$, we would get that there exists some $c$ in $(a,b)$ such that $$-R_{n,f,a}(b) = -\frac{f^{(n+1)}(c)}{n!}(b-c)^{n}(b-a)$$ which is known as Cauchy's residue formula. This expression is very ugly, so it's better to use auxiliary functions to get a simpler formula. For example, the one you wrote there.

I particularly learned the function $h(x)= (b-x)^{n+1}$, which is used among $g$ to apply Cauchy's mean value theorem: there exists $c \in (a,b)$ such that: $$ h'(c)(g(b)-g(a)) = g'(c)(h(b)-h(a)) $$ $$ -(n+1)(b-c)^n(-R_{n,f,a}(b)) = -\frac{f^{(n+1)}(c)}{n!}(b-c)^{n}(-(b-a)^{n+1})$$ From here, it's deduced: $$R_{n,f,a}(b) = \frac{f^{(n+1)}(c)}{(n+1)!}(b-a)^{n+1}$$

And this is what we wanted to prove.

I hope it may help you.