Jim Lambers
MAT 169
Fall Semester 2009-10
Lecture 3 Examples
This example corresponds to Section 8.1 in the text.
Example The sequence
an= (−1)n, n ≥0,
has terms that alternate between 1 and −1. That is, a0= 1, a1=−1, a2= 1, a3=−1,and so on.
Now, suppose we want to construct a formula for a sequence whose terms alternate signs in pairs.
That is,
a0=a1= 1, a2=a3=−1, a4=a5= 1,
and so on. We present two approaches to this.
The first approach makes use of the floor function bxc. For any real number x,bxcis defined
to be the greatest integer that is less than or equal to x. For example,
b1c= 1,bπc= 3,b14.9c= 14.
The floor function is also known as the greatest integer function, and is an example of a step
function, since its graph, shown in Figure 1, consists of line segments that are arranged like steps.
To define our sequence, we use the floor function to obtain the sequence of exponents for −1,
b0= 0, b1= 0, b2= 1, b3= 1, b4= 2, b5= 2,
and so on. This is accomplished by setting bn=bx/2c,which, for odd x, results in a fraction, equal
to 0.5, which is eliminated by taking the floor, or rounding down. By using bnas the exponent to
−1, we obtain the sequence
an= (−1)bn= (−1)bx/2c,
which has the desired terms that alternate between 1 and −1 in pairs. In general, to obtain a
sequence that alternates between 1 and −1 every nterms, we can use the sequence with terms
(−1)bx/nc.
An alternative approach uses trigonometric functions. Figure 2 shows the graph of sinxfor
0≤x≤4π. The circles on the graph indicate the points corresponding to
x=π
4+kπ
2, k = 0,1,...,7.
Note that the values of sinxat these points alternates between ±√2/2 in pairs. Therefore, we can
define the sequence anby
an=√2 sin π
4+nπ
2.
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