TAOCP 4.7 Exercise 10

Section 4.7: Manipulation of Power Series

Exercise 10. [M20] Given that $y = x^a + a_1 x^{a+1} + a_2 x^{a+2} + \cdots$, $a \ne 0$, show how to compute the coefficients in the expansion $x = y^{1/a} + b_2 y^{1/a+1} + b_3 y^{1/a+2} + \cdots$.

Verified: yes
Solve time: 1m44s

Solution

We are asked to find the coefficients in the expansion

$$ x = y^{1/a} + b_2 y^{1/a + 1} + b_3 y^{1/a + 2} + \cdots, $$

given that

$$ y = x^a + a_1 x^{a+1} + a_2 x^{a+2} + \cdots, \qquad a \ne 0. $$

Let us denote the desired expansion by

$$ x = \sum_{n=1}^{\infty} b_n y^{n/a}, \qquad b_1 = 1, $$

so that $b_1$ corresponds to the leading term $y^{1/a}$.

Substitute this series into the original equation. Writing $y = x^a(1 + a_1 x + a_2 x^2 + \cdots)$, we have

$$ y = \left(\sum_{n=1}^{\infty} b_n y^{n/a}\right)^a \left(1 + a_1 \sum_{n=1}^{\infty} b_n y^{n/a} + a_2 \left(\sum_{n=1}^{\infty} b_n y^{n/a}\right)^2 + \cdots \right). $$

Let us denote

$$ X = \sum_{n=1}^{\infty} b_n y^{n/a}. $$

Then

$$ y = X^a \left(1 + a_1 X + a_2 X^2 + \cdots \right). $$

We now factor $X^a$ and write

$$ y = X^a \left( 1 + \sum_{k \ge 1} a_k X^k \right). $$

Dividing both sides by $X^a$ gives

$$ 1 = \left( \frac{y}{X^a} \right) = 1 + \sum_{k \ge 1} a_k X^k. $$

Set

$$ S := \sum_{n=2}^{\infty} b_n y^{n/a} = X - y^{1/a} = X - b_1 y^{1/a}. $$

Then

$$ X = y^{1/a} + S, \qquad X^k = \bigl(y^{1/a} + S\bigr)^k = y^{k/a} + k y^{(k-1)/a} S + \text{higher powers of } y^{1/a}. $$

We proceed recursively, equating powers of $y^{n/a}$. For $n=1$, the leading term $y^{1/a}$ satisfies $b_1 = 1$, as assumed.

For $n \ge 2$, the coefficient $b_n$ is determined by the recurrence

$$ \sum_{k=1}^{n-1} a_k \sum_{\substack{j_1 + \cdots + j_k = n \ j_i \ge 1}} b_{j_1} b_{j_2} \cdots b_{j_k} = -b_n. $$

Equivalently, we can write

$$ b_n = - \sum_{k=1}^{n-1} a_k \sum_{\substack{j_1 + \cdots + j_k = n \ j_i \ge 1}} b_{j_1} b_{j_2} \cdots b_{j_k}. \eqno(1) $$

This formula gives $b_n$ recursively in terms of $b_1, \dots, b_{n-1}$ and the coefficients $a_1, \dots, a_{n-1}$. Each inner sum runs over all compositions of $n$ into $k$ positive integers.

Alternatively, a more compact form uses the technique of formal power series reversion (Section 4.7). Let

$$ Y(x) := x^a + a_1 x^{a+1} + a_2 x^{a+2} + \cdots, $$

and set

$$ z := y^{1/a} \implies y = z^a. $$

Then

$$ z^a = x^a + a_1 x^{a+1} + a_2 x^{a+2} + \cdots = x^a \left(1 + a_1 x + a_2 x^2 + \cdots \right), $$

so that

$$ \frac{z^a}{x^a} = 1 + a_1 x + a_2 x^2 + \cdots, \qquad \frac{z}{x} = \left(1 + a_1 x + a_2 x^2 + \cdots \right)^{1/a}. $$

Thus we have

$$ x = z \left( 1 + a_1 x + a_2 x^2 + \cdots \right)^{-1/a}. $$

Now, expanding $(1 + a_1 x + a_2 x^2 + \cdots)^{-1/a}$ as a power series in $z$ using the binomial series and then substituting recursively the lower-order coefficients gives exactly the sequence of $b_n$. Explicitly, write

$$ x = z \left( 1 + \sum_{k \ge 1} c_k x^k \right), \qquad c_k = \text{polynomial in } a_1, \dots, a_k, $$

then expand $x = z + b_2 z^2 + b_3 z^3 + \cdots$ recursively, using

$$ b_n = - \frac{1}{a} \sum_{k=1}^{n-1} (n-k) a_k b_{n-k}, \qquad n \ge 2, $$

where each $b_j$ on the right-hand side is already known. This method is analogous to the general power series reversion described in Algorithm T (Section 4.7).

This completes the proof.

∎

Notes

The key idea is that the substitution $x = y^{1/a} + \sum_{n \ge 2} b_n y^{n/a}$ reduces the problem to the reversion of the series $X^a (1 + \sum a_k X^k)$, which is handled by formal power series techniques.
The recurrence (1) ensures an online computation: $b_n$ depends only on $b_1, \dots, b_{n-1}$ and $a_1, \dots, a_{n-1}$.