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Functional equations of the form $f(g(x))=(f(x))^p$, where $g(x)$ is known, is called Bottcher equation. Generally, we have only asymptotic formula for the solution $f(x)$ under certain conditions. In one of my study, I come across the following problem,

How to find a nontrivial (i.e., nonconstant) function $f(x)\in C^\infty(0 ,+\infty)$ satisfying the following functional equation $$f(\frac{2x^3+a}{3x^2})=f(x)^2,$$ where $a>0$ is a constant.

Or can you tell me finding a explicit form for $f(x)$ is impossible?

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2 Answers 2

Introduce the new unknown function $$g(t):=\log f(a^{1/3}(1+t)).$$ It satisfies a functional equation of the type $$g(t)=g(t^2(1+o(1))/2$$ for $t$ near 0, and $g(0)=0$. I guess from this you should be able to prove that $g(t)$ is identically $0$.

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Sorry, you solution is too sketch for me to follow. What is $g(t)$ here, what is $f(t)$ here? –  Sunni Mar 11 '10 at 12:00
    
Perhaps we don't have $g(0) = 0$ but $-\infty$, corresponding to $f(a^{1/3})=0$. Then we can have $2 g(t) \sim g(t^2)$ by taking $g(t) \sim c \log t$ and thus $f(x) \sim |(x-a^{1/3})/a^{1/3}|^c$ as $x \to a^{1/3}$. –  Gerald Edgar Mar 11 '10 at 20:11
    
@Edgar:$f(a^{1/3})$ should be zero, or else only constant solution can be found. See Theorem 8.3.4 from [M. Kuczma, B. Choczewski & R. Ger, Iterative functional equations, Cambridge University Press 1990] –  Sunni Mar 12 '10 at 14:30

In my answer $x \mapsto f(x)$ is the function satisfying your functional equation and $g: t \mapsto g(t)$ is defined explicitly in terms of $f$. After some computation you will see that $g$ satisfies a functional equation of the stated type. Begin with $$\exp(2 g(t))=(f(a^{1/3}(1+t)))^2= \ldots$$

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So you mean, no explicit form of $f(x)$ in my question can be obtained. –  Sunni Mar 12 '10 at 14:24

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