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


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 $I$.


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 $u'(x)=f'(x)g(x)+f(x)g'(x)$.


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

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

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