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I was reading a statistics paper and saw a formula with floor operators. I wondered how to solve for one of the variables in the formula, but realized that I did not know how to work with these things.

I have the following, where I am trying to solve for $a$.

$y = \lfloor log(a) - log(b) \rfloor + 1$

I am confused about the order of operations with the floor operators $\lfloor \rfloor$.

I see some options

  1. Ignore the floor and solve for $a$

$y = log(a) - log(b) + 1$

$y - 1 + log(b) = log(a)$

$a = exp[y - 1 + log(b) ]$

  1. Apply floor to each expression in the formula then use ceiling

$y = \lfloor log(a) - log(b) \rfloor + 1$

$y = \lfloor log(a) \rfloor - \lfloor log(b) \rfloor + 1$

$\lfloor log(a) \rfloor = y - 1 + \lfloor log(b) \rfloor$

$log(a) = \lceil y - 1 + \lfloor log(b) \rfloor\rceil$

$a = exp[\lceil y - 1 + \lfloor log(b) \rfloor\rceil]$

I could keep going with more I think, but I hope this gets the point across. I don't know how to do algebra with these ceiling and floor operators.

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  • $\begingroup$ There is a more generic question. How should such equations as $~\displaystyle f(x) = \left\lfloor g(x)\right\rfloor~$ or $~\displaystyle f(x) \geq \left\lfloor g(x)\right\rfloor~$ be attacked? My standard approach is to create the $2$ variables $A$ and $r$, where $~A \in \Bbb{Z},~ 0 \leq r < 1,~$ and $~\displaystyle x = A + r \implies A = \left\lfloor x\right\rfloor.$ $\endgroup$ Commented Apr 12, 2022 at 4:46

1 Answer 1

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You have that $m = \lfloor{x}\rfloor$ if and only if $m \leq x < m + 1$. Thus, $$ y - 1 = \lfloor \log(a) - \log(b) \rfloor \implies y - 1 \leq \log(a) - \log(b) < y $$ and so we get $$ y - 1 + \log(b) \leq \log(a) < y + \log(b) $$ Exponentiating $$ be^{y - 1} \leq a < be^y $$ and so on...

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  • $\begingroup$ This is nice. In practice, when there is a large difference between the left hand side and the right hand side and say there is a need to make a program that calculates $a$ and say that the assumption is to be on the conservative side of things how to people tend to proceed? Could someone say $a = be^y$ or $a = be^y - 1$? $\endgroup$
    – Alex
    Commented Apr 12, 2022 at 17:48

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