adding more chapter 5

Author: Jonathan-Birkey

src/com/github/jonathanbirkey/chapter05/Exercise21.java

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+/**
+ * @author : Jonathan Birkey
+ * @mailto : jonathan.birkey@gmail.com
+ * @created : 24Dec2021
+ *
+ * Exercise 5.21
+ * (Financial application: compare loans with various interest rates) Write a program
+ * that lets the user enter the loan amount and loan period in number of years, and
+ * displays the monthly and total payments for each interest rate starting from 5% to
+ * 8%, with an increment of 1/8. Here is a sample run:
+ *
+ * Loan Amount: 10000
+ * Number of Years: 5
+ *
+ * Interest Rate    Monthly Payment     Total Payment
+ * 5.000%           188.71              11322.74
+ * 5.125%           189.29              11357.13
+ * 5.250%           189.86              11391.59
+ * ...
+ * 7.875%           202.17              12129.97
+ * 8.000%           202.76              12165.84
+ **/
+package com.github.jonathanbirkey.chapter05;
+
+import java.util.Scanner;
+
+public class Exercise21 {
+    public static void main(String[] args) {
+        Scanner input = new Scanner(System.in);
+        System.out.print("Loan Amount: ");
+        double loanAmount = input.nextDouble();
+        System.out.print("Number of Years: ");
+        int numOfYears = input.nextInt();
+        input.close();
+
+        double annualInterestRate = 5;
+        double totalPeriods = numOfYears * 12;
+
+
+        System.out.printf("%-20s%-20s%s", "Interest Rate", "Monthly Payment", "Total Payment\n");
+        while (annualInterestRate <= 8.0){
+            double monthlyInterestRate = (annualInterestRate / 100) / 12;
+            double monthlyPayment = loanAmount * (monthlyInterestRate * Math.pow(1+monthlyInterestRate, totalPeriods))
+                    / (Math.pow(1+monthlyInterestRate, totalPeriods) - 1);
+            double totalPayment = monthlyPayment * totalPeriods;
+            System.out.printf("%-5.3f%-15s%-20.2f%.2f\n", annualInterestRate, "%", monthlyPayment, totalPayment);
+            annualInterestRate += 0.125;
+        }
+
+
+
+    }
+}

src/com/github/jonathanbirkey/chapter05/Exercise22.java

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+/**
+ * @author : Jonathan Birkey
+ * @mailto : jonathan.birkey@gmail.com
+ * @created : 27Dec2021
+ *
+ * Exercise 5.22
+ * (Financial application: loan amortization schedule) The monthly payment for a given
+ * loan pays the principal and the interest. The monthly interest is computed by multiplying
+ * the monthly interest rate and the balance (the remaining principal). The principal paid
+ * for the month is therefore the monthly payment minus the monthly interest. Write a program
+ * that lets the user enter the loan amount, number of years, and interest rate then displays
+ * the amortization schedule for the loan. Here is a sample run:
+ *
+ * Loan Amount: 10000
+ * Number of Years: 1
+ * Annual Interest Rate: 7
+ *
+ * Payment#     Interest    Principal       Balance
+ * 1               58.33       806.93        9193.07
+ * 2               53.62       811.64        8381.43
+ * ...
+ * 11              10.00       855.26         860.27
+ * 12              5.01        860.25           0.01
+ **/
+package com.github.jonathanbirkey.chapter05;
+
+import java.util.Scanner;
+
+public class Exercise22 {
+    public static void main(String[] args) {
+        Scanner input = new Scanner(System.in);
+        System.out.print("Loan Amount: ");
+        double principal = input.nextDouble();
+        System.out.print("Number of Years: ");
+        int numberOfYears = input.nextInt();
+        System.out.print("Annual Interest Rate: ");
+        double interest = input.nextDouble();
+        input.close();
+        double totalPeriods = numberOfYears * 12;
+        double monthlyInterestRate = (interest / 100) / 12;
+
+        double monthlyPayment = principal * (monthlyInterestRate * Math.pow(1+monthlyInterestRate, totalPeriods))
+                / (Math.pow(1+monthlyInterestRate, totalPeriods) - 1);
+        double balance = monthlyPayment * totalPeriods;
+
+        System.out.printf("\nMonthly Payment: %.2f\nTotal Payment: %.2f\n\n",monthlyPayment, balance);
+        System.out.printf("%-15s%-15s%-15s%s\n", "Payment#", "Interest", "Principal", "Balance");
+        for (int i = 1; i <= numberOfYears * 12; i++) {
+            interest = monthlyInterestRate * balance;
+            principal = monthlyPayment - interest;
+            balance = balance - principal;
+            System.out.printf("%-15d%-15.2f%-15.2f%.2f\n", i, interest, principal, balance);
+        }
+    }
+}

src/com/github/jonathanbirkey/chapter05/Exercise23.java

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+/**
+ * @author : Jonathan Birkey
+ * @mailto : jonathan.birkey@gmail.com
+ * @created : 27Dec2021
+ *
+ * Exercise 5.23
+ * (Demonstrate cancellation errors) A cancellation error occurs when you are
+ * manipulating a very large number with a very small number. The large number
+ * may cancel out the smaller number. For example, the result of 100000000.0
+ * + 0.000000001 is equal to 100000000.0. To avoid cancellation errors and
+ * obtain more accurate results, carefully select the order of computation. For
+ * example, in computing the following summation, you will obtain more accurate
+ * results by computing from right to left rather than from left to right:
+ *
+ * 1 + 1/2 + 1/3 + ... + 1/n
+ *
+ * Write a program that compares the results of the summation of the preceding
+ * series, computing from left to right and from right to left with n = 50000.
+ **/
+package com.github.jonathanbirkey.chapter05;
+
+public class Exercise23 {
+    public static void main(String[] args) {
+        double n = 50000;
+        double leftToRight = 1;
+        for(double i = 2; i <= n; i++){
+            leftToRight += 1/i;
+        }
+        System.out.printf("Summation from left to right: %f\n", leftToRight);
+
+        double rightToLeft = 0;
+        for(double j = n; j >= 2; j--){
+            rightToLeft += 1/j;
+        }
+        rightToLeft += 1;
+        System.out.printf("Summation from right to left: %f\n", rightToLeft);
+    }
+}

src/com/github/jonathanbirkey/chapter05/Exercise24.java

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+/**
+ * @author : Jonathan Birkey
+ * @mailto : jonathan.birkey@gmail.com
+ * @created : 27Dec2021
+ *
+ * Exercise 5.24
+ * (Sum a series) Write a program to compute the following summation:
+ *
+ * 1/3 + 3/5 + 5/7 + 7/9 + 9/11 + 11/13 + ... + 95/97 + 97/99
+ **/
+package com.github.jonathanbirkey.chapter05;
+
+public class Exercise24 {
+    public static void main(String[] args) {
+        double summation = 0;
+        for(int i = 1; i < 100; i += 2) {
+            summation += i/(i + 2.0);
+        }
+        System.out.printf("The summation is %f", summation);
+    }
+}

src/com/github/jonathanbirkey/chapter05/Exercise25.java

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+/**
+ * @author : Jonathan Birkey
+ * @mailto : jonathan.birkey@gmail.com
+ * @created : 27Dec2021
+ *
+ * Exercise 5.25
+ * (Compute Pi) You can approximate p by using the following summation:
+ *
+ * Pi - 4(1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ... + -^i+1/2i-1)
+ * Write a program that displays the Pi value for i = 10000,20000,...,and100000.
+ **/
+package com.github.jonathanbirkey.chapter05;
+
+public class Exercise25 {
+    public static void main(String[] args) {
+        double sum1 = 0;
+        for(int i = 2; i <= 10000; i++){
+            sum1 += (Math.pow(-1, i + 1)) / (2.0 * i - 1.0);
+        }
+        double Pi1 = 4 * (1.0 - sum1);
+        System.out.printf("Pi value when i = 10000 is %f\n", Pi1);
+
+        double sum2 = 0;
+        for(int j = 2; j <= 20000; j++){
+            sum2 += (Math.pow(-1, j + 1)) / (2.0 * j - 1.0);
+        }
+        double Pi2 = 4 * (1.0 - sum2);
+        System.out.printf("Pi value when i = 20000 is %f\n", Pi2);
+
+        double sum3 = 0;
+        for(int k = 2; k <= 100000; k++){
+            sum3 += (Math.pow(-1, k + 1)) / (2.0 * k - 1.0);
+        }
+        double Pi3 = 4 * (1.0 - sum3);
+        System.out.printf("Pi value when i = 100000 is %f\n", Pi3);
+    }
+}