# Proving that a Polynomial Equals the $m$-th Degree Taylor Polynomial of a Function based on a Given Limit Expression

Let $$f \in C^{m+1}(\mathbb{R}^n, \mathbb{R})$$ be a function, and let $$P: \mathbb{R}^n \rightarrow \mathbb{R}$$ be a polynomial of degree less than or equal to $$m$$. Suppose $$a \in \mathbb{R}^n$$ satisfies $$\lim_{{x \to a}} \frac{\left| f(x) - P(x) \right|}{\|x - a\|^m} = 0$$ Show that $$P$$ must necessarily be equal to the $$m$$-th degree Taylor polynomial of $$f$$ at $$a$$.

My attempt: To show that $$P$$ must be equal to the $$m$$-th degree Taylor polynomial of $$f$$ at $$a$$, we'll utilize the Taylor's theorem with the remainder term in the Lagrange form. The theorem states that for a function $$f$$ which is $$(m+1)$$ times differentiable on an interval containing $$a$$ and a point $$x$$ in that interval, there exists a point $$c$$ between $$a$$ and $$x$$ such that: $$\begin{split}f(x) &= f(a) + f'(a)(x - a) + \frac{2!}{2!}f''(a)(x - a)^2 + \ldots + \frac{m!}{m!}f^{(m)}(a)(x - a)^m\\ &\quad+ \frac{(m+1)!}{(m+1)!}f^{(m+1)}(c)(x - a)^{m+1}\end{split}$$ In this case, we want to show that $$P(x)$$, which is a polynomial of degree less than or equal to $$m$$, agrees with the Taylor polynomial of $$f$$ up to the $$m$$-th degree terms.

Is my way of thinking correct? How can I continue this proof?

Take $$Q$$ your $$m$$-degree Taylor polynomial of $$f$$, you get $$(f-Q) (x-a) = o (|x-a|^m)$$ Just use the difference: $$\frac{|(P-f +f -Q)(x-a)|}{|x-a|^m} \leqslant \frac{|(P-f)(x-a)|}{|x-a|^m} + \frac{|(f -Q)(x-a)|}{|x-a|^m} \rightarrow 0$$ With that you get $$R=(P-Q)(X-a)$$ a polynomial of degree $$\leqslant m$$ that verifies $$\frac{R(x)}{x^m} \rightarrow 0$$ for $$x \rightarrow 0$$. Which means $$R=0$$ hence $$P=Q$$.
• $(f-Q)(x-a) = o(|x-a|^m)$ what do you mean by $o$? Jun 21, 2023 at 10:43
• If $g$ does not vanish near a point $a$ (but may vanish in $a$, $f = o_a (g)$ if $f/g \rightarrow 0$ in $a$ Jun 21, 2023 at 10:55
• I must add a warning, you used multidimensional variables, the Taylor polynomials don’t exist for them, it is much more complicated. You could define it with an alternative “polynomial” on the bar resolution $\bigcup E^{\otimes n}$ with product defined by distributivity on the tensor product… But you better stick to the one dimensional thing Jun 22, 2023 at 13:04