# Proof Check: The Product rule via Taylor Series

I've been working on the following proof of the product rule for derivatives using Taylor series. Are there any holes? Thanks.

Setup:

Let $$I\subseteq \Bbb R$$ be some interval. Let $$f,g:\Bbb R\mapsto\Bbb R$$ both be continuous and differentiable on $$I$$. Suppose that $$\forall n\in A=\{0,1,2,...\}$$, $$f^{(n)}(x)$$ and $$g^{(n)}(x)$$ (the $$n$$-th derivatives of $$f(x)$$ and $$g(x)$$ respectively) are continuous on $$I$$.

Edit:

Suppose that the Taylor series of $$f(x)$$ converges to $$f(x)$$ on $$I$$, and that the Taylor series of $$g(x)$$ converges to $$g(x)$$ on $$I$$.

Theorem: If $$u(x)=f(x)g(x)$$, then $$u'(x)=f'(x)g(x)+f(x)g'(x)$$.

Proof:

Let $$m\in I$$ be some constant. $$\therefore f(x)=\sum_{n\in A}\frac{f^{(n)}(m)}{n!}(x-m)^{n}, g(x)=\sum_{n\in A}\frac{g^{(n)}(m)}{n!}(x-m)^{n}$$ Let $$f_n=\frac{f^{(n)}(m)}{n!}$$, $$g_n=\frac{g^{(n)}(m)}{n!}$$, which are constants for all $$n\in A$$. $$\therefore f(x)=\sum_{n\in A}f_n\cdot(x-m)^n, g(x)=\sum_{n\in A}g_n\cdot(x-m)^n$$ $$\therefore f'(x)=\sum_{n\in A}nf_n\cdot(x-m)^{n-1}, g'(x)=\sum_{n\in A}ng_n\cdot(x-m)^{n-1}$$ Let $$u(x)=f(x)g(x)$$ $$\therefore u(x)=(\sum_{n_1\in A}f_{n_1}\cdot(x-m)^{n_1})(\sum_{n_2\in A}g_{n_2}\cdot(x-m)^{n_2})$$ $$\therefore u(x)=\sum_{n_1\in A}\sum_{n_2\in A}f_{n_1}g_{n_2}(x-m)^{n_1+n_2}$$ $$\therefore u'(x)=\sum_{n_1\in A}\sum_{n_2\in A}f_{n_1}g_{n_2}(n_1+n_2)(x-m)^{n_1+n_2-1}$$ $$\therefore u'(x)=\sum_{n_1\in A}\Biggl((\sum_{n_2\in A}f_{n_1}g_{n_2}n_1(x-m)^{n_1+n_2-1})+(\sum_{n_3\in A}f_{n_1}g_{n_3}n_3(x-m)^{n_1+n_3-1})\Biggl)$$ $$\therefore u'(x)=\sum_{n_1\in A}\Biggl([n_1f_{n_1}(x-m)^{n_1-1}\sum_{n_2\in A}g_{n_2}(x-m)^{n_2}]+[f_{n_1}(x-m)^{n_1}\sum_{n_3\in A}n_3g_{n_3}(x-m)^{n_3-1}]\Biggl)$$ $$\therefore u'(x)=\sum_{n_1\in A}\Biggl(n_1f_{n_1}(x-m)^{n_1-1}g(x)+f_{n_1}(x-m)^{n_1}g'(x)\Biggl)$$ $$\therefore u'(x)=\Biggl(g(x)\sum_{k_1\in A}k_1f_{k_1}(x-m)^{k_1-1}\Biggl)+\Biggl(g'(x)\sum_{k_2\in A}f_{k_2}(x-m)^{k_2}\Biggl)$$ $$\therefore u'(x)=g(x)f'(x)+g'(x)f(x)$$ $$\therefore u'(x)=f'(x)g(x)+f(x)g'(x)$$ QED

• It is possible that $f$ be differentiable yet its Taylor series converge but not to $f$. – Randall Oct 5 '18 at 1:25
• @Randall honestly I have no idea. I hadn't considered that. Do such functions exist? – clathratus Oct 5 '18 at 1:26
• Yes, something like $e^{-1/x^2}$ made continuous at 0. – Randall Oct 5 '18 at 1:29
• @Randall Thanks. I'll update the proof then. – clathratus Oct 5 '18 at 1:31
• @Randall how about now? I just added more to the setup. – clathratus Oct 5 '18 at 1:34