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Solve the functional equation $$\log f(x,y) = f(\log x, \log y) $$ The motivation comes from generalizing the arithmetic and geometric means, but I have no idea how to find a simpler expression for the function $f(x,y)$ with $f(x,y) = f(y,x)$

What I have tried:

  1. $f(ax,y) = e^{f(\log x +\log a, y)}$
  2. $f(1,x) = e^{f(0,\log x)}$
  3. $f(x,y) \equiv \infty $ solves the equation, so I am looking for non-trivial solutions.
  4. Letting $$f(1+x,1+y) = \sum_{m,n=0}^\infty \frac {a_{m,n}}{x^my^n} = a_{0,0} + \frac {a_{1,1}}{xy} + \frac {a_{2,0}}{x^2}+ \frac {a_{0,2}}{y^2}+\cdots$$ Using the functional equation $f(1+x,1+y) = e^{f(\log (1+x), \log (1+y))}$ we have $$ a_{0,0} + \frac {a_{2,0}}{x^2}+ \frac {a_{0,2}}{y^2}+\cdots = 1+f\left(x-\dfrac{x^2}2 + \dfrac{x^3}3 - \cdots, y-\dfrac{y^2}2 + \dfrac{y^3}3 - \cdots\right) + \cdots$$ which does not give anything pleasant.
  5. If I consider a recursive definition $f_{n+1} (x,y) = e^{f_n (\log x, \log y)}$, then modulo convergence issues, $f_n$ converges to $f$. Motivated by AM and GM, I consider $f_0 (a,b) = \frac{a+b}2$. Then $f_1 (a,b) = \operatorname{GM}(a,b)$.
  6. Test: $f_0 = \dfrac{x+y}2, f_1 = \sqrt {xy}, f_2 = e^{\operatorname{GM}(\log x,\log y)} = \exp\sqrt{(\log x)(\log y)}$, $f_3 = \exp(\exp\sqrt{(\log\log x)(\log\log y)})$. I guess the pattern is $$\exp(\exp\cdots(\exp\sqrt{(\log\log\cdots \log x)(\log\log\cdots \log y)})$$
  7. Note that by AM$\geq$ GM, $f_n (x,y) \leq \underbrace{\exp(\exp\cdots(\exp}_n\dfrac{\underbrace{\log\log\cdots \log}_n x+\underbrace{\log\log\cdots \log}_n y}2)$ = $ \underbrace{\exp(\exp\cdots(\exp}_{n-1}\sqrt{(\underbrace{\log\log\cdots \log}_{n-1} \,x)(\underbrace{\log\log\cdots \log}_{n-1}\,y)}) = f_{n-1}(x,y)$.
  8. I have tested out some explicit values: for $x = 2000, y= 8000$, $f_0 = 6000, f_1 = 4000, f_2 = 3832.6491..., f_3 = 3762.9980...$ by Wolfram Alpha. Some larger $n$'s give complex numbers. If $f_n$ is real, it is a priori apparent that $f_n\geq 0$.

But I have no idea how to go on further.

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    $\begingroup$ Did you see this thread that I found on SearchOnMath? $\endgroup$ Mar 17, 2022 at 12:21
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    $\begingroup$ Oh, you ask the same question! I think I have provided a (partial) answer by infinite interations of exp and log $\endgroup$
    – wilsonw
    Mar 17, 2022 at 12:29

1 Answer 1

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Take any function $g(x,y):A\to \Bbb R$ where $A = \{(x,y) \mid x \le 0 \text{ or } y \le 0\}$.

Now define $f(x,y)$ as follows: $$f(x,y) = \begin{cases} g(x,y) &\text{if } (x,y)\in A\\ \exp(f(\log x, \log y))& \text{otherwise} \end{cases}$$

This function satisfies the equation. Moreover, all solutions of the equation can be constructed this way.

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  • $\begingroup$ I think this is not explicit enough for a definition of f to be accepted as a solution $\endgroup$
    – wilsonw
    Mar 17, 2022 at 12:37
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    $\begingroup$ @wilsonw Note this one is a recursive definition as well, but the sequence "stops" when one of the logarithms is <0. I like this answer, it's the best way to solve functional equations IMO $\endgroup$
    – Compacto
    Mar 17, 2022 at 12:51
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    $\begingroup$ I wonder what happens when one logarithm is zero and the other one is positive (I think infinities appear) . Also, I think you meant $f(\log x, \log y)$. $\endgroup$
    – Compacto
    Mar 17, 2022 at 12:51
  • $\begingroup$ Thank you @Compacto. $\endgroup$
    – jjagmath
    Mar 17, 2022 at 12:53
  • $\begingroup$ I think it would be better if you write $f_{n+1}$ on the left hand side and $f_n$ on the right for easier reading. $\endgroup$
    – wilsonw
    Mar 17, 2022 at 12:54

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