created 08/09/99, addition: 04/08/2007

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 monthy 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.

Click here view the solution for this question.

Click here view the solution for the Improved Program of this question.

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 DISCARDEDClick here to go back to the main menu.

Click here view the solution for this question.

One of the more amazing facts from calculus is that the following
sum gets closer and closer to the value e^{x}
the more terms you add in:

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^{x}.
Notice that each term is:

x^{ N}/N!

for some N. This is the same as:

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

This is the same as 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: 7.38905609893065Click here to go back to the main menu.

Click here view the solution for this question.

Another amazing fact from calculus is that
division can be done without using a divide operation.
Say that you wish to divide `A`

by `B`

to produce `A/B`

.
You could do this by multiplying `A`

by the reciprocal of `B`

:

A * (1/B)

Of course, now you need the reciprocal of `B`

, but
this, too, can be calculated without division.
Say that `x`

is your best guess about 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 applying the same formula to it.
This formula is yet another application of Newton's Method.

For example, say that you want to calculate 1/4 and your first guess for this value is 0.1 (see below for a better first guess). Then an improved guess is:

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

Then the next guess 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
`A`

and `B`

and then writes out
`A/B`

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

.

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

.
It sounds like this might call for division,
but this can be avoided by starting with a
very tiny first guess.
Make the first guess negative if `B`

is negative,
and positive if `B`

is positive.
For some large values of `B`

, your first guess might not
be tiny enough, so protect your code with an if statement
that determines when `B`

is too large.
Also, the reciprocol of zero is not defined, so use
an if statement to test for this, too.

Click here view the solution for this question.