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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.

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  • $\begingroup$ In short: the constant $c$ might depend in $x$, which is not clear in the version of the theorem you cited. $\endgroup$ – M. Winter Mar 25 at 15:39
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You need to specify the interval $I$, the function $f$, the degree $n$, the value of $a$, and (what's most counter-intuitive because of how often we use the symbol), we have to fix a value of $x \in I$. Only after you have specified all of these, the theorem tell you there exists a $c$ between $a$ and $x$ (it may be clearer if you call it $c_x$) such that \begin{align} R_{n,a}(x) = \dfrac{f^{(n+1)}(c_x)}{(n+1)!}(x-a)^{n+1} \end{align}

But of course, everything depends on a pre-chosen value for $x$. If you change $x \in I$, you will have to choose a different value for the $c$.


Edit:

Here's how I'd phrase the theorem (just adding in a few adjectives to make it explicit what is being fixed etc)

Let $I \subset \Bbb{R}$ be a given open interval, let $n \in \Bbb{N}$ be given, and let $f: I \to \Bbb{R}$ be a given $\mathcal{C}^{n+1}$ function. Fix a number $a \in I$; now we denote $P_{n,a,f}$ and $R_{n,a,f}$ to be the $n^{th}$ order Taylor polynomial for $f$ about the point $a$, and the $n^{th}$ order Remainder about the point $a$.

Now, fix a particular number $x \in I$. Then, there exists a number $c$ between $a$ and $x$ such that \begin{align} R_{n,a,f}(x) &= \dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}. \end{align}

Notice that the number $c$ in the theorem depends on several things: it depends on $f,n,a,x$, but of course, we don't explicitly mention all of these in the notation. It is pretty much only with practice that you'll be able to recognize which quantities depends on which.

Here's another way of phrasing the same theorem:

Let $I \subset \Bbb{R}$ be a given open interval, let $n \in \Bbb{N}$ be given, and let $f: I \to \Bbb{R}$ be a given $\mathcal{C}^{n+1}$ function. Then, for every $a \in I$ (we let $R_{n,a,f}$ mean the $n^{th}$ order Taylor remainder) and any $x \in I$, there exists $c \in I$, lying between $a$ and $x$, such that \begin{align} R_{n,a,f}(x) &= \dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}. \end{align}

The number of "for all"'s and "there exists" in quick succession may be confusing, but it is very important to recognize which is a bound variable and which is not. I think part of your confusion in the theorem stems from the fact that in the quoted theorem, the author has tried to give a definition of $R_{n,a}$ (namely $R_{n,a} := f - P_{n,a}$) in the same sentence as the actual conclusion of the theorem (which is the final formula for $R_{n,a}(x)$ in terms of $f,n,a,x$ and some number $c$).


Edit 2: Some Additional Remarks

Assuming you have understood my remarks above, let me address your 2nd last paragraph

"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 $1+x+\dfrac{x^2}{2}+ \dfrac{x^3}{3!}$ in the interval $(−5,5)$. Here is a link to a Desmos page where I tried to find such a $c$."

This is in fact not a coincidence. There is actually no such value of $c$. The proof that there is no single $c$ is actually a very simple proof by contradiction. Let us suppose for simplicity that the interval $I$ is the whole real line $\Bbb{R}$. Suppose, for the sake of contradiction, there exists a $c$, such that for all $x \in \Bbb{R}$ \begin{align} e^x &= \left( 1 + x + \dfrac{x^2}{2!} + \dots + \dfrac{x^n}{n!} \right) + \dfrac{e^c}{(n+1)!}x^{n+1} \quad \text{(for all $x \in \Bbb{R}$)} \tag{$\ddot{\smile}$} \end{align} Notice that the RHS is a polynomial, while the LHS is an exponential, hence can't be a polynomial. This is a contradiction.

If you want to be more explicit about where the contradiction lies, here's one approach: Suppose as a first case, that $n$ is even. Then, the RHS is a polynomial with odd degree; therefore it has a root (this is a simple exercise using the intermediate value theorem). However, the exponential function has no roots. This is a contradiction.

If on the other hand $n$ is odd, then the RHS will be an even degree polynomial. Now, since I wish to stay within the realm of real numbers and not invoke the fundamental theorem of algebra, here's a simple trick: let's integrate both sides of $(\ddot{\smile})$. Then, you'll find that \begin{align} \text{exponential} =\text{polynomial of odd degree} \qquad \text{(everywhere on $\Bbb{R}$)} \end{align} Thus, we're back to case 1. This completes the proof that there is no hope of finding a value of $c$ as you suggested.

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  • $\begingroup$ First of all, I believe you're missing $(x-a)^{n+1}$ in your definition of the remainder. I'm still a bit skeptical here. The first statement of the theorem is that "For all $x$ in $I$, $$f(x) = p_n(x) + R_n(x)$$ Is this statement just untrue? The part that really gets me is that it says "for all $x$ in $I$." $\endgroup$ – BSplitter Mar 25 at 2:28
  • $\begingroup$ @BSplitter I indeed missed that power. sorry, that's a typo. I'll edit my answer to include a version of how I'd phrase the theorem, so it hopefully makes it clearer what is being fixed vs varied, and how the universal and existential quantifiers are being used $\endgroup$ – peek-a-boo Mar 25 at 2:30
  • $\begingroup$ @BSplitter But it ends, "for some point $c$ between $x$ and $a.$" So even the range from which you can select $c$ is dependent on $x.$ $\endgroup$ – David K Mar 25 at 2:30
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    $\begingroup$ My previous comment was before the edits to this answer, which show that actually you can write $f(x) = p_n(x) + R_n(x)$ as an identity if you make that your definition of $R_n(x)$; then the rest of the theorem statement gives you something else that $R_n(x)$ happens to equal. What you can't do is to plug that expression back into the identity while treating $c$ as a constant. $\endgroup$ – David K Mar 25 at 3:01
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    $\begingroup$ @BSplitter btw, I actually thought of a simpler proof to show an exponential cannot equal a polynomial on any open interval $(-\epsilon, \epsilon)$ (so this is a simpler proof, and also a stronger statement than what I initially proved). If $e^x = \text{polynomial}$ for all $x \in (-\epsilon, \epsilon)$, then by differentiating sufficiently many times (greater than the degree of the polynomial), you'll get the absurd contradiction that $e^x = 0 $ for all $x \in (-\epsilon, \epsilon)$. $\endgroup$ – peek-a-boo Mar 26 at 11:20
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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 $x.$

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  • $\begingroup$ I think I like the final statement of the theorem you provide. That, to me, is the clearest way to write it. Thank you for the thoughtful answer! $\endgroup$ – BSplitter Mar 25 at 3:23
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    $\begingroup$ I understand why the author wanted to avoid the customary $\xi$, but $c$ seems to be a poor choice for something that is decidedly not constant but depends on $x$. $\endgroup$ – Carsten S Mar 25 at 12:29

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