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Given the equation $f(ax) = bf(x)$, with $a, b > 0$, demonstrate that the solution is: $$f(x) = g(\log x)x^{\frac{\log b}{\log a}}$$

where $g(x) = g(x + \log a)$ is an arbitrary periodic function with period $\log(a)$.

By the method of induction I arrived in the particular solution:

$f(x) = Cx^{\frac{\log b}{\log a}}$, where $C$ is an arbitrary constant (case $g(x) =$ constant).

But I could not demonstrate how to get in the generic solution with the associated periodic function $g(x)$. Can someone help me?

JaberMac
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  • You probably want to specify $x > 0$ and $a > 0$. – Robert Israel May 19 '19 at 15:40
  • Express $g$ in terms of $f$ and show that the functional equation for $f$ translates into a the functional equation for $g$ that expresses periodicity. – Hagen von Eitzen May 19 '19 at 16:07
  • Expressing g in terms of f is from the assumption that there already exists a g(x). But this is not known at the beginning of the problem. That would be a check proof. The problem is how from the equation to arrive at the solution with g(x)? – JaberMac May 19 '19 at 16:35
  • @JaberMac is my answer no good for you? – dcolazin May 19 '19 at 18:47
  • @dcolazin Thank you! this is a brilliant check of an assigned function applied in the equation itself. But I did not understand how to set the $g(x) = f(e^x)/e^{Ax}$ definition? did you get the solution itself? we start from the equation that the solution is not known. – JaberMac May 19 '19 at 19:32

2 Answers2

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Assuming $x > 0$

$$ f(a^{\log_a(ax)})-bf(a^{\log_a x}) = 0 $$

or

$$ F(\log_a x + 1)-bF(\log_a x) = 0 $$

or calling $u = \log_a x$

$$ F(u+1)-b F(u) = 0 $$

This is a difference functional equation with solution

$$ F(u) = \Phi(u)b^u $$

where $\Phi(u)$ is any periodic function with period $1$

hence

$$ f(x) = \Phi(\log_a x)b^{\log_a x}=\Phi\left(\frac{\ln x}{\ln a}\right)b^{\frac{\log_a x}{\log_b x}\log_b x} = g(\ln x)x^{\frac{\ln b}{\ln a}} $$

NOTE

Regarding the periodicity of $\Phi(u)$ we have

$$ \Phi(u) = \Phi(u+1)\to \Phi(\log_a x) = \Phi(\log_a x+1) \to \Phi\left(\frac{\ln x}{\ln a}\right) = \Phi\left(\frac{\ln x}{\ln a}+1\right) $$

and finally

$$ \Phi(\ln x) = \Phi(\ln x+\ln a) $$

Cesareo
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  • Thank You! perfect explanation! now I see everything more clearly. sorry, I just did not understand something at the end: $\theta (lnx/lna) \equiv \theta´ (lnx)$, what does that mean? – JaberMac May 19 '19 at 20:53
  • Better and cleaner would be to make $\Phi\left(\frac{\ln x}{\ln a}\right) = g(\ln x)$ – Cesareo May 19 '19 at 21:00
  • you understood perfectly what I needed. Thank you very much. But sorry for my ignorance. I'm still intrigued by the $\phi (lnx/lna)$ function. This function is equivalent to $g(lnx)$? By the proof $\phi (u)$ has period equal to 1. The function $g(lnx)$ has period equal to $ln(a)$ (by the solution of the problem). I am confused by the equivalence of these functions and their periods. Can you help me with that? – JaberMac May 20 '19 at 21:39
  • @JaberMac Please. See attached note. – Cesareo May 20 '19 at 22:09
  • Perfect! now I understand. If $\phi(lnx) = \phi(lnx + lna)$ then we can assume that $\phi(x) = \phi(x + lna)$, what is the function shown in the problem solution. Beautiful demonstration! Thank you so much again. – JaberMac May 20 '19 at 22:30
  • @JaberMac It is my pleasure. – Cesareo May 21 '19 at 07:04
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Hints can be useful if you are willingful to hear.

  • Let $f$ be a solution of the equation, then define $g(x) = \frac{f(e^x)}{e^{Ax}}$, where $A = \frac{\log(b)}{\log(a)} = \log_a(b)$.

    $g(x)$ is periodic of period $\log(a)$: $$g(x+\log(a)) = \frac{f(e^{x+\log(a)})}{e^{Ax+A\log(a)}} = \frac{f(ae^{x})}{e^{Ax+\log(b)}} = \frac{bf(e^{x})}{be^{Ax}} = g(x)$$

So every solution can be written as $f(x) = g(\log(x))x^A$.

  • Viceversa, an $f$ of the kind $f(x) = g(\log(x))x^A$ is a solution of the equation:

$$f(ax) = g(\log(ax))(ax)^A = g(\log(x)+\log(a))x^Aa^A = a^A f(x) = bf(x)$$

So all the $f(x) = g(\log(x))x^A$ are solutions.

dcolazin
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  • Thank You! This is a brilliant check of an assigned function applied in the equation itself. But I did not understand how to set the definition $g(x) = f(e^x)/e^{Ax}$? Did you get the solution itself? we start from the equation that the solution is not known. – JaberMac May 19 '19 at 19:13