If I had the table of $x$ , $f(x)$ pairs, and this is the rule:

\begin{array} {|r|r|} \hline 1 &1 \\ \hline \hline x_i & f(x_i) \\ \hline \hline 2x_i & 1 +f(x_i) \\ \hline \end{array}

Basically that means that for every $x$ the function value will increment at the $2x$ point.

I'm struggling to find out the exact function, in which i can input the $x$, and it will give me the correct output. :)

I know that the function will look something like this.

So far I constructed this table \begin{array} {|r|r|} \hline x & f(x) \\ \hline \hline 1 &1 \\ \hline \hline 2 &2 \\ \hline \hline 3 & k \\ \hline \hline 4 & 3 \\ \hline \hline 5 & \\ \hline \hline 6 & k+1 \\ \hline \hline ... & ...\\ \hline \hline 8 & 4\\ \hline \hline ... & ...\\ \hline \hline 12 & k+2\\ \hline \hline ... & ...\\ \hline \hline 16 & 5\\ \hline \hline ... & ...\\ \hline \end{array}

where $k$ is the number between 2 and 3, obviously.

I tried interpolating that table in wolfram, but it can't work. I think that there is simple solution to this function. Or is there?

Feel free to edit the question if you think that it can be improved.


1 Answer 1


Another way of phrasing this would be that $f(2x)=1+f(x)$.

Note that $x=0$ can not be in the domain of this function as if it was there $f(2\times0)=1+f(0)\implies0=1$

A function which fits your data is:


Edit: How I arrived at this: take the table you have create and instead of looking at how $x$ links to $f(x)$ look for the reverse link. When $f(x)$ increases by one then $x$ doubles.



(the $-1$ is just there to match the powers up)

Then from this rearrange to get:



  • $\begingroup$ Tested it, and it works, but I'm struggling to figure out how did you come to this solution. $\endgroup$
    – user334473
    Jun 9, 2016 at 15:26
  • $\begingroup$ Updated with my how I got the answer. $\endgroup$
    – Ian Miller
    Jun 9, 2016 at 15:42

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