Why does $\lim_{x \to 0} \frac{f(x) - Q(x)}{x^{2n + 5}} = 0$ prove that $Q$ is the Taylor polynomial of $f$?

Question

Today I posted this question about finding degree-$n$ Taylor polynomial of $\frac{x^5}{1+x^2}$. I just saw my teacher's solution to this question and this post is about something I don't understand on that solution.

So we have that, for $r \in \mathbb R$:

$$1 + r + r^2 + ... + r^n = \frac{1 - r^{n + 1}}{1-r}= \frac{1}{1 - r} - \frac{r^{n + 1}}{1 - r}$$

Now, note that:

$$\frac{1}{1+x^2} = \frac{1}{1 - (-x)^2} = 1-x^2 + x^4 +...+(-1)^nx^{2n} + \frac{(-1)^{n +1} x^{2n + 2}}{1 + x^2}$$

So this means that:

$$f(x) = \frac{x^5}{1 + x^2}= \underbrace{x^5-x^7 + x^9 +...+(-1)^nx^{2n + 5}}_{Q(x)} + \frac{(-1)^{n +1} x^{2n + 7}}{1 + x^2}$$

My teacher asked us if we think that $Q(x)$ is the degree $2n + 5$ Taylor polynomial of $f(x)$. We said yes, and she proved it in the following way:

For $Q(x)$ to be the degree $2n + 5$ taylor polynomial, we just need to show that:

$$\lim_{x \to 0} \frac{f(x) - Q(x)}{x^{2n + 5}} = 0$$

And this is my question: Why does this imply that $Q(x)$ is the Taylor polynomial of $f(x)$?

Does this answer your question? – Paramanand Singh Apr 18 '21 at 15:42 — Paramanand Singh, Apr 18 '21 at 15:42

José Carlos Santos · Answer 1 · 2021-04-18T15:07:29.570

1

Your teacher is using a standard Calculus theorem:

Theorem: Let $f$ be $n$-times differentiable at $a$, and suppose that $P$ is a polynomial on $(x-a)$ of degree $\leqslant n$, which equals $f$ up to order $n$ at $a$. Then $P$ is the Taylor polynomial of $f$ of order $N$ at $a$.

For instance, this appears as a corollary in Spivak's Calculus, chapter 20.

edited Apr 18 '21 at 15:07

answered Apr 18 '21 at 15:04

José Carlos Santos

427,504

What does it mean in this context for a polynomial to "equal $f$ up to order $n$ at $a$"? – Eduardo Magalhães Apr 18 '21 at 15:06
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It means that$$\lim_{x\to a}\frac{f(x)-P(x)}{(x-a)^n}=0.$$ – José Carlos Santos Apr 18 '21 at 15:10
@EduardoMagalhães: you may have a look at the proof of the theorem mentioned here: https://math.stackexchange.com/a/1925036/72031 – Paramanand Singh Apr 18 '21 at 15:17
@ParamanandSingh: Doesn't that qualify as a duplicate target? – Martin R Apr 18 '21 at 15:22
@MartinR: I did not think that the question requested a proof, but perhaps it would not hurt the asker if the explanation includes a formal proof as well. – Paramanand Singh Apr 18 '21 at 15:26
@MartinR Would you mind trying to find a duplicate of this question? I tried and failed. – José Carlos Santos Apr 18 '21 at 15:29
@JoséCarlosSantos: The claim in that question is wrong as stated, therefore I am reluctant to dupe-hammer. But I have left a comment with a link to what the author probably means. – Martin R Apr 18 '21 at 15:37
@MartinR Thank you. I missed the lack of conjugation sign in the statement. – José Carlos Santos Apr 18 '21 at 15:39
@MartinR: I have put a comment on this question to get feedback and then I can use dupe hammer. – Paramanand Singh Apr 18 '21 at 15:47

Why does $\lim_{x \to 0} \frac{f(x) - Q(x)}{x^{2n + 5}} = 0$ prove that $Q$ is the Taylor polynomial of $f$?

1 Answers1