Let $x_0 \in \mathbb{R}$ and let $T_n(x)$ be a Taylor polynomial for $p(x) = a_0 + a_1x + ... + a_{k-1}x^{k-1} + a_kx^k$ of order n about $x_0$. Then I have to argue that $$ T_n(x) = p(x) \ \forall \ x \in \mathbb{R} \ \text{whenever} \ n \geq k $$ I have tried to solve this several times but my TA has told me that I was not right in my argumentation and has asked me to use a sentence in the my Analysis book.

From the definition, in my book, I know that the remainder is defined as

$$ R_nf(x) = \frac{1}{n!} \int_{x_0}^x f^{(n+1)}(t)(x-t)^n dt $$ but to use this formula I need to ensure that $I \subseteq \mathbb{R}$ is an open interval, that $x_0 \in I$ and lastly that $f \in C^\infty(I)$. Is it correct that every polynomial f satisfies $f \in C^\infty(I)$? I think it is but I am not sure. Can you verify? And if so to argue for that $T_n(x) = p(x)$ which must mean that $R_nf(x) = 0$ is the following calculation correct? $$ R_nf(x) = \frac{1}{n!} \int_{x_0}^x f^{(n+1)}(t)(x-t)^n dt = \frac{1}{n!} \int_{x_0}^x 0(x-t)^ndt = 0 $$ As I have calculated that $f^{(n)} = n!a_k$ which must mean that $f^{(n+1)} = 0$ and thus the above integral must equal to zero. Is this approach alright?

Note: I got this answer last time I asked Argue for that the Taylor polynomial is equal to an arbitrary polynomial whenever $n \geq k$ but my TA said that this was not enough.

  • $\begingroup$ You must first show that $T_n(x) $and $p(x) $ are equal upto order n whenever $n\ge k$. For that, show that $\lim_{x\to x_o} \frac{R_n(f)} {(x-x_o)^n} =0$, which implies $T_n(x) =p(x)$. $\endgroup$
    – Koro
    Apr 21, 2020 at 21:20
  • $\begingroup$ So I am not just able to calculate the above integral? My TA told me to specifically use a sentence in my book and the above is the only I can find which makes any sense in terms of what I want to show .. $\endgroup$
    – Mathias
    Apr 21, 2020 at 21:34
  • $\begingroup$ I think he might be referring to " Taylor polynomial coefficients $a_i$'s are 0 for$ i\gt k$" $\endgroup$
    – Koro
    Apr 21, 2020 at 22:10

1 Answer 1


Note that, whenever $n\ge k$, we have $\lim_{x\to x_o} \frac{R_n(f)} {(x-x_o)^n} =0 \tag{1}$ ,This is because of Taylor polynomial coefficients
$a_i= p^{(i)} /i! =0$ for $i\gt k$.
If $p(x) \ne T_n(x) $, let $R(x) = p(x) - T_n(x) \;\text{[note that degree of p(x) is $\le$ k and that of $T_n$ is $\le n$] } $
Therefore assume that, $R(x)=c_o+c_1(x-x_o) +c_2(x-x_o)^2+...+c_n(x-x_o)^n$
Using (1), $c_n=... =c_2=c_1=0$. Hence, $p(x) = T_n(x) $


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