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I have a function $f:\mathbb{R}^{>0}\rightarrow \mathbb{R}$ that satisfies the following equation: $$f(10\cdot x)-f(x)=1 \quad\quad (1)$$

From looking at this equation I can easily see that $f(x)=\log_{10}(x)$ works out nicely.

However, if the equation would have been $$f(10\cdot x)-5f(x)=1 \quad\quad (2)$$ then the logarithm would not have worked out.

Two questions:

  1. How can I prove that the logarithm is the only solution to equation (1)?
  2. Is there a general way to solve these kind of equations, for example, how would one solve equation (2)?
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    $\begingroup$ Do you require your function to be continuous? If not I suspect that logarithm might not be the only solution to (1). $\endgroup$ – Keen Sep 5 '17 at 20:48
  • $\begingroup$ No, does not have to be continuous. $\endgroup$ – Merlin1896 Sep 5 '17 at 20:58
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For the first question, $\log_{10}$ is not the only solution: let $g$ be any function defined on $[1, 10)$. Define $$f(x) = k + g(x \cdot 10 ^{-k})$$ when $10^k \le x \lt 10^{k+1}$. Then $f$ is a solution of the functional equation. Note that $k = \lfloor\log_{10}(x)\rfloor$

For the second equation, you can use a similar definition, I suggest $$f(x) = 5^k \left( g(x \cdot 10 ^{-k}) + \frac{1}{4}\right) - \frac{1}{4}$$

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  • $\begingroup$ What exactly is your k? The integer part of the log or what does this notation mean? Okay, its clearer now! $\endgroup$ – Merlin1896 Sep 5 '17 at 21:02
  • $\begingroup$ It is the floor function of the $\log_{10}(x)$. $\endgroup$ – Gribouillis Sep 5 '17 at 21:04
  • $\begingroup$ I guess you found this solution by a mixture of "educated guess" and trying a few things out. I believe that there is no more general way to solve this kind of equations? $\endgroup$ – Merlin1896 Sep 5 '17 at 21:09
  • $\begingroup$ The point is I don't know what is "this kind of equations". Here, for a given $x$, if you define $u_n = f(10^n x)$, you see that the relation reduces to a recurrence relation $u_{n+1} = u_{n} + 1$ or $u_{n+1} = 5 u_{n} + 1$. You can use your knowledge about numerical sequences to solve this: if you know one term, you know the whole sequence. Then it only remains to realize that only one of the $10^n x$ is in the interval $[1, 10)$. $\endgroup$ – Gribouillis Sep 5 '17 at 21:17
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    $\begingroup$ Yes, the name is functional equation. There is indeed no universal approach. $\endgroup$ – Gribouillis Sep 5 '17 at 21:30
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As Gribouillis points out, there are no unique solutions to equations like this. The equation only connects the values of $f(x)$ for values of $x$ that differ by factor ten, so it says nothing about $f(x)$ and $f(y)$ when $x/y$ is not an integer power of ten.

However, there is a way to find a "nice" solution to the kind of equation. If you want the general solution instead of pretty ones, this is not your answer. By pretty I mean something that satisfies a natural generalization of your condition.

Consider the second problem. Take any $x\in(0,\infty)$ and $n\in\mathbb N$. We have $$ \begin{split} f(10^nx) &= 1+5f(10^{n-1}x) \\&= 1+5(1+5f(10^{n-2}x)) \\&= (1+5)+5^2f(10^{n-2}x) \\&= \dots \\&= (1+5+25+\dots+5^{n-1})+5^nf(x) \\&= \frac14(5^n-1)+5^nf(x) \\&= 5^n(\frac14+f(x))-\frac14 . \end{split} $$ This formula can be used to find the general solution if you want to. To find a pretty solution, we add the additional restricting assumption that this holds for all values of $n\in\mathbb R$. For $x=1$ and $f(1)=a-1/4$ (the shift is for convenience) this leads to $$ f(y) = f(10^{\log_{10}y}) = %5^{\log_{10}y}(\frac14+a)-\frac14 5^{\log_{10}y}a-\frac14 . $$ It is easy to check that this indeed satisfies $f(10^tx) %=5^{t+\log_{10}x}a-\frac14=5^t5^{\log_{10}x}a-\frac14 =5^t(f(x)+1/4)-1/4$ for all $x>0$ and $t\in\mathbb R$.

You are free to choose the parameter $a$. For $a=0$ you get the constant function $f(x)=-1/4$.

Similarly, for the first problem you would get $f(y)=a+\log_{10}y$ with this method.

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$$f(10\cdot x)-f(x)=1 \quad\quad (1)$$

By induction it's easy to prove $ f(10^n) = n + f(1) $

So any function $g(s) \longrightarrow g(1) + \log_{10} $(s) satisfy $(1)$

With $ s \ne 0 $ ....

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    $\begingroup$ For the first equation, it is indeed easy to see that if $f(x)$ is a solution, then $f(x) + C$ is a solution for any constant $C$. $\endgroup$ – Gribouillis Sep 5 '17 at 21:39

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