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There's the floor and ceiling functions article.

Specifically, for $m,n \in \Bbb{Z}$ and $n\gt 0$ we have:

$$ \left\lfloor\frac{\lfloor x \rfloor -m}{n} \right\rfloor = \left\lfloor \frac{x - m}{n}\right\rfloor $$

I thought this would be a useful property to help prove properties of finite sums of such fractionally-defined floored expressions.

Consider functions $f(x) = \left\lfloor\frac{x - r}{d}\right\rfloor$ where $r,d \in \Bbb{Z}$, $0 \leq r \lt d$ and $d \geq 1$, $f: \Bbb{R} \to \Bbb{Z}$.

Then the trend line for single $f$ is defined to be $f^*(x) = \dfrac{x - r}{d}$. This line has the property that $f^*(x) \geq f(x)$ and $f^*(x) - f(x) \lt 1$. That last quantity is called the deviation of the original quantity from its trend line, and it's a real number.

Thus in general, a trend line $f^*(x)$ for a sum $f(x) =\displaystyle \sum_{i = 1}^n C_i f_i(x)$ of such floor-of-fraction functions is defined to be the "least line" in other words, an affine function such that $f^*(x) - f(x) \lt C$ is constant and $C \geq 0$ minimal.

Anyway, I want to compute the trend line and its deviation from $f$ for such finite sums $f$ of these floor-of-fraction functions.

Conjecture. The slope of $f^*$ is equal to $\displaystyle \sum_{i=1}^{n}\frac{C_i}{d_i}$ where $\displaystyle f = \sum_{i = 1}^n C_i\left\lfloor\frac{x - r_i}{d_i}\right\rfloor$.

Proof Attempt. It's true for $n = 1$ and $C_1 = 1$ clearly. Now break up the floor modularly: $$ C_1\left \lfloor\frac{x - r_1}{d_1} \right\rfloor = C_1\frac{x - r_1 - (x-r_1)_{\pmod {d_1}}}{d_1} $$ We get:

$$ f(x) = \frac{C_1}{d_1}x + b(x) $$

where $|b(x)|$ is clearly bounded bounded by $2d_1$. Therefore the slope of any trend line must be "parallel" or $f^*(x)$'s slope has to be $\dfrac{C_1}{d_1}x$.

Question. How does this genearlize to $n \gt 1$?

Daniel Donnelly
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1 Answers1

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The slope of the affine linear function $f^*(x)=mx+b$ is given by $$m = \lim_{x\to\infty}\frac{f^*(x)}{x}$$

Now, since $f(x)-f^*(x)$ is bounded, $$m = \lim_{x\to\infty} \frac{f^*(x)}{x} = \lim_{x\to\infty} \frac{f(x)}{x}$$ and finally, if $f(x) = \sum_{i=1}^n f_i(x)$ where each $f_i(x)/x \to m_i$ as $x\to\infty$, then we have $$m = \lim_{x\to\infty} \frac{f(x)}{x} = \lim_{x\to\infty}\sum_{i=1}^n \frac{f_i(x)}{x} = \sum_{i=1}^n\lim_{x\to\infty}\frac{f_i(x)}{x} = \sum_{i=1}^nm_i$$

In your particular case of $f_i(x) = C_i\left\lfloor\frac{x-r_i}{d_i}\right\rfloor$, you've already shown that $m_i = \frac{C_i}{d_i}$, so the slope of $f^*(x)$ is $\sum_{i=1}^n \frac{C_i}{d_i}$.

Brian Moehring
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