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I'm looking for a continuous solution to the functional equation

$$f(2x) = N - \frac{2x}{f(x)^2}$$

where $N$ is a constant natural number and $x \in \mathbb{R}$ is nonnegative. I don't have much experience with functional equations so I haven't tried anything yet. If it helps I'm mostly interested near $x=0$. Any ideas?

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  • $\begingroup$ Well, you could start by expanding around N, so taking the Ansatz $f(x)=N+ax+bx^2+cx^3+...$, plugging in, and determining a few a,b,c,.. recursively to get an impression of the function. $\endgroup$ Mar 9, 2018 at 23:11
  • $\begingroup$ It seems that if $n > 1$ then $f$ defines an entire function. $\endgroup$ Mar 10, 2018 at 0:50

1 Answer 1

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This is a simple study of $f(x)$ as $x\to0$.

Let $N>0$.

First case, if $f(0)=0$, then $$\lim_{x\to0}\frac{2x}{f(x)^2}=N\implies f(x)=\sqrt\frac{2x}{N}+o(\sqrt x)$$

Second case, if $f(0)=N$,

  • Assuming $f(x)=N+ax+o(x)$, then \begin{align} f(x)^2 = \frac{2x}{N-f(2x)}&\implies N^2+o(1)=-\frac{2x}{2ax+o(x)}\\ &\implies a=-N^{-2} \end{align}

  • Assuming $f(x)=N-N^{-2}x+bx^2+o(x^2)$, then \begin{align} f(x)^2 = \frac{2x}{N-f(2x)}&\implies N^2-2N^{-1}x+o(x)=\frac{1}{N^{-2}-2bx+o(x)}\\ &\implies b=-N^{-5} \end{align}

  • And so on...

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  • $\begingroup$ Thanks, I got the same, as well as $c = \frac{-5}{4}N^{-8}$ for the cubic coefficient. Any chance you recognize that series form? $\endgroup$
    – Brady Gilg
    Mar 10, 2018 at 0:31
  • $\begingroup$ @BradyGilg I don't think there exists a closed form. $\endgroup$
    – Aforest
    Mar 10, 2018 at 0:38

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