1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ I believe the exponent inside the parentheses should be $2-4k$. $\endgroup$ – Ross Millikan Apr 28 at 15:41
0
$\begingroup$

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.
enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.