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Consider the following functional equation $$A(x)=c\frac{A\big(\sqrt{x}\big)+A\big(-\sqrt{x}\big)}{2} + x^k,$$ where $c\in\mathbb R$ and $k\in\mathbb N$ are constants, and $c$ is chosen to make $A(x)$ analytic (if I am not mistaken, we can find through the root test that the radius of convergence of $A$ is fine for $1<c$). The question is: are there known "nice" solutions? Of course I am looking for a closed formula or solutions in integral form, infinite products, and so on.

Notice that the coefficients $\{a_n\}_{n\ge 0}$ in $A(x)=\sum_{n\ge 0}a_nx^n$ verify the nice relation $$a_n = c\cdot a_{2n} +\delta(n-k),$$ where $\delta(m)$ is equal to $1$ if $m=0$ and $0$ otherwise. Then, writing $k=o\cdot 2^h$ with $o$ an odd number and $h\ge 0$, the only nonzero coefficients are $a_{o\cdot 2^i}$ for $i=0,1,\dots$ and they are related in a straightforward way, that depends on $i$ being less or greater than $h$, to $a_o$.

I have briefly checked standard references on functional equations without success.

Thanks for your patience.

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1 Answer 1

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There is an easy to check polynomial solution. For $k=2^ho$ as above,
$$A(x):=\sum_{j=0}^h c^{h-j}x^{2^jo}=x^k+cx^{k/2}+c^2x^{k/4}+\dots+c^hx^{k/2^h}$$ Note that the only odd-degree term is the last one, $c^hx^o$.
$$\sim *\sim $$ If $\bf |c|<1$, this is also the only locally bounded solution on $\mathbb C$; more generally, for any $r\ge1$ and any bounded $\mathbb C$-valued function $f$ on the disk $B(0,r)$ of radius $r$, the linear functional equation $$A(x)=c\frac{A(\sqrt x)+A(-\sqrt x)}{2}+f(x)$$ has a unique bounded solution on $B(0,r)$ given inverting the operator $I-cH$ with $$(Hg)(x):=\frac{g(\sqrt x)+g(-\sqrt x)}{2}=\frac12\sum_{z^2=x}g(z), $$ the mean value of $g$ on the square roots of $x$. So the iterates of $H$ are given by $$H^n g(x)=\frac1{2^n}\sum_{z^{2^n}=x}g(z), $$ and give the mean value of $g$ on the $2^n$-th roots of $x$. Since $\|H\|\le 1$ one has the Neumann expansion $(I-cH)^{-1}=\sum_{n=0}^\infty H^n$ and finds $$A(x):=\sum_{n\in\mathbb N, z^{2^n}=x} (c/2)^n f(z).$$ If $f$ is a locally bounded function on $\mathbb C$, the latter is the only locally bounded solution on $\mathbb C$.

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