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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.

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    $\begingroup$ How are Taylor series expansions proved in Calculus? $\endgroup$ – Somos Jul 27 '18 at 19:31
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A general reference for formal power series (f.p.s.) in one and several variables would be Wikipedia. But substitution in the context of f.p.s in several variables is not treated there.
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.

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