# Argue for that the Taylor polynomial is equal to an arbitrary polynomial whenever $n \geq k$

Given an arbitrary polynomial

$$p(x) = a_0 + a_1x + a_2x^2 + ... + a_kx^k$$ I had to find the Taylor series centered around $$0$$. After a few calculations, I realized that the Taylor series for an arbitrary polynomial it just the polynomial itself which intuitively makes sense as the idea behind the Taylor series is to find the polynomial which best approximates a given function.

Then I afterwards had to argue for that the Taylor polynomial $$T_n(x) = p(x)$$ for all $$x \in \mathbb{R}$$ whenever $$n \geq k$$. At first I thought it was enough to just argue given what I said above but my TA has told me that I need to show that the difference between the Taylor series and $$p(x)$$ is zero.

I know that

$$p(x) = T_n(x) + Rnf(x)$$

which is equal to

$$Rnf(x) = p(x) - T_n(x)$$ so to show that this is zero have

$$Rnf(x) = p(x) - T_n(x) = a_0 + a_1x + a_2x^2 + ... + a_kx^k - (a_0 + a_1x + a_2x^2 + ... + a_kx^k) = 0$$ Is this all I need? I am just not sure whether or not I have assumed what I want to show which can be hard for me sometimes to understand.

That is circular: you are saying that $$p(x)=T_n(x)$$ because their diference is $$0$$ and that their difference is $$0$$ because $$p(x)=T_n(x)$$.

Consider the difference $$R_n(x)=p(x)-T_n(x)$$. Then $$R_n(x)$$ is a polynomial function whose degree is at most $$n$$. You can therefore write it as $$a_0+a_1x+a_2x^2+\cdots+a_nx^n$$. But:

• $$a_0=p(0)-T_n(0)=0$$;
• $$a_1=p'(0)-T_n'(0)=0$$;
• -$$\vdots$$
• $$\displaystyle a_n=\frac{p^{(n)}(0)}{n!}-\frac{T_n^{(n)}(0)}{n!}=0$$.

So, $$R_n(x)=0$$.

• Thanks for the help! :) Apr 1, 2020 at 14:10
• I'm glad I could help. Apr 1, 2020 at 14:16