# Taylor expansion for formal power series

Let $P = p_1x + p_2x^2 + \dots$, $Y = y_1x + y_2x^2 + \dots$, and $V = v_1x + v_2x^2 + \dots$ all be formal power series with indeterminate $x$ and coefficients in some field $\mathbb{F}$, satisfying $p_1,y_1,v_1 \neq 0$. How can we prove that

$$P(Y) = P(V) + P'(V)(Y-V) + \frac{1}{2!}P''(V)(Y-V)^2 + \frac{1}{3!}P'''(V)(Y-V)^3 + \dots$$

holds formally, i.e. the coefficients of $x^k$ on both sides are equal for all $k$? This is the Taylor expansion for formal power series, but I can't seem to find any reference proof of this result.

• How are Taylor series expansions proved in Calculus? – Somos Jul 27 '18 at 19:31

So, following Somos' ideas, we can do the following. We start with two independent indeterminates $u,v$ to get $$P(v)=P(u+(v-u))=\sum_n a_n (v-u)^l=\sum_na_n\sum_l\binom{n}{l}u^{n-l}(v-u)^l=\\\sum_l\left(\frac{1}{l!}\sum_n (n)_l a_n u^{n-l}\right)(v-u)^l=\sum_l\frac{P^{(l)}(u)}{l!}(v-u)^l,$$ where $(n)_l:=\prod_{j=0}^{l-1} (n-j)$.
Then the result follows by substituting $V$ for $u$ and $Y$ for $v$ in this formula.