TAOCP 4.7 Exercise 24

Let $U=(u_{nk})$ be the power matrix of $U(z)$.

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Section 4.7: Manipulation of Power Series

Exercise 24. [**] [HM35] Given the power series $U(z) = U_1 z + U_2 z^2 + \cdots$, where $U_1$ is not a root of unity, let $U = (u_{nk})$ be the power matrix of $U(z)$.

a) Explain how to compute a matrix $\ln U$ so that the power matrix of $U^{[\alpha]}(z)$ is $\exp(\alpha \ln U) = I + \alpha \ln U + (\alpha \ln U)^2/2! + \cdots$.

b) Show that if $W(z)$ is not identically zero and if $U(W(z)) = W(U_1 z)$, then $W(z) = U^{[\alpha]}(z)$ for some complex number $\alpha$.

Verified: yes
Solve time: 5m59s

Let $U=(u_{nk})$ be the power matrix of $U(z)$. Since $U_1$ is not a root of unity, the diagonal entries of $U$ are

$$ U_1,\ U_1^2,\ U_1^3,\ \ldots, $$

and are pairwise distinct. Write

$$ D=\operatorname{diag}(U_1,U_1^2,U_1^3,\ldots). $$

Then

$$ N=D^{-1}U-I $$

is strictly upper triangular. Hence the matrix logarithm is defined entrywise by the finite series

$$ \ln U=\ln D+\ln(I+N) =\operatorname{diag}(n\log U_1){n\ge1} +\sum{m\ge1}\frac{(-1)^{m-1}}{m}N^m . $$

For every complex $\alpha$ we may therefore define

$$ U^{[\alpha]} $$

to be the power series whose power matrix is

$$ \exp(\alpha\ln U). $$

Since $\exp((\alpha+\beta)\ln U)

=\exp(\alpha\ln U)\exp(\beta\ln U)$, these matrices form a one-parameter group of fractional iterates.

Let $W(z)\not\equiv0$ satisfy

$$ U(W(z))=W(U_1z). $$

If $r$ is the least index with $[z^r]W(z)\neq0$, comparison of the coefficients of $z^r$ gives

$$ U_1=[U_1]^r. $$

Since $U_1$ is not a root of unity, $r=1$; therefore $W_1\neq0$, and the power matrix $C$ of $W$ is invertible. The functional equation becomes

$$ UC=CD. $$

Hence

$$ U=CDC^{-1}, \qquad \ln U=C(\ln D)C^{-1}. $$

Because the eigenvalues of $D$ are distinct, the eigenspace corresponding to $U_1$ is one-dimensional. Therefore every nonzero solution of

$$ UX=XD $$

is determined by its first coefficient. Choose $\alpha$ so that

$$ W_1=U_1^\alpha . $$

The power series $U^{[\alpha]}$ has linear coefficient $U_1^\alpha$ and its power matrix satisfies the same conjugacy relation with $D$. Consequently $W$ and $U^{[\alpha]}$ are nonzero solutions of the same Schröder equation and have the same first coefficient. By the one-dimensionality just established, they are identical:

$$ W(z)=U^{[\alpha]}(z). $$

This completes the proof.

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