0
$\begingroup$

So, there's this equation that I've been pondering upon:(to slove for n) $\left\lfloor N\alpha ^{n} \right\rfloor = \left\lfloor N\alpha ^{n+1}\right\rfloor+1$
I can't remove the floors for approximation purposes because that would distort the answer too much. But, I thought of using the definition of floors.That is if
$\left\lfloor k\right\rfloor = m$
Then:
$m\leq k < m+1$.
Thus, here we can say that we've
$\left\lfloor N \alpha ^{n+1}\right\rfloor+1 \leq N\alpha ^{n} < \left\lfloor N \alpha ^{n+1}\right\rfloor+2$
But that would only complicate the matters more.
Thus, I'm stuck down here. Any help appreciated.

$\endgroup$
1
$\begingroup$

Hint: you are on the right track with thinking at the definitions. Write it as:

$$N \alpha^{n+1} \lt \lfloor N \alpha^n \rfloor \le N \alpha^n \lt \lfloor N \alpha^n \rfloor + 1 = \lfloor N \alpha^{n+1} \rfloor + 2 \le N \alpha^{n+1} + 2$$

$$ \implies \quad N \alpha^{n+1} \lt N \alpha^n \lt N \alpha^{n+1} + 2 \quad \iff \quad \alpha \lt 1 \lt \alpha + \frac{2}{N \alpha^n} $$

$\endgroup$
  • 1
    $\begingroup$ I am not said downvoter, but just to check, I think that should be a $2$ on the right hand side? Or a $\leq$ $\endgroup$ – Badam Baplan Mar 1 '17 at 4:32
  • 1
    $\begingroup$ @BadamBaplan Thank you for the correction, edited and fixed. $\endgroup$ – dxiv Mar 1 '17 at 4:44
  • 1
    $\begingroup$ Then finding the value of $n$ using the inequality $ 1<\alpha+\frac{2}{N\alpha ^{n}}$ ? @dxiv $\endgroup$ – Mooncrater Mar 1 '17 at 4:54
  • 1
    $\begingroup$ @Mooncrater Right, and $\,\alpha^n \lt \frac{2}{N(1-\alpha)}\,$ will give a lower bound for $n\,$. $\endgroup$ – dxiv Mar 1 '17 at 4:58
  • 1
    $\begingroup$ @Mooncrater No, that inequality actually gives a lower bound since $\alpha \lt 1$ and once the inequality is satisfied for $n_0$ it will automatically be satisfied for $\forall n \ge n_0\,$. But I realize that my hint doesn't address the upper bound. Is there any additional known relation between $N$ and $\alpha$ in that context? $\endgroup$ – dxiv Mar 1 '17 at 5:07

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.