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How can I prove that a Taylor series represents a function $f(x)$ in $\mathbb{R}$ given only the form of the derivatives and the Taylor series of $f$?

$f(x)$ is infinitely differentiable, I know the nth derivation, as well as the form of the Taylor series $T(x)$. I have to show that the remainder converges to zero. The form of the function $f(x)$ is not given, so considering the limit of the remainder $R_{n,f,a}(x)=f(x)-T_n(x)$ is not possible.

My idea:

In such a case, does it make sense to consider the limit of the Lagrangian form of the remainder: Let $x\in \mathbb{R}$ be arbitrary. There is a $\xi\in (a,x)$ such that: $R_{n,f,a}(x)=\frac{f^{n+1}(\xi)}{(n+ 1)!}(x-a)^{n+1}$

Can one conclude from $\frac{f^{n+1}(\xi)}{(n+1)!}(x-a)^{n+1}\rightarrow 0$ that $T(x)$ the representation of $f(x)$ for all $x\in \mathbb{R}$?

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  • $\begingroup$ There is a result by Borel that states that for any sequence $a_n$ in $\mathbb{R}$ there is $\phi\in\mathcal{C}_\infty$ such that the coefficients of the Taylor polynomial of any degree fo $\phi$ are given by $a_n$: $\phi^{(k)}(0)=a_k k!$. That does not mean that $\sum^\infty_{n=0}a_nx^k=\phi(x)$ in an interval around $0$. The assignation of sequence to function in $\phi$ is not one to one. $\endgroup$
    – Mittens
    Oct 24, 2022 at 18:45

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But this just isn't true. It is known that the class of smooth functions (=infinitely differentiable) is not the same as the class of (real) analytic functions.

The function: $$\Bbb R\ni x\mapsto\begin{cases}\exp(-x^{-2})&x\neq0\\0&x=0\end{cases}\in\Bbb R$$Is smooth and its Taylor expansion at zero is identically zero. Yet, this function is not identically zero on any neighbourhood of zero. Thus it is not analytic.

Anyway - you need more information, such as an bound on the remainder terms. For example, having bounded derivatives is a sufficient condition.

Proof that the function's Taylor coefficients are all zero:

Denote this function as $f$. $$\lim_{h\to0}\frac{f(h)-f(0)}{h}=\lim_{h\to0}\frac{1}{h}\exp(-h^{-2})=\lim_{|t|\to\infty}t\exp(-t^2)=0$$

And $f'(x)=2x^{-3}\exp(-x^{-2})=P_1(x^{-1})\exp(-x^{-2})$ for some polynomial $P_1$, on $x\neq0$.

Given that $f^{(n)}(x)$ exists and equals $P_n(x^{-1})\exp(-x^{-2})$ for some polynomial $P_n$, $x\neq0$, and $f^{(n)}(0)=0$, then I claim $f^{(n+1)}$ exists and equals $P_{n+1}(x^{-1})\exp(-x^{-2})$ for $x\neq0$, and is zero at $x=0$.

$$f^{(n+1)}(0)=\lim_{h\to0}\frac{1}{h}P_n(h^{-1})\exp(-h^{-2})=\lim_{|t|\to\infty}tP_n(t)\exp(-t^2)\overset{L.H.}{=}0$$

And $f^{(n+1)}(x)$, $x\neq0$, exists and equals: $$2x^{-3}P_n(x^{-1})\exp(-x^{-2})-x^{-2}P_n(x^{-1})\exp(-x^{-2})=P_{n+1}(x^{-1})\exp(-x^{-2})$$For some new and obviously defined polynomial $P_{n+1}$.

It follows by induction that $f^{(n)}(0)=0$ for all $n$.

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  • $\begingroup$ Thank you! So it would be sufficient if I found a null sequence $(a_n)$, such that for any $x\in\mathbb{R}$ and for all $n\in\mathbb{N}$: $$R_{n,f,a}(x)=\frac{f^{n+1}(\xi)}{(n+ 1)!}(x-a)^{n+1}\leq a_n$$ Because of the monotony of the limit $R_{n,f,a}(x)\rightarrow 0$ follows directly from $a_n\rightarrow 0$ . Since $x$ was chosen arbitrarily, it follows that $T(x)=f(x)$ in $\mathbb{R}$. Correct? Sry, for the many edits of my comment. $\endgroup$ Oct 24, 2022 at 18:02
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    $\begingroup$ It's impossible to get this for any $x\in\Bbb R$, unless the coefficients are identically zero,... but it is necessary to get this for all $x$, $|x-a|<r$ where $r$ is fixed (you should also have an absolute value sign) @distortedPicture $\endgroup$
    – FShrike
    Oct 24, 2022 at 18:05
  • $\begingroup$ 1)If $a=0$, $r>0$ is fixed and $|x-a|=|x|<r$, then $r$ must be "equal" to infinity if I want to show $f(x)=T(x)$ for all $x$. Otherwise one could always find a $x'\in \mathbb{R}$ with $|x'|>r$. Is $r=\infty$ allowed? $$$$ 2) Why do I need an absolute value signs? $|R_{n,f,a}(x)|=|\frac{f^{n+1}(\xi)}{(n+ 1)!}(x-a)^{n+1}|\leq |a_n|$ $$$$ 3) "It's impossible to get this for any $x\in \mathbb{R}$ " I don't understand why $|R_{n,f,a}(x)|\leq |a_n|$ is not possible for all $x$? $\endgroup$ Oct 24, 2022 at 18:26
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    $\begingroup$ @distortedPicture 1) Not quite. I appreciate my $r$ might look like a radius of convergence but it isn’t really. Suppose, for any finite fixed $r>0$, you can find a sequence $a_n$. Then $f(x)=T(x)$ for all $x$, because you can let $r>|x-a|$, consider $a_n$, and show the Taylor polynomial converges to $f(x)$. 2) You need an absolute value sign because the error needs to be bounded in magnitude. If you bound it only from above, it can still diverge to $-\infty$. 3) Let me rewrite the inequality as $|c_n(x-a)^n|<a_n$ for all $x$. As $|(x-a)^n|\to\infty$ as $x$ does ($n$ fixed) this cannot be true $\endgroup$
    – FShrike
    Oct 24, 2022 at 18:35
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    $\begingroup$ @distortedPicture To help you with $(1)$ and $(3)$. The Taylor series for $e^x$ converges everywhere. However, your inequality means to say it converges uniformly everywhere - this is false. It does converge uniformly on bounded subsets though, ie when one fixes an $r$. Oh, and I should have said for $(3)$ that $c_n\neq0$ is needed. $\endgroup$
    – FShrike
    Oct 24, 2022 at 18:37

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