Chapter 2.3, Problem 2E

To determine

To enter the given program and run on computer by changing the line 50 .

Explanation of Solution

Given:

The line 50 will be 50 LET $P\left(N\right)=P(N-1)\times \left(-1/2\right)\uparrow (N-1)$

Calculation:

The following is the required BASIC Program-

Let variable P which locate the position of points. Bee starting from point P0 which flies to point P1 and lands where 0 and 1 are the coordinates of points. Then bee returns half of the way to P0, landing at P2. From P2, the bee returns half of the way to P1, landing on P3 and so forth. Thus through BASIC program compute and print the bee’s location after 2 to 10 trip.

Program:

Let P and N are two variables where P is use for the position of bee and N is for number of position change by bee and $P\left(N\right)$ is the distance between the initial positions of bee to actual position of bee.

10 dim $P\left(50\right)$ //P(50) shows the bee position can change 50 times.

20 LET $P\left(0\right)=0$

30 LET $P\left(1\right)=1$

40 For $N=2to10$

50 LET $P\left(N\right)=P(N-1)\times \left(-1/2\right)\uparrow (N-1)$

60 PRINT $N,P(N)$

70 NEXT $N$

80 END

Sample Output:

2 0.5

3 0.75

4 0.625

5 0.6875

6 0.65625

7 0.671875

8 0.6640625

9 0.66796875

10 0.66601562

The result is same as the program 1.

Output Explanation:

When executed program asked for values which is going to executed in loops. N assigned value $2,3,4,5,6,7,8,9,10$ when loop runs. And P(N) as its result that is $0.5,0.75,0.625,0.6875,0.65625,0.671875,0.6640625,0.66796875,0.66601562$ .

This result is as same as program 1.

Consider the equation given below that is as like formula $P\left(N\right)=P\left(N-1\right)+{\left(-1/2\right)}^{n-1}$

  $1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\mathrm{.....}+{\left(-\frac{1}{2}\right)}^{n-1}$

Thus each term series reflect the bee’s return half of the way from $P_{n-1}$ to $P_{n-2}$ .

This equation shows as same as the program been written above.