Series reversion with constant term

Question

Suppose we have $y=k+ax+bx^2+cx^3...$ as an infinite series. In order to reverse this I have been taught that we can assume $y-k=z$

Let's suppose $x=Az+Bz^2+Cz^3...$ Where $A,B,C...$ are determined by their usual formulas. By substituting $y-k=z$ back we get $x=A(y-k)+B(y-k)^2+C(y-k)^3...$. Rearranging this we obtain $$x=(-Ak+Bk^2-Ck^3....)+y(A-2Bk+3Ck^2...)+y^2(B-3Ck...)$$ Now since each coefficient is an infinite series in itself and for many examples these series may diverge, how do we determine them?

Yup, even I thought of that seeing this https://math.stackexchange.com/questions/2440923/reverse-of-series-with-a-0-neq0 — , Jun 27 '18 at 07:27

score 2 · Answer 1 · answered Jun 27 '18 at 07:21

The terms are not an infinite series.

Suppose that we write $$y=k+\sum_{n=1}^{p>5}a_i x^i$$ and we make a Taylor expansion of it up to $O(x^5)$. Now, for more clarity, let $t=\frac{y-k}{a_1}$ and making the series reversion, we should get $$x=t-\frac{a_2 }{a_1}t^2+\frac{\left(2 a_2^2-a_1 a_3\right) }{a_1^2}t^3+\frac{\left(-5 a_2^3+5 a_1 a_3 a_2-a_1^2 a_4\right) }{a_1^3}t^4+O\left(t^5\right)$$

Series reversion with constant term

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