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I am simplifying an equation and I would like to known if there is any approximation for this value $\log_2(1-2^x)$

EDIT

I make this question because I am simplifying a big formula to find the big notation of a certain algorithm. The expression is:

$$\log_2(1-l)/\log_2(1-2^{nH(m/n)-n})$$

where $0<l< 1$, $m,n>0$ and $H$ is the binary entropy function.

In practice, I known that the algorithm is exponential, but I do not known how simplify that to get the exponential expression. Could you help me please?

juaninf
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2 Answers2

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Since $x<0$, let $x=-y$ and use composition of Taylor series around $y=0$

$$2^{-y}=e^{-y \log(2)}=1-y \log (2)+\frac{1}{2} y^2 \log ^2(2)+O\left(y^3\right)$$ $$1-2^{-y}=y \log (2)-\frac{1}{2} y^2 \log ^2(2)+O\left(y^3\right)=y\log(2)\left( 1-\frac{1}{2} y \log (2)+O\left(y^2\right)\right)$$ $$\log(1-2^{-y})=\log(y)+\log(\log(2))+\log\left( 1-\frac{1}{2} y \log (2)+O\left(y^2\right)\right)$$ $$\log(1-2^{-y})=\log(y)+\log(\log(2))-\frac{1}{2} y \log (2)+O\left(y^2\right)$$

$$\log_2(1-2^{-y})=\frac{\log \left(1-2^{-y}\right)}{\log (2)}=\frac{\log (\log (2))}{\log (2)}+\frac{\log (y)}{\log (2)}-\frac{y}{2}+O\left(y^2\right)$$ which seems to be quite good $$\left( \begin{array}{ccc} y & \text{exact} & \text{approximation} \\ 0.05 & -4.87562 & -4.87569 \\ 0.10 & -3.90041 & -3.90069 \\ 0.15 & -3.34008 & -3.34073 \\ 0.20 & -2.94954 & -2.95069 \\ 0.25 & -2.65196 & -2.65377 \\ 0.30 & -2.41313 & -2.41573 \\ 0.35 & -2.21480 & -2.21834 \\ 0.40 & -2.04608 & -2.05069 \\ 0.45 & -1.89993 & -1.90577 \\ 0.50 & -1.77155 & -1.77877 \\ 0.55 & -1.65754 & -1.66626 \\ 0.60 & -1.55535 & -1.56573 \\ 0.65 & -1.46307 & -1.47525 \\ 0.70 & -1.37922 & -1.39334 \\ 0.75 & -1.30259 & -1.3188 \\ 0.80 & -1.23226 & -1.25069 \\ 0.85 & -1.16743 & -1.18823 \\ 0.90 & -1.10745 & -1.13077 \\ 0.95 & -1.05180 & -1.07777 \end{array} \right)$$

If more accuracy is required, you could continue the process and get $$\log_2(1-2^{-y})=\frac{\log (\log (2))}{\log (2)}+\frac{\log (y)}{\log (2)}-\frac{y}{2}+\frac{1}{24} y^2 \log (2)+O\left(y^4\right)$$ For illustartion purposes $$\int_0^1 \log_2(1-2^{-y})\,dy=-\frac{1}{2}-\frac{\pi ^2}{12 \log ^2(2)}\approx -2.21186$$ while the last approximation to $O\left(y^4\right)$ would give $$-\frac{1}{4}+\frac{\log (2)}{72}+\frac{\log (\log (2))-1}{\log (2)}\approx -2.21183$$

Claude Leibovici
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HINT.- Obviously the domain is for $x\lt 0$. You do have $$y=\log_2(1-2^x)\iff 2^y=1-2^x\iff x=\log_2(1-2^y)$$ so the function is symmetric respect the diagonal $y=x$.

Besides the intersection, given by $2^{x+1}=1$, is the point $(-1,-1)$. Consequently you do have to approximate just on the interval $-1\le x\lt0$.There are several methods to do it.

Personally, I would first choose to deal with the equivalent function with the Neperian logarithm $$\frac{\log(1-2^x)}{\log 2}$$

Piquito
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