2017年8月18日 · $\begingroup$ I think "simply" isn't quite the right word, maybe "effectively." IEEE 754 format (in normal form) stores numbers as $\pm 1.bbb...bb \times 2^n$, so to compute the floor it would first need to figure out where the decimal place is, and then replaces the corresponding digits (if any) to 0, and if the result would be 0, then the format switches to …

so clearly the floor of x divided by x must be less then or equal to 2/3; or x divided by the floor of x is greater then or equal to 3/2; Of course there is another constraint that I have left out (3⌊x⌋ ≤ 2x < 3⌊x⌋+1) but I am sure it is simpler this way

2012年1月25日 · Is there a convenient way to typeset the floor or ceiling of a number, without needing to separately code the left and right parts? For example, is there some way to do $\\ceil{x}$ instead of $\\lce...

The correct answer is it depends how you define floor and ceil. You could define as shown here the more common way with always rounding downward or upward on the number line. OR. Floor always rounding towards zero. Ceiling always rounding away from zero. E.g floor(x)=-floor(-x) if x<0, floor(x) otherwise

2013年6月8日 · \floor is not defined in amsmath. The \DeclaredPairedDelimiter' is good, but in comparison to the \newcommand` above it mostly provides an easy way to change the code when a different size is required.

2022年8月11日 · Alternatively, I tried to circumvent the floor function by using the identity $$ x \bmod y = x - y \left\lfloor \frac{x}{y} \right\rfloor $$ namely using $$ \left\lfloor \frac{x}{y} \right\rfloor = \frac{x - (x \bmod y)}{y} $$ but the modulo operation seems to be as worse as the floor function in terms for finding a solution for the sum $$ M(a ...

Again considering your first example, for $1 \leq x < 2$, the floor function maps everything to 1, so you end up with a rectangle of width 1 and height 1. It is the areas of these rectangles you need to add to find the value of the integral (being careful to understand that rectangles below the x-axis have "negative areas").

The definition of Floor is $\lfloor x \rfloor$ = Largest integer less than x. This is very similar to rounding down as $\lfloor 2.3 \rfloor = \lfloor 2.999 \rfloor = 2$. However, the subtlety is that for negative numbers it acts slightly differently, as $\lfloor -1.5 \rfloor = -2$ which might not be what you expect at first.

There are some threads here, in which it is explained how to use \\lceil \\rceil \\lfloor \\rfloor. But generally, in math, there is a sign that looks like a combination of ceil and floor, which means...

If you try asking Wolfram Alpha to differentiate the floor function, it will just output "Floor'(x)". If you force Wolfram Alpha to plot the derivative of the floor function, I think what Wolfram Alpha does is it as an infinite sum of dirac deltas, so that when you …