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For the function $f(x) = \begin{cases} \ e^{\frac{-1}{x^2}} & x\neq0,\\\\ 0 & x=0 \end{cases}$

Let $R_n(x)$ denote Lagrange's form of remainder. Is $\lim_{n\to\infty} R_n(x)=0$

Lagrange's form of remainder is $R_n(x)=f(x)-\sum_{k=0}^{n}\frac1{k!}f^{(k)}(a)(x-a)^k$.

But I am stuck on calculating the $n+1^{th}$ derivative of the give function and thereby applying limits.

Please help me in proving or disproving the given statement.

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  • $\begingroup$ I don't understand what you mean. What's $a$ here? Is $limR_n(x)=0$ supposed to be $\lim_{x\to a} R_n(x)=0$ ? Is it $\lim_{n\to\infty} R_n(x)=0$ ? Pointwise? How is "Lagrange's form" relevant? $R_n(x)=f(x)-\sum_{k=0}^{n}\frac1{k!}f^{(k)}(a)(x-a)^k$ regardless of whether or not you observe that there is some $\xi(x)$ such that yada yada. And none of the limits I've mentioned depend on $\xi(x)$. $\endgroup$ Nov 16, 2023 at 12:23
  • $\begingroup$ Lagrange's form is given in the question. We need to calculate it's limit pointwise. Sorry, I misunderstood the formula. I will make changes $\endgroup$ Nov 16, 2023 at 12:34
  • $\begingroup$ What you've written is not Lagrange's form of the remainder. $\endgroup$ Nov 16, 2023 at 13:12
  • $\begingroup$ I don't know much about Lagrange's remainder. I got these formulae from another site . $\endgroup$ Nov 16, 2023 at 18:39

1 Answer 1

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What you've written as $R_n(x)$ is Tailor ramainder. The Lagrange form of the remainder comes from a theorem that ensures there is a point $c$ between $a$ and $x$ such that: $$ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1} $$ Then you shall prove that the expression at the right converges to $0$, which requires finding a general bound for the term $f^{(n+1)}(c)$.

However, the function you defined as $f$ is a classic example in analysis as a function that is $C^n$ but the Taylor polynomial centered at $0$ doesn't converges to $f$ at any point other than $x=0$. This is proved by checking that, for every $n$, $f^{(n)}(x) = p_n(\frac{1}{x^2})e^{\frac{-1}{x^2}}$, for some polynomial $p_n$.

I hope my answer is something like what you wanted, and that it helped you.

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  • $\begingroup$ Thanks a lot. I got it $\endgroup$ Nov 18, 2023 at 3:31

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