Wednesday, 29 September 2004, at start of class.

We would like to see how many iterations it takes to get 15 digits of accuracy. You can use the following examples as models:

• Example 3: Loop Examples,   
• Example 4: Calculate Pi.

• There are 5 calculations below, so you should produce 5 short Java programs.
• In each case, at the end of the programs you should print the final approximation to the constant, either e, pi, or phi.
• Each of the formulas calls for infinitely many terms, but in each case you need to use just some finite number of terms.
• In each case you should use the smallest number of iterations needed to produce roughly 15 digits of accuracy in the constant. This "smallest number" is an interesting result of each part, since it shows how rapidly the method converges to the constant.
• Each program should print the intermediate results obtained at each interation of the algorithm.

1. e -- the base of natural logarithms:
   • Use the following formula:

   • Hint:

2. pi -- the ratio of circumference to diameter of a circle:
   • Use the following formula:

   • Hint: Pretty much the same as the previous one. However, it doesn't converge nearly as fast as I thought it would. For this series, you should determine how many terms are needed to get pi accurate to 3 significant digits, that is: 3.14.

3. pi -- the ratio of circumference to diameter of a circle:
   • Use the following formula:

   • Hint: Don't panic; it's not really very complicated.

4. phi -- the golden ratio = (1/2)(1 + sqrt(5)):
   • Use the following formula:

   • Hint: It looks hard, but you should write a short finite version. This can be calculated from the bottom up to the top. At each stage just add 1 and divide into 1.
   • Super Hint: Look at the following definitions:

     
     p0 =                          1
     p1 =                      1 + 1
     p2 =               1 + 1/(1 + 1)
     p3 =        1 + 1/(1 + 1/(1 + 1))
     p4 = 1 + 1/(1 + 1/(1 + 1/(1 + 1)))
     

     Now write p4 in terms of p3, that is p4 = some expression involving p3.
     Write p3 in terms of p2, that is p3 = some expression involving p2.
     Write p2 in terms of p1, that is p2 = some expression involving p1.
     Write p1 in terms of p0, that is p1 = some expression involving p0.
     Do you see the simple pattern?

5. pi -- the ratio of circumference to diameter of a circle:
   • Use the following formula:

   • Hint: Similar to the previous one.
     How can you get sqrt(2 + sqrt(2)) from sqrt(2)?
     Then how can you get sqrt(2 + sqrt(2 + sqrt(2))) from sqrt(2 + sqrt(2))? Just repeat this.

     Also notice that here you are multiplying successive terms together, rather than adding, and the final value of pi is buried in the denominator.

1. Source listings (the Java code) for each of the 5 programs.

2. Results of runs for each of the 5 programs.

   Note: You should use netBeans, though it is permissible to use Java from another source. In netBeans, the simplest way to make up the material to print is to just copy from netBeans (using ctrl-C) and paste into Word or WordPad (using ctrl-V). Then print the resulting document. You can assemble all 5 programs into a single document to print.

   In case you have written a program for one part, but it doesn't produce output or doesn't produce the correct output, you should still turn in a listing for part credit.

Here is another equation for e (here the pattern continues as 2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, 1, 1,...):