4.3 Transformation of graphs by greatest integer function  (Page 18/2)

Drawings of graphs resulting from transformation applied by greatest integer function (GIF) follow the same reasoning and steps as deliberated for modulus operator. Before, we proceed to draw graphs for different function forms, we need to recapitulate the graph of greatest integer function (GIF) and also infer thereupon few of the values of GIF around zero.

For clarity, we apply circular symbols : solid circle to denote inclusion of point and empty circle to denote exclusion of point. Using solid circle is optional, but helps to identify points on the graph. The values of GIF around zero are (we can write these expressions by observing graph. A bit of practice to write down these intervals helps.) :

$$\left[x\right]=-3;-3\le x<-2$$ $$\left[x\right]=-2;-2\le x<-1$$ $$\left[x\right]=-1;-1\le x<0$$ $$\left[x\right]=0;0\le x<1$$ $$\left[x\right]=1;1\le x<2$$ $$\left[x\right]=2;2\le x<3$$ $$\left[x\right]=3;3\le x<4$$

Important point to note is that lower integer in the interval is included and higher integer is excluded. For negative interval like $-3\le x<-2$ , note that -3 is lower end whereas -2 is higher end. Yet another feature of this function is that domain of the function is continuous in R. It means, there need to be a function value corresponding to all x.

A function like y=f(x) has different elements. We can apply GIF to these elements of the function. There are following different possibilities :

1 : y= f([x])

2 : y=[f(x)]

3 : [y]=f(x)

Greatest integer operator applied to independent variable

The form of transformation is depicted as :

$$y=f\left(x\right)\Rightarrow y=f\left(\left[x\right]\right)$$

The graph of y=f(x) is transformed in y=f([x]) by virtue of changes in the argument values due to operation on independent variable. The independent variable of the function is subjected to greatest function operator. This changes the normal real value input to the function. Instead of real numbers, independent variable to function is rendered to be integers – depending on the value of x and interval it belongs to. A value like x= - 2.3 is passed to the function as -3 in the interval $-3\le x<-2$ .

Clearly, real values of “x” are truncated to integer values in the interval of unity i.e. [-1,0), [0,1), [1.2) etc. It means that values of the function y=f(|x|) will remain same as that of its value corresponding to integral value of “x” till value of “x” changes to next interval. We need to apply modification to the curve to reflect this effect. Knowing that truncation takes place for successive integral values of x, we divide graph of y=f(x) to correspond to 1 unit segments of x-axis. For this, we draw lines parallel to y-axis at integral points along x-axis. From intersection point of lines drawn and function graph, we draw lines parallel to x-axis for the whole interval which extends for a unit value. This ensures that function values remain same to that of function value for the lower integral value of x in a particular interval of one.

From the point of construction of the graph of y=f([x]), we need to modify the graph of y=f(x) as :

1 : Draw lines parallel to y-axis (vertical lines) at integral values along x-axis to cover the graph of y=f(x).