I've encountered the following statement: Let $f$ and $g$ be analytical functions, then define $k(x) = f(x)g(x)$. Let $g(x) =a_0+ a_1x + a_2\frac{x^2}{2!} +R_2(x)$ be its Taylor expansion at $x=0$. Now they claim that $$ k(x)= f(x)a_0+ f(x)a_1x + f(x)a_2\frac{x^2}{2}! +f(x)R_2(x) = T_n(x)+R_n(x). $$ My question is it true (and why?) that $T_n(x)$ is Taylor expansion of order $n$ of $k(x)$?

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    $\begingroup$ Well the first equality should be rather obvious. You just multiply $g$ by $f$ directly. The last equality is a bit hardy and follows from Cauchy products. $\endgroup$ – Simply Beautiful Art May 16 '17 at 22:23
  • $\begingroup$ Thank you. I can see why the RHS is a polynomial representation of $k(x)$ but how do I know that it is Taylor expansion and not some other series? $\endgroup$ – V. Vancak May 16 '17 at 22:27
  • $\begingroup$ Merten's theorem holds your answer. $\endgroup$ – Simply Beautiful Art May 16 '17 at 22:52
  • $\begingroup$ This doesn't make sense as stated. What is this $n$ you speak of? $\endgroup$ – zhw. May 16 '17 at 23:10

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