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I have come up with a proof of Taylor's formula but I am not sure that it is correct. Can you check for mistakes? Or give suggestion on my proof-writing?

Theorem: Let $f:]a,b[\rightarrow \mathbb{R}$ be a function that is differentiable $k\in \mathbb{N}$ times on all of its domain and let there be $x_0\in ]a,b[$. Then there exists one and only one polynomial $P(x)$ of degree at most $k$ such that $$\lim_{x\to x_0} \frac{f(x)-P(x)}{(x-x_0)^k}=0$$, and this polynomial is $$P(x)=\sum_{n=0}^{k}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n $$.

Proof: First we prove that if there is such polynomial, there it must be the Taylor one. Then that it in fact verifies the condition.

Let $P(x)=a_0+a_1(x-x_0)+...+a_k(x-x_0)^k$ be a polynomial of degree at most $k$ which satisfies the condition $f(x)-P(x)=o((x-x_0)^k)$. Since by hypothesis $f^{(k)}{(x)}$ exists and polynomials are differentiable infinitely many times, the ratio $$\frac{f^{(k)}(x)-P^{(k)}(x)}{k!}$$ approaches a finite limit as $x\to x_0$, namely $\frac{f^{(k)}(x_0)-P^{(k)}(x_0)}{k!}$. By L'Hopital rule, $$\lim_{x\to x_0}{\frac{f^{(k-1)}(x)-P^{(k-1)}(x)}{k!(x-x_0)} }$$ exists as well and is equal to the above limit. Applying the rule iteratively, we infer that $$ \lim_{x\to x_0}{\frac{f^{(n)}(x)-P^{(n)}(x)}{\frac{k!}{(k-n)!}[(x-x_0)^{k-n}]}} $$ exists for all $0\leq n \leq k$ and that all the limits are equal; but since by hypothesis $$\lim_{x\to x_0} \frac{f(x)-P(x)}{(x-x_0)^k}=0$$, then all the limits are equal to $0$. This implies that $f^{(n)}(x_0)-P^{(n)}(x_0)=0$, because, were it another real number, all the limits would be infinite. On the other hand, it is clear that $$P(x_0)=a_0, P'(x_0)=a_1, P''(x_0)=2a_2,...,P^{(k)}(x_0)=k!a_k$$, and thus $a_n=f^{(n)}(x_0)/n!$ for $n=0,1,...,k$. Hence $P(x)$ is the Taylor polynomial.

Now we must prove that $P(x)$ actually satisfies the condition $f(x)-P(x)=o((x-x_0)^k)$. Since $$\lim_{x\to x_0}\frac{f^{(k)}(x)-P^{(k)}(x)}{k!} =\frac{f^{(k)}(x_0)-P^{(k)}(x_0)}{k!}=\frac{f^{(k)}(x_0)-f^{(k)}(x_0)}{k!}=0 $$. By applying l'Hopital repeatedly again, we infer that $$\lim_{x\to x_0}\frac{f(x)-P(x)}{(x-x_0)^k}=0 $$, hence the thesis.

I'm not just afraid that this might be wrong, but also that it's poorly written mathematically. Can you give me tips/feedback?

Paramanand Singh
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Adrian
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1 Answers1

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Your proof looks almost correct but it has subtle flaws! However kudos if you have managed to come with the proof yourself without any external help. (+1 goes for your question). Your assertion that the limit $$\lim_{x \to x_{0}}\frac{f^{(k)}(x) - P^{(k)}(x)}{k!}$$ exists is not true unless we assume that $f^{(k)}$ is continuous. Next we come to the assertion that $$\lim_{x \to x_{0}}\frac{f^{(k - 1)}(x) - P^{(k - 1)}(x)}{k!(x - x_{0})}$$ exists and is equal to the limit $$\lim_{x \to x_{0}}\frac{f^{(k)}(x) - P^{(k)}(x)}{k!}$$ is also incorrect unless we have $f^{(k - 1)}(x_{0}) = P^{(k - 1)}(x_{0})$. Same problem exists in your iterative limits. You have failed to check the conditions of applicability of L'Hospital's Rule at each stage.

Note that the statement of the Taylor's theorem actually requires only the existence of $f^{(k)}(x)$ and not its continuity. In fact even much less hypotheses are needed. It is sufficient to just assume that $f^{(k)}(x_{0})$ exists.

There is another proof of this theorem which is not widely known and avoids the use of L'Hospital's Rule. See this related question. Also uniqueness of Taylor's polynomial is discussed in this answer.

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Paramanand Singh
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