Solve functional equation $f(z)=c+zf(z^2)$ with series expansion?

Question

Let the functional equation $(1)$ be given as

$$ f(z)=c+zf(z^2) \tag{1}$$

where $c \in\mathbb R$ and $c \neq 0$.

1. How can this functional equation be solved with series expansion (power, Taylor or Laurent)?

2. Are there any criteria for the constant $c$?

Thank you!

Answer 1

There are a few things that would be very important for us to know, if we're to approach this functional equation. In particular, we'd want to have some idea of the domain of definition. Without that, attempting to do a series expansion could be very misleading. I will assume herein that $f$ should be defined in a (punctured) neighborhood of $0.$

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The results will differ if we use Laurent expansion versus Taylor expansion, since the Taylor expansion will necessarily be defined at $0,$ whereas the Laurent expansion need not be.

Let's suppose that $$f(z)=\sum_{n\in\Bbb Z} a_n z^n,$$ so that $$zf\left(z^2\right)=z\sum_{n\in\Bbb Z} a_n \left(z^2\right)^n=\sum_{n\in\Bbb Z}^\infty a_n z^{2n+1}.$$ Noting that we can rewrite $$f(z)=\sum_{k\in\Bbb Z} a_{2k} z^{2k}+\sum_{k\in\Bbb Z} a_{2k+1} z^{2k+1},$$ we find that $$f(z)-zf\left(z^2\right)=\sum_{k\in\Bbb Z} a_{2k}z^{2k}+\sum_{k\in\Bbb Z} (a_{2k+1}-a_k)z^{2k+1}.$$ Since we need this to be equal to $c,$ then we must have $a_0=c,$ $a_{2k}=0$ for all non-zero $k,$ and $a_{2k+1}=a_k$ for all $k.$

I claim moreover that $a_n=c$ if $n=2^j-1$ for some nonnegative integer $j$; for $n=-1,$ we cannot determine $a_n$ without more information; for all other $n,$ $a_n=0.$ Since $a_0=c$ and $a_{2n+1}=a_n$ for all $n,$ then we can easily prove by induction that $a_n=c$ whenever $n=2^j-1$ for some nonnegative integer $j.$ (I leave that to you.) Likewise, since $a_{2k}=0$ for all nonzero $k,$ then readily, $a_n=0$ whenever $n$ is a non-zero even integer, and in particular when $n=2^j$ or $n=-2^j$ for some positive integer $j.$ (Again, I leave the proof to you.) To prove that $a_n=0$ otherwise (except possibly when $n=-1$), we proceed by induction as follows:

Suppose there is some nonnegative integer $k$ such that, for all integers $m$ with $-2^{k+1}\le m\le-2^k-1,$ we have $a_m=0.$ (This clearly holds in the $k=0$ case.) Take any odd integer $m$ with $-2^{k+2}\le m\le-2^{k+1}-1.$ Since $m$ is odd, then in fact, $-2^{k+2}+1\le m\le-2^{k+1}-1.$ Moreover, $m=2j+1$ for some integer $j,$ so we have: $$-2^{k+2}+1\le 2j+1\le-2^{k+1}-1$$ $$-2^{k+2}\le 2j\le-2^{k+1}-2$$ $$-2^{k+1}\le j\le-2^k-1$$ Thus, $a_j=0$ by inductive hypothesis, so $a_m=a_{2j+1}=a_j=0.$

Now, suppose there is a positive integer $k$ such that, for all integers $m$ with $2^k\le m\le 2^{k+1}-2,$ we have $a_m=0.$ (This clearly holds for $k=1.$) Take any odd integer $m$ such that $2^{k+1}\le m\le 2^{k+2}-2.$ Since $m$ is odd, then in fact, $2^{k+1}+1\le m\le 2^{k+2}-3,$ and letting $m=2j+1,$ we have $$2^{k+1}+1\le 2j+1\le 2^{k+2}-3$$ $$2^{k+1}\le 2j\le 2^{k+2}-4$$ $$2^k\le j\le 2^{k+1}-2$$ Thus, $a_j=0$ by inductive hypothesis, so $a_m=a_{2j+1}=a_j=0.$

Letting $a_{-1}=k,$ we consequently find that $$f(z)=\frac{k}{z}+c\sum_{j=0}^\infty z^{2^j-1}.$$ The series $\sum_{j=0}^\infty z^{2^j-1}$ converges for $|z|<1,$ so $f(1)$ will be defined if and only if $c=0,$ in which case we would simply have $$f(z)=\frac{k}{z},$$ which is defined for all $z\ne 0.$

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On the other hand, if we require $f$ to be defined at $0,$ we can either just use the Taylor expansion, say $$f(z)=\sum_{n=0}^\infty a_n z^n,$$ then we'd instead have deduced that either $f$ is identically zero or $$f(z)=c\sum_{j=0}^\infty z^{2^j-1},$$ which is defined for $|z|<1.$ However, the Laurent approach still works there: we simply determine that we need $k=0.$

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Upshot: For any constants $c$ and $k,$ $$f(z)=\frac{k}{z}+c\sum_{j=0}^\infty z^{2^j-1}$$ satisfies the given functional equation, and is defined for all $z$ such that $0<|z|<1.$ It will be defined at $z=0$ if and only if $k=0,$ and it will be defined for $z$ with $|z|\ge 1$ if and only if $c=0.$

Answer 2

Setting $z = 1$ forces $c = 0$. Putting $g(z) = z f(z)$, we then have $g(z) = g(z^2)$. You haven't put any restrictions on $f$; but if it, say, has a Laurent series expansion near $z = 0$, then it's easy to see that $g$ must be constant, i.e., $f(z) = \alpha /z$ for some constant $\alpha$.