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I am trying to find a continuous curve that will go trough all the points in the graph created if you were to plot the 2-adic valuation of x which is defined as how many times you can divide a number by two before the result gives an odd number. This can be defined by the function: $$f(x)=\left\{\begin{array} &0\text{ when x is odd}\\ f(\frac{x}{2})+1\text{ when x is even}\\ \end{array}\right. $$ The problem with this is that I cant differentiate since I need a continuous curve, and right now this only gives me what the values go like: 0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,5,etc. $$$$ What I am currently trying is to find a trigonometric function that represents it. For example $\cos^2{\left(\frac{\pi}{2}x\right)}$ works for some values (just the alternating 0's and 1's) but I dont know how I could modify this to fit all values.

r00r
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  • This question explains how to do it with the Mittag-Leffler theorem, if you're familiar with that. While the procedure given there is explicit, I don't think one could write down the formula in a very convenient way. – saulspatz Dec 17 '20 at 19:10
  • @reuns this looks promising, however what does $v_2(n)$ denote? – r00r Dec 17 '20 at 21:17

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$v_2(n)$ is the 2-adic valuation of $n$. Then look at things like

$$g(z)=e^{z^2}\frac{\sin(\pi z)}{\pi}+\sum_{n\ne 0} \frac{e^{-n^2} v_2(n)}{z-n}$$ which is entire.

It gives $g(0)=0$, add $\frac{\sin(\pi z)}{z^3}$ if you want a meromorphic function such that $f(n)=v_2(n)$ for all $n\in \Bbb{Z}$

reuns
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