created 08/09/99, additions: 04/08/2007, 09/06/10, 11/04/2012

Modify the DollarsAfterForty program in the chapter so that it asks the user for the interest rate, the initial investment, and the annual contribution. The program prints out the current value of the investment at the end of each year and continues until it reaches or exceeds one million dollars.

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Say that you owe the credit card company $1000.00. The company charges you 1.5% per month on the unpaid balance. You have decided to stop using the card and to pay off the debt by making a monthly payment of N dollars a month. Write a program that asks for the monthly payment, then writes out the balance and total payments so far for every succeeding month until the balance is zero or less.

Enter the monthly payment: 100 Month: 1 balance: 915.0 total payments: 100.0 Month: 2 balance: 828.725 total payments: 200.0 Month: 3 balance: 741.155875 total payments: 300.0 Month: 4 balance: 652.273213125 total payments: 400.0 Month: 5 balance: 562.057311321875 total payments: 500.0 Month: 6 balance: 470.4881709917031 total payments: 600.0 Month: 7 balance: 377.54549355657866 total payments: 700.0 Month: 8 balance: 283.20867595992735 total payments: 800.0 Month: 9 balance: 187.4568060993263 total payments: 900.0 Month: 10 balance: 90.26865819081618 total payments: 1000.0 Month: 11 balance: -8.377311936321576 total payments: 1100.0

For each month, calculate the interest due on the unpaid balance. Then calculate the new balance by adding the interest and subtracting the payment.

**Improved Program:** Have the program prompt for the
beginning balance, the monthly interest, and the payment amount.
Also, when the balance falls below the amount of the monthly payment,
write out the final payment that will bring the balance to exactly zero.

**Further Improved Program:** The output of the program
is somewhat ugly, with the columns not aligned and too many digits printed
in the decimal fractions.
This can be improved.
Place the following lines in the appropriate places in your program.
You will need to modify the last line to fit with the rest of your program.

import java.text.*; DecimalFormat numform = new DecimalFormat(); System.out.print( "balance: " + numform.format(balance) );

For details about how this works, see Chapter 24.

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A certain drug looses 4% of its effectiveness every month it is in storage. When its effectiveness is below 50% it is considered expired and must be discarded. Write a program that determines how many months the drug can remain in storage.

month: 0 effectiveness: 100.0 month: 1 effectiveness: 96.0 month: 2 effectiveness: 92.16 month: 3 effectiveness: 88.47359999999999 month: 4 effectiveness: 84.93465599999999 month: 5 effectiveness: 81.53726975999999 month: 6 effectiveness: 78.27577896959998 month: 7 effectiveness: 75.14474781081599 month: 8 effectiveness: 72.13895789838334 month: 9 effectiveness: 69.253399582448 month: 10 effectiveness: 66.48326359915008 month: 11 effectiveness: 63.82393305518407 month: 12 effectiveness: 61.27097573297671 month: 13 effectiveness: 58.82013670365764 month: 14 effectiveness: 56.46733123551133 month: 15 effectiveness: 54.20863798609088 month: 16 effectiveness: 52.04029246664724 month: 17 effectiveness: 49.95868076798135 DISCARDED

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One of the more amazing facts from calculus is that the following
sum gets closer and closer to the value `e`

the more terms you add in:
^{x}

e^{ x }= 1 + x + x^{ 2 }/2! + x^{ 3 }/3! + x^{ 4 }/4! + x^{ 5 }/5! + x^{ 6 }/6! + . . . .

Remember that `n`

means **!**`n`

factorial,
`n*(n-1)*(n-2)* ... *1.`

For example, if `x`

is 2 then

e^{2}= 1 + 2 + 2^{2}/2! + 2^{3}/3! + 2^{4}/4! + 2^{5}/5! . . . .

e^{2}= 1 + 2 + 4/2 + 8/6 + 16/24 + 32/120 + . . . .

e^{2}= 1 + 2 + 2 + 1.3333 + 0.6666 + 0.2666 + . . . .

e^{2}~ 7.266

More exactly, `e`

^{2} = 7.38907...

Write a program that
asks the user to enter `x`

,
then calculates `e`

using a loop to add up successive terms until the
current term is less than 1.0E-12.
Then write out the value ^{x}`Math.exp(x)`

to see how your value compares.

To do this program sensibly,
the loop will add in a `term`

each iteration.

sum = sum + term;

Look carefully at the first equation for `e`

.
Notice that each term in the series is:
^{x}

x^{ N}/N!

for some `N`

. This is the same as:

x^{(N-1)}/(N-1)! * x/N

This is the __previous term__ times `x/N`

.
So each iteration of the loop merely
has to multiply the previous term by `x/N`

and add it to
the accumulating sum.

Don't let the math scare you away! This is actually a fairly easy program, and is typical of a type of calculation that computers are often used for.

Enter x: 2 n:1 term: 2.0 sum: 3.0 n:2 term: 2.0 sum: 5.0 n:3 term: 1.3333333333333333 sum: 6.333333333333333 n:4 term: 0.6666666666666666 sum: 7.0 n:5 term: 0.26666666666666666 sum: 7.266666666666667 n:6 term: 0.08888888888888889 sum: 7.355555555555555 n:7 term: 0.025396825396825397 sum: 7.3809523809523805 n:8 term: 0.006349206349206349 sum: 7.387301587301587 n:9 term: 0.0014109347442680777 sum: 7.3887125220458545 n:10 term: 2.8218694885361555E-4 sum: 7.388994708994708 n:11 term: 5.130671797338464E-5 sum: 7.389046015712681 n:12 term: 8.551119662230774E-6 sum: 7.3890545668323435 n:13 term: 1.3155568711124268E-6 sum: 7.389055882389215 n:14 term: 1.8793669587320383E-7 sum: 7.3890560703259105 n:15 term: 2.5058226116427178E-8 sum: 7.389056095384136 n:16 term: 3.1322782645533972E-9 sum: 7.389056098516415 n:17 term: 3.6850332524157613E-10 sum: 7.389056098884918 n:18 term: 4.094481391573068E-11 sum: 7.389056098925863 n:19 term: 4.309980412182177E-12 sum: 7.3890560989301735 n:20 term: 4.309980412182177E-13 sum: 7.389056098930604 my e^x: 7.389056098930604 real e^x: 7.38905609893065

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Another amazing fact from calculus is that
you can calculate `1/B`

without doing division.
(Assume that `B`

is a positive real number.)

Say that `x`

is your first guess for the value `1/B`

.
Then your guess can be improved by using the formula:

x' = x*(2-B*x)

Of course, now __that__ guess can be further improved
by using the same formula.
This formula is yet another application of Newton's Method.

For example, say that `B`

is 4.0.
You want to calculate 1/4.
Say that your first
guess for this value is 0.1 ;
then an improved guess is:

x' = 0.1 * (2 - 4*0.1) = 0.1 * (2 - .4) = 0.1*(1.6) = 0.16

The next estimate is:

x' = 0.16 * (2 - 4*0.16) = 0.16 * (2 - 0.64) = 0.16 * 1.36 = 0.2176

A further improvement is:

x' = 0.2176 * (2 - 4*0.2176) = 0.2176 * (2 - 0.8704) = 0.2176 * 1.1296 = 0.24580096

Repeat the formula to further improve the guess.

Write a program that asks the user for
`B`

and then writes out
`1/B`

without using a division.
You will need to think of an appropriate ending condition
for the loop that calculates `1/B`

.

Unfortunately, the first guess for this method must
be between zero and the true value of `1/B`

.
It sounds like you need division to make the first guess,
but this can be avoided by always starting with a
hard-coded (a constant that is part of the program)
very tiny first guess.

For very large values of `B`

your first guess might not
be tiny enough, because `1/B`

is so small.
Protect your code with an if statement
that checks if `B`

is too large for the program to handle.
If the first guess is hard-coded, then you know the maximum value for `B`

.

Of course,
`1/0`

is not defined, so test for this, too.

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Write a program that asks the user for a positive integer `N`

and then
calculates a new value for `N`

based on whether it is even or odd:

if N is even, the new N is N/2. (Use integer division.) if N is odd, the new N is 3*N + 1.

Repeat with each new `N`

until you reach the value 1.

For example, say the initial value is 12. Then the successive values are:

12 (even, next value is 12/2) 6 (even, next value is 6/2) 3 (odd, next value is 3*3+1) 10 (even, next value is 10/2) 5 (odd, next value is 3*5+1) 16 (even, next value is 16/2) 8 (even, next value is 8/2) 4 (even, next value is 4/2) 2 (even, next value is 2/2) 1 (stop calculation)

The sequence of integers is called a *hailstone series* because
the integers increase and decrease, sometimes many times, before eventually reaching 1.
(This is supposed to be like hailstones blowing up and down in a storm before reaching the ground.)

For all integers that have been tested (up to 10^{48}) the series eventually
does reach 1.
But no one knows for sure if this is true for all integers.

**Additional Things to Calculate:**
In addition to calculating and printing out the sequence for a particular initial `N`

,
keep track of the length of the sequence and the maximum integer in the sequence and print
them out when the sequence reaches its end.

A variation on this program is to generate each sequence for initial numbers from 1 to 1000, and to determine the length and starting number of the longest of those sequences. You probably don't want to print out each sequence, just keep track of each length and compare it to the current maximum.

**Hard:** Are there any integers from 1 to 1000 that are not in any hailstone sequence
other than the one they start?
To write a sensible program that determines this, you will need to use an array,
which is the subject of chapter 46.
You may wish to return to this problem after you have covered that chapter.

Note: it is natural to investigate sequences that start with initial numbers in the millions or higher.
The terms of these sequences get very large; larger than can be held in an `int`

variable.
Use data type `long`

to avoid these problems.
That will work for initial values up to at least five billion.

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