How do I solve for $P$ in $ A=\frac{\left \lfloor (n+1)^{P} \right \rfloor}{\left \lfloor n^{P} \right \rfloor} $

Question

I have the following equation:

$$ A=\frac{\left \lfloor (n+1)^{P} \right \rfloor}{\left \lfloor n^{P} \right \rfloor} $$

How to solve for $P$ using $A$ and $n$, where $n \in \Bbb N \setminus \{ 0 \}$ and $A, P \in \Bbb R$?

Answer 1

From the definition of the floor function

$x-1 < \lfloor{x}\rfloor \le x \tag{1}$

Hence if we let

$f(P)=\frac{\left \lfloor (n+1)^{P} \right \rfloor}{\left \lfloor n^{P} \right \rfloor}$

then we can apply inequality (1) to both the numerator and denominator to get

$\frac{(n+1)^P-1}{n^P} < f(P) < \frac{(n+1)^P}{n^P-1} \tag{2}$

or $g(P) < f(P) < h(P)$ with $g(P)=\frac{(n+1)^P-1}{n^P}$ and $h(P)=\frac{(n+1)^P}{n^P-1}$

Since $g(P)$ is an increasing function (for $n\ge1$), solve $g(P)=A$ to get a single solution $P_g$. Similarly solve $h(P)=A$ to get a single solution $P_h$. Then the original equation can have no solutions $P \ge P_g$ as we would then have $f(P) > g(P) \ge g(P_g) = A$. Nor can it have solutions with $P \le P_h$ as then $f(P) < h(P) \le h(P_h) = A$.

So possible solutions will lie in the interval $(P_h , P_g)$ - it can be shown that $P_h < P_g$.

Not all of these will be solutions, because the value of each floor term in $f(P)$ will, for increasing P, change every time it attains an integer value, so the value of $f(P)$ will change every time either $(n+1)^P$ or $n^P$ becomes integral. You will get a positive discontinuity if $(n+1)^P$ is an integer but $n^P$ isn't, and a negative discontinuity wherever $n^P$ is an integer.

Answer 2

First $n \geq 1$, because otherwise the denominator may be $0$ for $0 \leq n < 1$.

Using basic Floor and ceiling functions rules and properties:

Given $n \in \mathbb{N}, n \geq 1$ and $A \in \mathbb{R}$, if $P \in \mathbb{N}$ then $P = \lfloor \log _{\frac {n+1} n} A \rfloor$, and if $P \in \mathbb{R} \setminus \mathbb{N}$ then $P = \log _{\frac {n+1} n} A$.