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Suppose that $f$ is $n$ times differentiable at $x_0$ and $p$ is a polynomial: $$p(x) = a_0 +a_1(x-x_0)+...+a_n(x-x_0)^n$$ of degree $\leq n$ such that $$\lim\limits_{x \rightarrow x_0} \frac{f(x)-p(x)}{(x-x_0)^n} =0$$ Show that $a_r= \frac{f^{(r)}(x_0)}{r!}$ if $0 \leq r \leq n$, that is $p=T_n$, the $n^{th}$ Taylor polynomial of $f$ about $x_0$.

First, $f^{(r)}(x_0)$, with $0 \leq r \leq n$, does exist according to the hypothesis.

If n=1, let's check that $a_1 =\frac{f'(x_0)}{1!}=f'(x_0)$

$$\lim\limits_{x \rightarrow x_0} \frac{f(x)-p(x)}{(x-x_0)} =0$$

$$\lim\limits_{x \rightarrow x_0} \frac{f(x)-p(x)}{(x-x_0)} \cdot \lim\limits_{x \rightarrow x_0} (x-x_0)=0 \cdot \lim\limits_{x \rightarrow x_0} (x-x_0)$$

$$\lim\limits_{x \rightarrow x_0} f(x)-p(x) = 0$$ $$\lim\limits_{x \rightarrow x_0} f(x) = \lim\limits_{x \rightarrow x_0} p(x)$$ $$\lim\limits_{x \rightarrow x_0} f(x) = \lim\limits_{x \rightarrow x_0} [a_0 +a_1(x-x_0)]$$ Taking the limit with $f(x)$ and $p(x)$ continuous $$f(x_0) = a_0 +a_1(x_0-x_0)$$ $$f(x_0) = a_0$$ Finding $a_1$: $$\lim\limits_{x \rightarrow x_0} \frac{f(x)-p(x)}{(x-x_0)} =0$$ $$\lim\limits_{x \rightarrow x_0} \frac{f(x)-a_0-a_1(x-x_0)}{(x-x_0)} =0$$ $$\lim\limits_{x \rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0} =\lim\limits_{x \rightarrow x_0} \frac{a_1(x-x_0)}{x-x_0}$$ $$\lim\limits_{x \rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0} =a_1 \cdot \lim\limits_{x \rightarrow x_0} \frac{(x-x_0)}{x-x_0}$$ $$\lim\limits_{x \rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0} =a_1 \cdot \lim\limits_{x \rightarrow x_0} 1$$ $$\frac{f'(x_0)}{1!} =a_1 $$

Assumption $P_{r-1}$: $a_{r-1}= \frac{f^{(r-1)}(x_0)}{(r-1)!}$ to be true such that $$g_{r-1}=\lim\limits_{x \rightarrow x_0} \frac{f(x)-p_{r-1}(x)}{(x-x_0)^{r-1}} =0$$

Finding $a_r$

$$\lim\limits_{x \rightarrow x_0} \frac{f(x)-p_r(x)}{(x-x_0)^r} =0$$ $$\lim\limits_{x \rightarrow x_0} \frac{f(x)-p_{r-1}(x)-a_r(x-x_0)^r}{(x-x_0)^r} =0$$ $$\lim\limits_{x \rightarrow x_0} \frac{f(x)-p_{r-1}(x)}{(x-x_0)^r} = a_r \lim\limits_{x \rightarrow x_0} \frac{(x-x_0)^r}{(x-x_0)^r}$$ $$\lim\limits_{x \rightarrow x_0} \frac{f(x)-p_{r-1}(x)}{(x-x_0)^r} = a_r $$

Given $f(x) - p(x) = R(x)$ with $R(x)$ a polynomial of degree $r$, we have $R(x)=o((x-x_0)^r)$. Therefore as $x \rightarrow x_0$, $R(x) \rightarrow 0$.

Given $f(x) - p_{r-1} =R_{r-1}$ with $R_{r-1}(x)$ a polynomial of degree $r-1$, we have $ \lim\limits_{x \rightarrow x_0} R(x) = \lim\limits_{x \rightarrow x_0} R_{r-1}(x-x_0) = 0$

$$\lim\limits_{x \rightarrow x_0} \frac{f(x)-p(x)}{(x-x_0)^r} = a_r $$ $$ \lim\limits_{x \rightarrow x_0} \frac{[f(x)-p(x)]^{(r-1)}}{[(x-x_0)^r]^{(r-1)}} = a_r $$ $$ \frac{1}{r!} \lim\limits_{x \rightarrow x_0} \frac{ f^{(r-1)}(x)-p^{(r-1)}(x) } {x-x_0} = a_r $$ $$ \frac{1}{r!} \lim\limits_{x \rightarrow x_0} \frac{ f^{(r-1)}(x)-f^{(r-1)}(x_0) } {x-x_0} = a_r $$ $$ \frac{1}{r!} \cdot f^{(r)}(x_0) = a_r $$

I understood that you are supposed to apply L'hospital rule and do the $(r-1)^{th}$ derivative, but I unsure of the correctness of argumentations of the last 5 lines above (or if it is correct, what would the justifications be?) leading to the end result.

$$ \frac{f^{(r)}(x_0)}{r!}=a_r$$

?

gegu
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  • You forgot $n=0$. – Arthur Jul 18 '17 at 05:49
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    I rarely advocate for it, but this seems like a perfect opportunity for repeated l'Hopital's. The proof is about two or three sentences long. – adfriedman Jul 18 '17 at 06:25
  • @adfriedman Thx for the input. At which level would you apply l'Hospital? – gegu Jul 18 '17 at 06:36
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    This is basically Taylor's Theorem. See this link. – Fimpellizzeri Jul 18 '17 at 06:58
  • @Fimpellizieri No, it's not. Let $T_n(x)$ denote the $nth$ order Taylor approximation of $f$ at $x_0$. Taylor's theorem asserts that $f(x) - T_n(x) \in o((x-x_0)^n)$ whereas this problem asserts that if there is a degree $\leq n$ polynomial $p$ such that $f(x) - p(x) \in o((x-x_0))^n$, then $p(x) = T_n(X)$. Subtle difference. – MathematicsStudent1122 Jul 18 '17 at 20:59
  • @MathematicsStudent1122 Taylor's theorem asserts that $T_n$ is asymptotically unique, in a sense. Perhaps you're not familiar with the full version? – Fimpellizzeri Jul 18 '17 at 21:05
  • @Fimpellizieri I did a new edit for the last 5 lines. am i going in the right direction? Trying to read the "proof on the uniqueness .." i do not see the relation with this proof. – gegu Jul 18 '17 at 21:11
  • Have you checked the answer in the linked duplicate? It has a simple proof of what you're asking. – Fimpellizzeri Jul 18 '17 at 21:28
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    Ok your argument needs to be polished a bit with some minor corrections. And it is best that you try to understand the answer I gave at https://math.stackexchange.com/a/1925036/72031 this answer also works in the same manner like yours but it includes all the fine details and subtleties and these are necessary for a correct proof. If you have any problem understanding the answer you can post a comment for more information. – Paramanand Singh Jul 19 '17 at 02:10
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    @MathematicsStudent1122: you can see that under the given conditions the polynomial $p(x) - T_{n} (x) $ is also $o((x-x_{0})^{n})$ and my answer to the linked question proves that any polynomial of degree $n$ or less with $o((x-x_{0})^{n})$ behavior has to be the zero polynomial so that $p(x) =T_{n} (x) $. So you see there is no subtle difference. The way I have written the proof it relies on Taylor's theorem instead of L'Hospital's Rule. Here OP is using L'Hospital's Rule which is also correct. – Paramanand Singh Jul 19 '17 at 02:22
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    @ParamanandSingh Yes but you still need to do extra work. You need to prove a "lemma" that if $R(x)$ is a degree $\leq n$ polynomial such that $R(x) \in o((x-x_0)^n)$ then $R(x) = 0$ identically. Once you have proved this, uniqueness follows from Taylor's theorem. – MathematicsStudent1122 Jul 19 '17 at 02:36
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    @MathematicsStudent1122: it's like proving $A=B$ starting with $A$ and manipulating it to get $B$ versus showing $A-B=0$. The approaches are same and none is more difficult than the other. But one may like one approach better based on personal preference. – Paramanand Singh Jul 19 '17 at 04:19
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    Related: Is this proof of Taylor's theorem correct and mathematically well written? – Winther Jul 19 '17 at 20:50

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