Nearest integer

Algebra Level pending

I have a function called $n(x)$ such that:

$\begin{cases}~\text{if}~\{x\}<0.5~\text{then}~n(x)=\lfloor x \rfloor \\ \text{if}~\{x\}=0.5~\text{then}~n(x)=x \\ \text{if}~\{x\}>0.5~\text{then}~n(x)=\lfloor x \rfloor+1 \end{cases}$

What is a good approximation of $n(x)$ ?

Note

$\{ \cdot \}$ is the fractional part function
$\lfloor \cdot \rfloor$ is the floor function

\lfloor x-0.5 \rfloor

x+0.5

\lfloor x+0.5 \rfloor

1 solution

Henry U
Oct 14, 2018

The cases $(x) < 0.5$ and $(x) > 0.5$ mean that $x$ is closer to the next integer smaller (for $(x) < 0.5$ ) or larger (for $(x) > 0.5$ ) than x. Rounding down in both cases, but adding 1 in the second case is the same as rounding to the nearest integer.

$\lfloor x+0.5 \rfloor$ does exactly that:

$0 \leq (x) \leq 1 \Leftrightarrow 0.5 \leq x+0.5 \leq 1.5$ . After rounding down, $\lfloor x+0.5 \rfloor = \begin{cases} \lfloor x \rfloor + 0, & \text{if } (x)<0.5 \\ \lfloor x \rfloor + 1, & \text{if } (x)>0.5 \end{cases}$ .

So $\boxed{\lfloor x+0.5 \rfloor}$ only differs at one point from the given function which makes it a good aproximation.

Great, but it should be $\lfloor x+0.5 \rfloor=\begin{cases} \lfloor x \rfloor~\text{if}~(x)<0.5 \\ \lfloor x \rfloor+1~\text{if}~(x)>0.5 \end{cases}$ because every problem that I made if I say nothing then that means it applied to real numbers, not just $0<x<1$ .

Gia Hoàng Phạm - 2 years, 7 months ago

Oh right, I forgot about that. Now I've corrected it.

Henry U - 2 years, 7 months ago

Nearest integer

1 solution

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