TAOCP 4.7 Exercise 3

Section 4.7: Manipulation of Power Series

Exercise 3. [M15] Does formula (9) give the right results when $n = 0$? When $n = 1$?

Verified: yes
Solve time: 1m44s

Solution

We examine formula (9) from Section 4.7, which expresses the coefficients of $W(z) = V(z)^\alpha$, with $V_0 = 1$, as

$$ W_n = \frac{(\alpha + 1 - n)V_1 W_{n-1} + (2\alpha + 2 - n)V_2 W_{n-2} + \cdots + n \alpha V_n W_0}{n}, \qquad n \ge 1. $$

We are asked to consider the cases $n = 0$ and $n = 1$.

Case $n = 0$. Formula (9) involves a sum over $k = 1$ to $n$, and a division by $n$. If we formally set $n = 0$, the upper limit of the sum becomes zero, so the sum is empty, and the denominator is $0$. Hence (9) is undefined for $n = 0$.

On the other hand, by the definition of the power series in (5), we have $W_0 = 1^\alpha = 1$. Therefore formula (9) does not apply to $n = 0$, and the correct value must be taken directly from the initial term of the series: $W_0 = 1$.

Case $n = 1$. Substituting $n = 1$ into (9) gives

$$ W_1 = \frac{(\alpha + 1 - 1) V_1 W_0}{1} = \frac{\alpha V_1 W_0}{1} = \alpha V_1 W_0. $$

Since $W_0 = 1$ by definition, we have

$$ W_1 = \alpha V_1. $$

This agrees with the first derivative method from (6), where $W'(0) = \alpha V_1$. Hence formula (9) is valid for $n = 1$.

Conclusion. Formula (9) is undefined for $n = 0$ and must be replaced by the initial term $W_0 = 1$. For $n = 1$, formula (9) produces the correct coefficient $W_1 = \alpha V_1$.

This completes the proof.

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