# How to solve this inequality exactly?

I am trying to solve the inequality $$1-3\cdot 2^{1-4k^2}+3\cdot2^{3-(2k+1)^2} >x$$ exactly for $$k$$. In fact, I am looking for the smallest integer $$k\geq1$$ for which this inequality holds. I am particularly interested in the resulting $$k$$ for $$x=0.5927$$ (which is an approximation of the critical value for site percolation on the square lattice), but to keep things general we could take $$x \in (0,1)$$. I've tried to solve it, but I rapidly run into problems. My instinct is to take the $$\log_2$$ on the left side, but since it's a sum, things do not get a lot prettier. Rewriting to $$2^{1-4k^2}(2^{1-4k}-1)>\frac{x-1}{3}$$ also hasn't helped me a lot. Does anyone know how to tackle such an equation? Is it even possible to solve it exactly? I don't have (access) to Mathematica or similar software that solves equations exactly, maybe someone could get it to work?

• I believe the exponent inside the parentheses should be $2-4k$. – Ross Millikan Apr 28 at 15:41

## 1 Answer

If you plot $$2^{1-4k^2}(2^{2-4k}-1)$$ you find it rises rapidly from $$-\frac 3{64}$$ at $$k=1$$ to very near $$0$$ at $$k=2$$ (it is $$-\frac {31}{2^{20}}$$ there) so unless $$x$$ is very close to $$1$$ the answer will be $$k=1$$. The term in parentheses approaches $$-1$$ rapidly, so for $$x$$ very close to $$1$$ I would take it to be $$-1$$ and solve the rest, which you can do explicitly $$1-4k^2=\log_2\left(\frac {1-x}3\right)\\k=\frac 12\sqrt{1-\log_2\left(\frac {1-x}3\right)}$$ Now round up and check that it is large enough in the original equation. If not add $$1$$ and you will be there. 