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Let us consider Taylor's Theorem in the case of one real variable. For simplicity let us assume that $f$ is infinitely differentiable about $a$. Then we have that

$$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(k)}(a)}{k!}(x-a)^k + h_k(x)(x-a)^k$$

My question is why does the remainder term converge to $0$? In particular, why do we have that

$$ \lim_{x \to a} h_k(x) = 0$$

as well as that

$$ \lim_{k \to\infty} h_k(x) = 0?$$

If I understood this last statement, in particular, then I would understand why

$$f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!} (x-a)^k$$

which has also been a source of confusion for me.

davidlowryduda
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user1770201
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  • Take $f(x)=\exp(x)$. Obviously, the corresponding $h_{k}$ evaluated at $x=a$ is never zero. The last equality is true if the function you consider is an analytical function. If it is just $C^{\infty}$ there are counter examples showing that $f(x)$ is not equal its power series even for a very small neighbourhood around $x=a$.... – Alex Aug 24 '16 at 00:04
  • Your first result $\lim_{x \to a}h_{k}(x) = 0$ is true provided that $f^{(k)}(a)$ exists (and no more conditions are needed for that). This result is called Taylor's Theorem with Peano's form of remainder. See http://math.stackexchange.com/q/1809293/72031 The second result may or many not be true depending on the specific function $f$ and the specific points $a$ and $x$. – Paramanand Singh Aug 24 '16 at 08:16

3 Answers3

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There are two questions here. Firstly, you ask why $h_k(x) \to 0$ as $x \to a$.

I think the easiest way to see this is to use l'Hopital's rule. So consider $$ \frac{f(x) - (f(a) + f'(a) (x-a) + \cdots + f^k(a) (x-a)^k / k!)}{(x-a)^k} = h_k(x).$$ Let us evaluate the left hand side as $x \to a$, and in so doing evaluate the limit $\lim_{x \to a} h_k(x)$.

Clearly the left hand side appears like $0/0$, and differentiating once gives another limit in the indeterminate form $0/0$. But after applying l'Hopital's rule $k$ times, the left hand side becomes $$ \lim_{x \to a} \frac{f^k(x) - f^k(a)}{k!} = 0.$$ This answers your first question.

Your second question is why $h_k(x) \to 0$ as $k \to \infty$. And the short answer is that this is not always true, even if $f$ is infinitely differentiable. Sometimes it's true, but sometimes it is not. Functions for which this is true for every $a$ are called analytic and are very special --- in many ways, they are the nicest functions.

As an aside, I would point out that I wrote a note for my students some years ago in an attempt to give a somewhat better look at what Taylor series and polynomials really are. In it, I discuss the concept of the error terms in ways that some of my students have found very helpful.

davidlowryduda
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Here, we present a straightforward approach to proving that the Peano form of the remainder, $h_k(x)$, approaches zero as $x\to a$. To that end, we proceed.

Let $f(x)$ be $k$ times continuously differentiable and let $h_k(x)$ be defined as

$$\bbox[5px,border:2px solid #C0A000]{h_k(x)=\frac{f(x)-\sum_{j=0}^k \frac{f^{(j)(a)}}{j!}(x-a)^j}{x^k}}$$

Then, applying L'Hospital's Rule successively $k$ times reveals

$$\begin{align} \lim_{x\to a}h_k(x)&=\lim_{x\to a}\frac{f(x)-\sum_{j=0}^k \frac{f^{(j)(a)}}{j!}(x-a)^j}{x^k}\\\\ &=\lim_{x\to a}\frac{f'(x)-\sum_{j=1}^k \frac{f^{(j)(a)}}{(j-1)!}(x-a)^{j-1}}{x^k}\\\\ \vdots\\\\ &=\lim_{x\to a}(f^{k}(x)-f^{k}(a))\\\\ &=0 \end{align}$$

Therefore, we find that

$$\bbox[5px,border:2px solid #C0A000]{f(x)=\sum_{j=0}^k \frac{f^{j}(a)}{j!}(x-a)^j+h_k(x)(x-a)^k}$$

where $\lim_{x\to a}h_k(x)=0$.

Mark Viola
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Do you understand that this is not always true? There exist infinitely differentiable function such that the Taylor series exist and converges for all x but not to the original function!

$f(x)= e^{-\frac{1}{x^2}}$ if x is not 0, f(0)= 0, is such a function. It is not difficult to show that every derivative is $e^{-\frac{1}{x^2}}$ times a polynomial in x, so every derivative has value 0 at x= 0. That is, the Taylor's series, evaluated at 0, is identically 0 but the function is clearly not always 0.

user247327
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    This is half right, half wrong: you do always have $h_k(x) \to 0$ as $x \to a$. –  Aug 24 '16 at 00:30