/sci/ - Science & Math

>>14859357
Divide both sides by [math]2^t[/math].
For [math]N[/math] sufficiently large you can assume that the largest possible value on the left is assumed with [math](C - 1)C/2^{t + 1} \geq C[/math], so for Landau notation purposes you can discard the [math]C 2^t[/math] term then square root both sides.