# On the Definition of Taylor Polynomials

One definition of Taylor Polynomials proceeds, as follows.

Definition:

Let $$a \in \mathbb{R}$$. Let $$f$$ be a function continuous at $$a$$. Let $$n \in \mathbb{N}$$. The $$n$$th Taylor Polynomial for $$f$$ at $$a$$ is the polynomial, $$P_n$$ of the smallest possible degree, which is an approximation of $$f$$ near $$a$$ of order $$n$$. That is, $$\lim _{x \rightarrow a} \frac{f(x)-P_n(x)}{(x-a)^n}=0$$

I wish to ask why it was necessary to state of the continuity of $$f$$ at $$a$$.

Edit: Most namely, we know this limit to be satisfied when the first $$n$$ derivatives of $$f$$ and of $$P_n$$ agree. Yet, if we merely presume of $$f$$'s continuity at a point, how may guarantee that $$f$$ is $$n$$-times differentiable? Is it not more reasonable to presume of $$f$$'s being $$C^n$$ (by which I mean that all $$n$$ derivatives exist and are continuous)?

• If $f$ is not continuous at $a$, then $P_n(a)$ is not an approximation of $f(a)$. Jul 28, 2020 at 14:08
• It is not clear how approximation is defined in this context. Jul 28, 2020 at 14:13
• More specifically, we want $P_n(a)=f(a)$, otherwise $P_n$ would be a pretty bad approximation of $f$ around $a$. Jul 28, 2020 at 14:16
• What is the source for this definition? Jul 28, 2020 at 16:05

Let's assume that we don't know if $$f$$ is continuous at $$a$$. We only assume that $$\lim\limits_{x\to a}\frac{f(x)-P_n(x)}{(x-a)^n}=0$$. Then for an arbitrary $$\frac{\epsilon}{2}>0$$ we find a $$\delta'>0$$ (which depends on $$a$$) such that for all $$x\in\mathbb{R}$$ with $$|x-a|<\delta'$$ we have $$\Bigl|\frac{f(x)-P_n(x)}{(x-a)^n}\Bigl|<\frac{\epsilon}{2}$$. Let's define $$\delta:=\min\{\delta',1\}$$. Then it holds that:
$$\Bigl|\frac{f(x)-P_n(x)}{(x-a)}\Bigl|<\Bigl|\frac{f(x)-P_n(x)}{(x-a)^n}\Bigl|<\frac{\epsilon}{2}.$$
As polynomials are $$n$$-times differentiable we find a $$\delta''>0$$ for the above given $$\frac{\epsilon}{2}$$ such that for all $$x\in\mathbb{R}$$ with $$|x-a|<\delta''$$, we can conclude $$\Bigl|\frac{P_n(x)-P_n(a)}{(x-a)}\Bigl|<\frac{\epsilon}{2}.$$
If we the set $$|x-a|<\min\{\delta,\delta''\}$$ it yields: $$\Bigl|\frac{f(x)-P_n(a)}{(x-a)}\Bigl|\leq \Bigl|\frac{f(x)-P_n(x)}{(x-a)}\Bigl|+\Bigl|\frac{P_n(x)-P_n(a)}{(x-a)}\Bigl|<\epsilon.$$ If we then assume that approximation means $$P_n(a)=f(a)$$ then $$f$$ would be differentiable at $$a$$ with $$f'(a)=0$$. Hence, $$f$$ is always continuous at $$a$$ - you don't have to assume it. Without any further assumptions one cannot say if $$f$$ has higher order derivatives at $$a$$.
For one, if $$f$$ is continuous at $$a,$$ then $$\lim_{x \to a} f(x) = f(a).$$ But also, the statement $$\lim_{x \to a} \frac{f(x) - P_n(x)}{(x - a)^n} = 0$$ says that the difference function $$e(x) = f(x) - P_n(x)$$ tends to $$0$$ more quickly than the polynomial $$(x - a)^n$$ as $$x$$ approaches $$a.$$ Observe that the function $$e(x)$$ gives the error in approximating $$f(x)$$ via the polynomial $$P_n(x),$$ hence this limit gives a criterion for the approximation to be "good enough" in some rigorous sense.
• If we assume that $f(x)$ is $n$-times continuously differentiable in a neighborhood of $a,$ then the Taylor polynomial would not be as powerful. What is important is the $f(x)$ can be quite general (e.g., it need only be continuous at $a$), so the Taylor polynomial is an extremely useful tool. Jul 28, 2020 at 14:49