Moving terms outside of a floor

Question

If I have $x^{\lfloor\frac{c}{a-b}\rfloor}$ is this equivalent to $(x^{\lfloor\frac{1}{a-b}\rfloor})^{\lfloor c \rfloor}$ if a,b,c are all integers and x is between 0 and 1?

I'm concerned with this expression in the context of a geometric series. Normally if we have $\sum_{i=0}^{\infty} r^{q}$ with |r| < 1 then this sum converges to $\frac{1}{1-q}$. So I'm wondering if I can do something like $\sum_{i=0}^{\infty} (x^{\lfloor\frac{1}{a-b}\rfloor})^{\lfloor i \rfloor}$

It seems that the answer is no, because we could have something like $\frac{2}{3}$ which is .75 but the floor is 0. Splitting it up we would end up with .75

Answer 1

No.   Witness:

$$\left\lfloor \tfrac{4}{5-2}\right\rfloor = 1 \\ \lfloor 4\rfloor\left\lfloor \tfrac 1{5-2} \right\rfloor=0$$

Answer 2

Unfortunately not. Floor is only guaranteed to be splittable around addition between an integer and a real number. Therefore, we are only able to do the following:

$$\lfloor \frac c {a-b}\rfloor = \lfloor \frac {c - ca + cb} {a-b}\rfloor + c$$

This is not particularly useful as an identity for your particular situation. since the addition of powers changes to multiplication, about all you can do is factor out terms somehow. If either a, b, or c is a constant then this rule might apply to every term in the sequence. I cannot really healp you with that as it isn't my strong suit. However, this is how you can reduce the expression. As for the original claim... well I couldn't turn it into multiplication could I? So there's your answer.

Answer 3

So I'm a bit late to the party but it seems to me that the other answers, while correct, are incomplete. Reading between the lines it seems like you're trying to evaluate the series

$$\sum_{k = 0}^{\infty} x^{\lfloor \frac{k}{a-b} \rfloor},\quad\text{integers }a,b \text{ and } 0 \leqslant x < 1.$$

You'd like to be able to pull out the index variable $k$ and then treat it like a regular geometric progression. (Note that $\lfloor k \rfloor = k$ for all integers $k$.) However, like others have stated, this is invalid. You can sometimes do it, but not here. The general rule (for real $r$) is

$$\lfloor nr \rfloor = n \lfloor r \rfloor \quad \iff \quad \{r\} < \tfrac 1 n,\quad \text{integer } n>0,$$

where $\{r\} := r - \lfloor r \rfloor$ is the fractional part of $r$.

Anyway, let's see why this doesn't work for our summand. First, we'll put $m = a-b$ to clean things up. We cannot have $m = 0$, and the series diverges for negative $m$, so we must have $m > 0$. Consider the terms after the first, those with $k > 0$. When is it true that $\lfloor k/m \rfloor = \lfloor 1/m \rfloor k$? Precisely when $\{1/m\} < 1/k$. If $m=1$ this certaintly holds, but if $m>1$ we have

$$ \{1/m\} < 1/k \quad \iff\quad 1/m < 1/k \quad \iff \quad k < m.$$

So it works for the first few terms, but once $k$ is large enough it's always wrong to pull it out! This makes sense: $\lfloor 1/m \rfloor$ = 0 for $m > 1$, so multiplying it by $k$ on the outside will never get us back up to $\lfloor k/m \rfloor$, unless $\lfloor k/m \rfloor$ is $0$.

In case you're still interested, here's the proper way to do the sum: realize that the first $m$ terms are equal to $x^0$, the next $m$ terms are equal to $x^1$, and so forth. So the sum is equal to

$$m\sum_{j=0}^{\infty} x^j = \frac{m}{1-x}.$$