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Few function expressions in different intervals are :
$$\text{For}\phantom{\rule{1em}{0ex}}-2\le x<-\mathrm{1,}\phantom{\rule{1em}{0ex}}f\left(x\right)=\left\{x\right\}=x-\left[x\right]=x-\left(-2\right)=x+2$$
$$\text{For}\phantom{\rule{1em}{0ex}}-1\le x<\mathrm{0,}\phantom{\rule{1em}{0ex}}f\left(x\right)=\left\{x\right\}=x-\left[x\right]=x-\left(-1\right)=x+1$$
$$\text{For}\phantom{\rule{1em}{0ex}}0\le x<\mathrm{1,}\phantom{\rule{1em}{0ex}}f\left(x\right)=\left\{x\right\}=x-\left[x\right]=x-0=x$$
$$\text{For}\phantom{\rule{1em}{0ex}}1\le x<\mathrm{2,}\phantom{\rule{1em}{0ex}}f\left(x\right)=\left\{x\right\}=x-\left[x\right]=x-1$$
$$\text{For}\phantom{\rule{1em}{0ex}}2\le x<\mathrm{3,}\phantom{\rule{1em}{0ex}}f\left(x\right)=\left\{x\right\}=x-\left[x\right]=x-2$$
The graph of the function is shown here :
We see that there is no restriction on values of x and as such its domain has the interval equal to that of real numbers. The fractional part function can only evaluate to non-negative values between 0≤y<1. Hence,
$$\text{Domain}=R$$
$$\text{Range}=0\le y<1$$
FPF is a periodic function. The values are repeated with a period of 1. Further, function is defined for all real x, but graph is not continous. It breaks at integral values of x.
We have seen that greatest integer function represents the integer, which can be considered to be the floor integral value of a real number. Correspondingly, we define a ceiling function called “least integer function (LIF)”, which returns the least integer greater than or equal to the number (x). We denote least integer function as “[x)” or "(x)". Some authors reserve "(x)" for near integer function. It is not important as we can always specify what we mean by qualifying the symbol explicitly. We interpret LIF as :
Clearly, least integer function returns a value, which is the integral “ceiling” of the number. For this reason, least integer function is also known as “ceiling” function. Working rules for finding least integer function are :
The value of f(x) is an integer (n) such that :
$$f\left(x\right)=n;\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}n-1<x\le n\phantom{\rule{1em}{0ex}}n\in Z$$
Few initial values of the functions are :
$$For\phantom{\rule{1em}{0ex}}-3<x\le -\mathrm{2,}\phantom{\rule{1em}{0ex}}f\left(x\right)=\left[x\right)=-2$$
$$For\phantom{\rule{1em}{0ex}}-2<x\le -\mathrm{1,}\phantom{\rule{1em}{0ex}}f\left(x\right)=\left[x\right)=-1$$
$$For\phantom{\rule{1em}{0ex}}-1<x\le \mathrm{0,}\phantom{\rule{1em}{0ex}}f\left(x\right)=\left[x\right)=0$$
$$For\phantom{\rule{1em}{0ex}}0<x\le \mathrm{1,}\phantom{\rule{1em}{0ex}}f\left(x\right)=\left[x\right)=1$$
$$For\phantom{\rule{1em}{0ex}}1<x\le \mathrm{2,}\phantom{\rule{1em}{0ex}}f\left(x\right)=\left[x\right]=2$$
We see that there is no restriction on values of “x” and as such its domain has the interval equal to that of real numbers. On the other hand, the least integer function evaluates only to integer values. It means that the range of the function is set of integers, denoted by "Z". Hence,
$$\text{Domain}=R$$
$$\text{Range}=Z$$
GIF is not a periodic function. Though function is defined for all real x, but graph is not continous. It breaks at integral values of x.
Certain properties of least integer function are presented here :
1: If and only if “x”is an integer, then :
$$\left[x\right)=x$$
2: If and only if at least either “x” or “y” is an integer, then :
$$[x+y)=\left[x\right)+\left[y\right)$$
For example, let x = 2.27 and y = 0.63. Then,
$$\Rightarrow [x+y)=[2.27+0.63)=\left[2.9\right)=3$$
$$\Rightarrow \left[x\right)+\left[y\right)=\left[2.27\right)+\left[0.63\right)=3+1=4$$
However, if one of two numbers is integer like x = 2 and y = 0.63, then the proposed identity as above is true.
4: If “x” belongs to integer set, then :
$$\left[x\right)+[-x)=0;\phantom{\rule{1em}{0ex}}x\in Z$$
For example, let x = 2.Then
$$\Rightarrow \left[2\right)+[-2)=2-2=0$$
We can use this identity to test whether “x” is an integer or not?
3: If “x” does not belong to integer set, then :
$$\left[x\right)+[-x)=+1;\phantom{\rule{1em}{0ex}}x\notin Z$$
For example, let x = 2.7.Then
$$\Rightarrow \left[2.7\right)+[-2.7)=3-2=+1$$
Nearest integer function, as the name suggests, returns the nearest integer. It is denoted by the symbol, "(x)".
The value of "(x)" is an integer "n" such that :
$$f\left(x\right)=\left(x\right)=n;\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}n\le x\le n+1/\mathrm{2,}\phantom{\rule{1em}{0ex}}n\in Z$$
$$f\left(x\right)=n+1;\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}n+1/2<x\le n+\mathrm{1,}\phantom{\rule{1em}{0ex}}n\in Z$$
Examples :
$$\left(2.3\right)=\mathrm{2,}\phantom{\rule{1em}{0ex}}\left(2.6\right)=3$$
$$\left(-2.3\right)=-\mathrm{2,}\phantom{\rule{1em}{0ex}}\left(-2.6\right)=-3$$
Find domain of the function :
$$f\left(x\right)=\sqrt{{x}^{2}-[x{]}^{2}}$$
We analyze given function using its properties to find domain. Subsequently, we shall use graphical solution, which is more elegant. Now, for radical function,
$$\Rightarrow {x}^{2}-[x{]}^{2}\ge 0$$
Evaluation of this expression for integer values of x is easy. We know that [x] evaluates to x for all integer values of x :
$$\left[x\right]=x;\phantom{\rule{1em}{0ex}}x\in Z$$
Squaring both sides,
$$[x{]}^{2}={x}^{2};\phantom{\rule{1em}{0ex}}x\in Z$$ $${x}^{2}-[x{]}^{2}=0;\phantom{\rule{1em}{0ex}}x\in Z$$
However, evaluation of expression is slightly difficult for other values of x. Now, consider positive interval 1≤x<2. Here, [x] evaluates to 1 and its square is 1, which is less than or equal to ${x}^{2}$ . On the other hand, in negative interval -2≤x<-1, [x] evaluates to -2 and its square is 4, which is equal to or greater than ${x}^{2}$ .
$$[x{]}^{2}\le {x}^{2};\phantom{\rule{1em}{0ex}}x>0$$ $$[x{]}^{2}\ge {x}^{2};\phantom{\rule{1em}{0ex}}x>0$$
Note that we have included “equal to sign” for both intervals of x. Equal to sign is appropriate when x is integer. For x=0, expression evaluates to 0. It means expression is non-negative for all non-negative x. But expression also evaluates to 0 for negative integers. Hence, domain of given function is :
$$\text{Domain}=\left(\mathrm{0,}\infty \right)\cup \{-n;n\in N\}$$
Graphical analysis
We draw $y=\left[x\right]$ and $y=[x{]}^{2}$ as in the first and second figures. Finally, we superimpose $y={x}^{2}$ on the graph $y=[x{]}^{2}$ as shown in the third figure. Noting values of x for which value of ${x}^{2}$ is greater than or equal to $[x{]}^{2}$ , the domain of the function is :
$$\text{Domain}=\left(\mathrm{0,}\infty \right)\cup \{-n;n\in N\}$$
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