Draw the graphs of some step functions

by RoRi

July 16, 2015

For all $x \in \mathbb{R}$ define

$f(x) = [x], \qquad g(x) = [2x]$

where $[x]$ denotes the greatest integer less than or equal to $x$ . Draw the function $h(x)$ for each of the following on the interval $[-1,2]$ :

$h(x) = f(x) + g(x)$ .
$h(x) = f(x) + g\left(\frac{x}{2} \right)$ .
$h(x) = f(x) g(x)$ .
$h(x) = \frac{1}{4} f(x) g(x).$

Graph of $h(x) = f(x) + g(x) = [x]+[2x]$ .
Graph of $h(x) = f(x) + g\left(\frac{x}{2}\right) = [x] + [x] = 2[x]$ .
The graph of $h(x) = f(x)g(x) = [x]\cdot[2x]$ .
The graph of $h(x) = \frac{1}{4}f(x)g(x) = \frac{1}{4}[x][2x]$ .

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6 comments

March 15, 2021 at 2:02 pm

Anonymous says:

I think you shall define the value of function when x = 2.

Reply
January 16, 2021 at 8:52 am

Artemii says:

a) is wrong. For example h(-0.5) = [-0.5] + [-1] = 0 – 1 = -1.
In reality, the drawing will be like this:
h( -1 ) = -3;
h( (-1, 0.5] ) = -1;
h( (-0.5, 0] ) = 0;
h( (0, 0.5] ) = 2;
h( (0.5, 1] ) = 3;
h( (1, 1.5] ) = 5;
h( (1.5, 2] ) = 6.
(I hope such a record does not create difficulties)

Reply
- January 16, 2021 at 9:26 am
  
  Artemii says:
  
  I’m sorry, it’s not. My function is the smallest integer >= x
  
  Reply
October 19, 2020 at 10:00 am

sigh says:

c) is wrong. Note that ceil(0) * ceil(anything) = 0. So the interval [-1,0] on graph c should be 0.

Reply
- March 15, 2021 at 2:07 pm
  
  Anonymous says:
  
  Let’s take x = -1/2 for instance. We have [x] = -1 and [2x] = -1. Therefore, h(x) = f(x)g(x) = 1
  
  Reply
May 2, 2019 at 10:54 am

Artem says:

There is a mistake: the interval is closed on the right: you should also include the point at x=2 in the plots

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