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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)?

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

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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 $a$.

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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.

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  • $\begingroup$ I thank you for your response. Might you please refer to the questions I have posed in my recent edit? $\endgroup$ Jul 28, 2020 at 14:27
  • $\begingroup$ 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. $\endgroup$ Jul 28, 2020 at 14:49

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