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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?

Bruno B
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Herrpeter
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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 $P=Q$.

What you did is express the Taylor polynomial but the goal is not to prove existence but unicity; your way of thinking is good though but maybe overkill?

julio_es_sui_glace
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  • $(f-Q)(x-a) = o(|x-a|^m)$ what do you mean by $o$? – Herrpeter Jun 21 '23 at 10:43
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    If $g$ does not vanish near a point $a$ (but may vanish in $a$, $f = o_a (g)$ if $f/g \rightarrow 0$ in $a$ – julio_es_sui_glace Jun 21 '23 at 10:55
  • Note that the proof will strongly depend on your definition of the Taylor polynomials – julio_es_sui_glace Jun 21 '23 at 10:56
  • 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 – julio_es_sui_glace Jun 22 '23 at 13:04