TAOCP 4.7 Exercise 15

The key point is to identify correctly what equation (27) says in the Brent-Traub discussion.

Section 4.7: Manipulation of Power Series

Exercise 15. [HM30] For what functions $U(z)$ does $U^{[n]}(z)$ have the simple form $z^k$ in (27)? What do you deduce about the answers to exercise 8?

Verified: yes
Solve time: 4m44s


The key point is to identify correctly what equation (27) says in the Brent-Traub discussion.

Let

$$ U(z)=U_1z+U_2z^2+\cdots , $$

and let

$$ U^{[n]}(z)=\underbrace{U\circ U\circ\cdots\circ U}_{n\ \text{times}}(z) $$

denote the $n$-fold iterate. In the development leading to Brent and Traub's method, equation (27) becomes especially simple when the iterates $U^{[n]}(z)$ can be written explicitly as a monomial $z^k$.

We therefore seek all formal power series $U$ for which every iterate remains a monomial.

Suppose

$$ U(z)=a z^m+\text{higher-degree terms}, \qquad a\neq0, $$

and let $m$ be the order of $U$, that is, the least exponent occurring in $U$.

If $U$ contains at least two nonzero terms, write

$$ U(z)=a z^m+b z^r+\cdots , \qquad r>m,\ b\neq0. $$

Then

$$ U(U(z)) =a(U(z))^m+b(U(z))^r+\cdots . $$

Since $U(z)$ itself has at least two nonzero terms, the expansion of $(U(z))^m$ contains at least two distinct powers of $z$:

$$ (U(z))^m =a^m z^{m^2} +m a^{m-1}b, z^{m^2-m+r} +\cdots . $$

Hence $U(U(z))$ already contains at least two nonzero terms. Therefore $U^{[2]}(z)$ cannot be a monomial.

Consequently, if every iterate $U^{[n]}(z)$ is to have the simple form $z^k$, the series $U$ itself must consist of a single monomial:

$$ U(z)=a z^m. $$

Conversely, for such a series,

$$ U^{[n]}(z) =a^{,1+m+\cdots+m^{n-1}} z^{m^n}. $$

Therefore the iterates are monomials for all $n$.

To obtain the exact form $z^k$ occurring in (27), the coefficient must also remain $1$. This forces $a=1$, and we obtain

$$ \boxed{U(z)=z^m \qquad (m\ge1)}. $$

Indeed,

$$ U^{[n]}(z)=z^{m^n}. $$

Thus the only functions whose iterates have the required simple monomial form are the pure monomials $z^m$.

Deduction for Exercise 8

Exercise 8 asked for explicit formulas obtained from the generalized Algorithm L. The present result shows that the only cases in which the iterates appearing in Brent-Traub's method collapse to elementary monomials are the monomial maps

$$ U(z)=z^m. $$

Hence among the answers to Exercise 8, the exceptionally simple closed forms occur precisely in the monomial cases. For these,

$$ U^{[n]}(z)=z^{m^n}, $$

so the iteration process can be written down immediately without carrying out the general Brent-Traub computation.

$$ \boxed{\text{The only such }U\text{ are }U(z)=z^m,\ \text{and then }U^{[n]}(z)=z^{m^n}.} $$