I have a quick question about Taylor's Theorem.


For an infinitely differentiable function, if the error term in Taylor's Theorem tends to zero, does the existance of a Taylor series follow from the fact the series and the partial sum to N can differ by at most $R_N$ (where this is the error from Taylor's Theorem), which tends to 0, meaning we can make the difference as small as we like and hence they are the same?

I hope that is clear. If not please let me know. Feel free to also post short comments as answers.

  • $\begingroup$ No, this is not clear. Error term tends to zero when $x$ tends to $x_0$ or when $N \to \infty$? Use mathjax to state your question clearly. $\endgroup$ Commented May 27, 2022 at 22:22

1 Answer 1


I agree with the comments that your question is not clear, but what I've written what I think you are asking below. If I am right, the answer is no. This page is helpful:

Non-analytic Smooth Function.

My understanding of your question is: Assuming that $f(x)$ is infinitely differentiable, and $f(x_0) - P_n(x_0) \to 0$ as $n \to \infty$, where $P_n$ is the $n^{th}$ Taylor polynomial for $f(x)$ at $x_0$, can we conclude that $f$ has a Taylor expansion at $x_0$?

The page linked above gives a classic example of an infinitely differentiable function all of whose derivatives are $0$ at $x = 0$, but the function is not equal to $0$ in any neighborhood of $x = 0$. All of the Taylor polynomials will therefore be $0$ at $x = 0$, so the difference $f(0) - P_n(0) = 0$ for all $n$. On the other hand, $f(x) \neq 0$ on any neighborhood of $0$, so $f(x)$ can't be equal to its Taylor expansion at that point.

  • $\begingroup$ Hi, thank you for the response. What I meant is that the error term tends to zero. These include error terms of the type of Lagrange's Error Term or Cauchy's Error Term. $\endgroup$ Commented May 28, 2022 at 12:15
  • $\begingroup$ Could you specify the exact question you would like to answer (by editing the post)? This might also be helpful for you in answering the question - as you write down the exact proposition you're hoping to prove/disprove, you might realize that you can actually answer it. $\endgroup$ Commented May 28, 2022 at 18:32

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