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I'm trying to understand a proof that the teacher gave. The course is Calculus I and we are just starting, topics like series and integration have not been covered yet. We started with linearization and then Taylor's polynomial.

Theorem: If a function $f$ is N-times differentiable in $a$, then it exists only one polynomial (and it's $P_n(f, a)$) that satisfies: \begin{equation*} f(x) = P_{n, a}(x) + r((x - a)^n) \end{equation*} Here, $P_{n, a}(x)$ is the Nth-order Taylor polynomial of $f(x)$ at $x = a$ and \begin{equation*} \lim_{x\to a} \tfrac{r((x - a)^n)}{(x - a)^n} = 0 \end{equation*} For $n = 1$ the theorem is already proved (Mean value theorem I think).

The problem is that I don't understand why (he said) in order to prove the theorem it's enough to prove this limit: \begin{equation*} \lim_{x\to a} \tfrac{r((x - a)^n)}{(x - a)^n} = 0 \end{equation*} Or: \begin{equation*} r_n(x) = f(x) - P_{n, a}(x) \end{equation*} Basically what the teacher did is:

1) He evaluate the function for $x = a$ and concluded $r_n(a) = 0$: \begin{equation*} f(a) = f(a) + \tfrac{f^{(1)}(a)(a - a)}{1!} + \tfrac{f^{(2)}(a)(a - a)^2}{2!} + \dots + \tfrac{f^{(n)}(a)(a - a)^n}{n!} + r((a - a)^n) \end{equation*} 2) Did the same as in step 1 but this time he evaluate the differentiated function (instead of the original function) for $x = a$. He concluded $r^{(1)}_n(a) = 0$.

3) Did the same as in step 2 but for $n$ higher than 1, and he concluded $r^{(n)}_n(a) = 0$. I think he showed that $f$ is N-times differentiable in $a$ and the remainder is zero.

4) Then with the help of L'Hôpital's rule he did:

$\lim_{x\to a} \tfrac{r_n(x)}{(x - a)^n} = \lim_{x\to a} \tfrac{r^{(1)}_n(x)}{n(x - a)^{n - 1}} = \lim_{x\to a} \tfrac{r^{(2)}_n(x)}{n(n - 1)(x - a)^{n - 2}} = \dots = \lim_{x\to a} \tfrac{r^{(n - 1)}_n(x)}{n!(x - a)}$

So: \begin{equation*} r^{(n)}_n(a) = r^{(n - 1)}_n(a) \end{equation*} With the derivative definition and the result of step 3 he did:

$r^{(n)}_n(a) = \lim_{x\to a} \tfrac{r^{(n - 1)}_n(x) - r^{(n - 1)}_n(a)}{x - a} = 0$

He knows that $r^{(n - 1)}_n(a) = 0$ (also step 3), so: \begin{equation*} \lim_{x\to a} \tfrac{r^{(n - 1)}_n(x)}{x - a} = 0 \end{equation*} 5) Finally, because of step 4:

$\lim_{x\to a} \tfrac{r_n(x)}{(x - a)^n} = \lim_{x\to a} \tfrac{r^{(n - 1)}_n(x)}{n!(x - a)} = \lim_{x\to a} \tfrac{1}{n!}\tfrac{r^{(n - 1)}_n(x)}{x - a} = \lim_{x\to a} \tfrac{1}{n!}r^{(n)}_n(a) = 0$

I follow the proof, it's easy, but as I said I don't understand why in order to prove the theorem it's enough to prove the limit. I've searched on the Internet, read books (like Calculus, Vol. 1 by Tom M. Apostol) and sincerely I didn't found a proof like the one the teacher gave.

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He exhibit one polynomial that works. It is not hard to see that there cannot be two polynomials of degree $\le n$, so that does it. The proof is usually done by integrating by parts, but this works too, maybe with a triflingly stronger differentiability condition than absolutely necessary. – André Nicolas Apr 9 '12 at 19:01
@André Nicolas: can you elaborate more on your comment? – David Robert Jones Apr 9 '12 at 23:57
Which one? The uniqueness proof has been dealt with by Christopher Creutzig. The L'Hospital's Rule argument shows that the error term is $o(x-a)^n$, which is what he was trying to prove. – André Nicolas Apr 10 '12 at 1:52
up vote 1 down vote accepted

If what you wrote is the complete proof, then, technically speaking, he did not prove uniqueness. But that part is relatively easy:

Assume that there is a polynomial $Q$ of degree $\leq n$ such that $\lim\limits_{x\to a}\frac{f(x)-Q(x)}{(x-a)^n} = 0$. Since we also know that $\lim\limits_{x\to a}\frac{f(x)-P_{n,a}(x)}{(x-a)^n} = 0$, we find $\lim\limits_{x\to a}\frac{P_{n,a}(x)-Q(x)}{(x-a)^n} = 0$.

Now, using l'Hôpital, we can show for each coefficient of the polynomial $P_{n,a}(x)-Q(x)$ that it is zero. (This is probably easiest to do when you write it as a polynomial in $x-a$, or, equivalently, consider only the case $a=0$ after showing that that does not lose generality. Start from the coefficient of $(x-a)^n$.) In other words, $Q=P_{n,a}(x)$.

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So, if he didn't prove uniqueness then what did he prove? That's what I don't understand, the implications this limit has in the proof: $\lim_{x\to a} \tfrac{r((x - a)^n)}{(x - a)^n} = 0$. – David Robert Jones Apr 9 '12 at 23:46
He proved that $P_{n,a}$ has the required property. Which also proves that such a polynomial does exist – that is a non-trivial result. – Christopher Creutzig Apr 10 '12 at 9:35
I wasn't understanding your answer because I got confused. You wrote: "he did not prove uniqueness", and you're right, in this proof he was proving existence, not uniqueness. He did prove uniqueness after this proof, but I need to focus in this proof first. What I need to understand is why this proof proves existence. You said "He proved that $P_{n, a}$ has the required property. Which also proves that such a polynomial does exist", obviously you understood the proof and you know he's proving existence, but I'm not, so how did you realize that this proof proves existence? Thanks. – David Robert Jones Apr 10 '12 at 13:19
He took a polynomial which we know exists (because it is explicitly given) and showed that it has the desired property. I mean, what else is there to prove? (I find the way the statement is written somewhat confusing, maybe that is your problem. Making $r$ a function of $(x-a)^n$ instead of $x$ is, in my eyes, rather pointless; I'm not even sure if it is strictly speaking correct. I'd go with $r(x)$ and $\lim_{x\to a}\frac{r(x)}{(x-a)^n}=0$.) – Christopher Creutzig Apr 10 '12 at 14:25

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