Question

Suppose that on each trip the bee returned one third of the way to the previous point instead of half of the way. How would the series in Exercise 2 be modified? How would line 50 of Exercise 2 be modified? RUN a modified program for 30 trips and determine what point the bee seems to be approaching.

Short answer

The values are converging towards the simple fraction$\frac{3}{4}=0.75$

Step-by-step solution

Step 1. Write BASIC Program.

Let variable P which locate the position of points. Bee starting from point $P_0$which flies to point $P_1$and lands where 0 and 1 are the coordinates of points. Then bee returns half of the way to $P_0$, landing at $P_2$. From $P_2$, the bee returns half of the way to $P_1$, landing on $P_3$and so forth. Thus, through BASIC program compute and print the bee’s location after 2 to 10 trips.

Step 2. Write Program for bees’ location after 2 to 10 trips.

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

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

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

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

40 For $N=2$to 10

50 LET $P\left(N\right)=P(N-1)\times \left(-1/3\right)\uparrow (N-1)$

60 PRINT $N,P\left(N\right)$

70 NEXT N

80 END

Sample Output:

2 0.66666667

3 0.77777778

4 0.74074074

5 0.75308642

6 0.74897119

7 0.75034294

8 0.74988569

9 0.7500381

10 0.7499873

11 0.75000423

12 0.74999859

13 0.75000047

14 0.74999984

15 0.75000005

16 0.74999998

17 0.75000001

18 0.75

19 0.75

20 0.75

21 0.75

22 0.75

23 0.75

24 0.75

25 0.75

26 0.75

27 0.75

28 0.75

29 0.75

30 0.75

Step 3. State the conclusion.

When N is 2 then $P\left(2\right)$is $\frac{2}{3}$ that is 0.666667. N is 3 then $P\left(3\right)$is $\frac{7}{9}$ that is 0.777778. N is 4 $P\left(4\right)$is $\frac{20}{27}$that is 0.74074074 and so on. You have many answers from loop. But when $N=17$then $P\left(17\right)=0.75000001$and $N=18$then and $N=19$ then again. Thus, N from 18 to 30 then you have the approximate answer that is$P\left(18\right),P\left(19\right),P\left(20\right),P\left(21\right),\dots ,P\left(30\right)=0.75$ that is bee seems to be approaching distance $\frac{3}{4}$from the initial point.

Therefore, from the output, it can be observed that the values are converging towards the simple fraction $\frac{3}{4}=0.75$.