TAOCP 4.7 Exercise 7
We wish to find the coefficients $W_n$ in the reversion of the series $z = t - t^n,$ that is, we seek the power series $t = z + W_2 z^2 + W_3 z^3 + \cdots$ such that substituting into the right-hand s...
Section 4.7: Manipulation of Power Series
Exercise 7. [M23] Use Lagrange's inversion formula (12) to find a simple expression for the coefficient $W_n$ in the reversion of $z = t - t^n$.
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Solution
We wish to find the coefficients $W_n$ in the reversion of the series
$z = t - t^n,$
that is, we seek the power series
$t = z + W_2 z^2 + W_3 z^3 + \cdots$
such that substituting into the right-hand side recovers $z$.
By Lagrange's inversion formula (12), for a series
$z = t + \sum_{k\ge 2} V_k t^k,$
the $n$th coefficient of the reversion is
$W_n = \frac{1}{n} [t^{,n-1}] \bigl(1 + V_2 t + V_3 t^2 + \cdots \bigr)^{-n}.$
In the present case, we have $V_k = 0$ for all $k \ne n$, and $V_n = -1$. Therefore
$1 + \sum_{k \ge 2} V_k t^{k-1} = 1 - t^{,n-1}.$
Hence
$W_m = \frac{1}{m} [t^{,m-1}] (1 - t^{,n-1})^{-m}.$
We expand $(1 - t^{,n-1})^{-m}$ using the binomial theorem for negative exponents: for any integer $m > 0$,
$(1 - t^{,n-1})^{-m} = \sum_{j=0}^{\infty} \binom{m + j - 1}{j} t^{,j(n-1)}.$
We are interested in the coefficient of $t^{,m-1}$. Comparing powers, we require
$j(n-1) = m - 1 \implies j = \frac{m - 1}{n - 1}.$
Since $j$ must be a nonnegative integer, we conclude that $W_m \ne 0$ if and only if $m \equiv 1 \pmod{,n-1}$, that is,
$m = 1 + j(n-1), \qquad j = 0,1,2,\dots.$
For such $m$, the coefficient is
$W_{1 + j(n-1)} = \frac{1}{1 + j(n-1)} \binom{m + j - 1}{j} = \frac{1}{1 + j(n-1)} \binom{1 + j(n-1) + j - 1}{j}.$
Simplifying the binomial coefficient in terms of $j$ and $n$, we have
$1 + j(n-1) + j - 1 = jn,$
so that
$\binom{jn}{j} = \frac{(jn)!}{j! , ((n-1)j)!}.$
Thus the coefficient $W_{1 + j(n-1)}$ is
$W_{1 + j(n-1)} = \frac{1}{1 + j(n-1)} \binom{jn}{j} = \frac{1}{(n-1) j + 1} \binom{jn}{j}.$
Equivalently, letting $m = 1 + j(n-1)$, we have
$\boxed{W_m = \frac{1}{m} \binom{(m-1) n / (n-1)}{(m-1)/(n-1)}, \quad m = 1, 1 + (n-1), 1 + 2(n-1), \dots}.$
All other coefficients vanish. This gives a complete and simple formula for the reversion of $z = t - t^n$.
This completes the proof.
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