# Show that the function is equal to its Taylor polynomial for all $x \in \mathbb{R}$ when $n \geq k$.

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.

• 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 $T_n(x) =p(x)$.
– Koro
Apr 21, 2020 at 21:20
• 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 .. Apr 21, 2020 at 21:34
• I think he might be referring to " Taylor polynomial coefficients $a_i$'s are 0 for$i\gt k$"
– Koro
Apr 21, 2020 at 22:10

## 1 Answer

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, $$p(x) = T_n(x)$$