Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

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, $Q=P_{n,a}(x)$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.