TITLE: Misunderstanding the Taylor Remainder Theorem
QUESTION [9 upvotes]: I believe I am misinterpreting the Taylor Remainder theorem somehow. The Taylor Remainder theorem is (taken from Briggs 3rd ed Calculus: Early Transcendentals) 

Let $f$ have continuous derivatives up to $f^{(n+1)}$ on an open interval $I$ containing $a$. For all $x$ in $I$, 
  $$f(x) = p_n(x) + R_n(x)$$
  where $p_n$ is the $n$th-order Taylor polynomial for $f$ centered at $a$ and the remainder is 
  $$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$$
  for some point $c$ between $x$ and $a$. 

Suppose that my function is $f(x)=e^x$, $a$ is set to $0$, and that I'm considering the 2nd-order Taylor polynomial for $e^x$. Namely, 
$$p_2(x) = 1+x +\frac{x^2}{2}$$
Then the remainder will be 
$$R_2(x) = \frac{e^c}{3!}x^3$$
Here's where I think I might be messing up. If I consider the interval $(-5,5)$, which is an open interval containing $0$, where $f(x)$ is $(n+1)$-times differentiable, I am unable to come up with a $c$ where the function $e^x$ is identical to $1+x +\frac{x^2}{2}+\frac{e^c}{3!}x^3$ in the interval $(-5,5)$. Here is a link to a Desmos page where I tried to find such a $c$. 
So I guess the main question here is this: Am I supposed to specify the interval $I$ from the beginning, or is the theorem stating that there is some interval $I$ containing $a$ where $f(x)=p_n(x)+R_n(x)$ for all $x\in I$? Or perhaps there's some other key idea that I'm missing here. Please let me know where I'm going wrong.

REPLY [11 votes]: Let's break this down a bit. The syntax of the theorem statement actually works like this:

Let $f$ have continuous derivatives up to $f^{(n+1)}$ on an open interval $I$ containing $a$. For all $x$ in $I$, the statement given below is true.

The "statement given below" consists of all the rest of the theorem, including both of the displayed formulas and the text that follows them:

$$f(x) = p_n(x) + R_n(x)$$
  where $p_n$ is the $n$th-order Taylor polynomial for $f$ centered at $a$ and the remainder is 
  $$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$$
  for some point $c$ between $x$ and $a$. 

In no way shape or form does the theorem say that
for all $x$ in $I$, $f(x) = p_n(x) + R_n(x).$
That would be meaningless, because $R_n(x)$ has not even been defined.
Instead, for some point $c$ between $x$ and $a$ the two displayed equations are satisfied, and the point $c$ may depend on $x$ (indeed it must depend on $x,$ given the "between" condition), just as $\delta$ may depend on $\epsilon$ in an epsilon-delta proof.
Here is another way the theorem could be stated:

Let $f$ have continuous derivatives up to $f^{(n+1)}$ on an open interval $I$ containing $a$. For all $x$ in $I$, there exists some point $c$ between $x$ and $a$ such that
  $$f(x) = p_n(x) + R_n(x)$$
  where $p_n$ is the $n$th-order Taylor polynomial for $f$ centered at $a$ and the remainder is 
  $$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}.$$

Or even more directly:

Let $f$ have continuous derivatives up to $f^{(n+1)}$ on an open interval $I$ containing $a$, and let $p_n$ be the $n$th-order Taylor polynomial for $f$ centered at $a$.
  Then for all $x$ in $I$, there exists some point $c$ between $x$ and $a$ such that
  $$f(x) = p_n(x) + \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}.$$

So this is actually the familiar "for all .. there exists ..." that occurs in epsilon-delta definitions as well.
The theorem statement in the book may be a little confusing because the "there exists" is worded as "for some" and is tucked away at the very end of the theorem statement instead of just after the "for all".
The way the theorem is stated is fairly typical for calculus textbooks (if my memory serves), and I think the reason is to introduce the notation $R_n(x)$ for the remainder term.
It might be a little less misleading if it were written
$$R_n(x) = \frac{f^{(n+1)}(c(x))}{(n+1)!}(x-a)^{n+1},$$
writing $c(x)$ rather than just $c$
to remind you that $c$ is not a constant over all $x.$