Reversion of Power Series

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Theorem

Let $\ds y = \sum_{n \mathop = 1}^\infty c_n x^n$ be a power series.


Then:

$\ds x = \sum_{n \mathop = 1}^\infty C_n y^n$

is also a power series, where:

\(\ds c_1 C_1\) \(=\) \(\ds 1\)
\(\ds {c_1}^3 C_2\) \(=\) \(\ds -c_2\)
\(\ds {c_1}^5 C_3\) \(=\) \(\ds 2 {c_2}^2 - c_1 c_3\)
\(\ds {c_1}^7 C_4\) \(=\) \(\ds 5 c_1 c_2 c_3 - 5 {c_2}^2 - {c_1}^2 c_4\)
\(\ds {c_1}^9 C_5\) \(=\) \(\ds 6 {c_1}^2 c_2 c_4 + 3 {c_1}^2 {c_3}^2 - {c_1}^3 c_5 + 14 {c_2}^4 - 21 c_1 {c_2}^2 c_3\)
\(\ds {c_1}^{11} C_6\) \(=\) \(\ds 7 {c_1}^3 c_2 c_5 + 84 c_1 {c_2}^3 c_3 + 7 {c_1}^3 c_3 c_4 - 28 {c_1}^2 c_2 {c_3}^2 - {c_1}^4 c_6 - 28 {c_1}^2 {c_2}^2 c_4 - 42 {c_2}^5\)
\(\ds {c_1}^{13} C_7\) \(=\) \(\ds 8 {c_1}^4 c_2 c_6 + 8 {c_1}^4 c_3 c_5 + 4 {c_1}^4 {c_4}^2 + 120 {c_1}^2 {c_2}^3 c_4 + 180 {c_1}^2 {c_2}^2 {c_3}^2 + 132 {c_2}^6 - {c_1}^5 c_7 - 36 {c_1}^3 {c_2}^2 c_5 - 72 {c_1}^3 c_2 c_3 c_4 - 12 {c_1}^3 {c_3}^3 - 330 c_1 {c_2}^4 c_3\)



A derivation of the explicit formula for the $n$th term is given by:

$\ds C_n = \dfrac 1 {n {c_1}^n} \sum_{s, t, u, \cdots} \paren {-1}^{s + t + u + \cdots} \dfrac {n \paren {n + 1} \cdots \paren {n - 1 + s + t + u + \cdots} } {s! \, t! \, u! \ldots} \paren {\dfrac {c_2} {c_1} }^s \paren {\dfrac {c_3} {c_1} }^t \cdots$

where $ s + 2 t + 3 u + \cdots = n - 1$.


Proof



Plugging:

$\ds x = \sum_{n \mathop = 1}^\infty C_n y^n$

into:

$\ds y = \sum_{n \mathop = 1}^\infty c_n x^n$

the following equation is obtained

$y = c_1 C_1 y + \paren {c_2 C_1^2 + c_1 C_2} y^2 + \paren {c_3 C_1^3 + 2 c_2 C_1 C_2 + c_1 C_3} y^3 + \paren {3 c_3 C_1^2 C_2 + c_2 C_2^2 + c_2 C_1 C_3} + \cdots$

Equating coefficients then gives the formula.




$\blacksquare$


Sources