Exercise #3, Functions Given a positive integer n and a floating point number x between 0.0 and 360.0. The objective of this exercise is to write a C++ program to compute the below math equations: y = xx π/180 yll y5 3! 5! y7 y9 7! 9! + 11! We decompose the eqations into 5 simplified expressions as follows: 1. y=r(x) = x x π/180 2. p(y, k) = yk 3. f(k)= k! 4. q(y,k) = P(y.k) f(k) 5. S(y,n)=y-q(y,3)+q(y, 5)-q(y, 7) + q(y, 9) -q(y, 11)+...+q(y,n) S(y, n) = y- + y3 where k is an integer smaller than or equal to n Write C++ functions to calculate each of the above expressions. In your main, you prompt the user to enter x and n, control the user input, then call function S(x, n) to compute expression 5. Function S(x, n) will call function r(x) to get y, then it calls other functions as needed to complete the computations and display your result at the end. e.g. Value of x: 30 Number of interations n: 12 S(30,12): 0.5 Method 1 int fact, sign=-1; float y = x*M_PI/180, p, sum = 0; for(int i=1;i

Exercise #3, Functions Given a positive integer n and a floating point number x between 0.0 and 360.0. The objective of this exercise is to write a C++ program to compute the below math equations: y = xx π/180 yll y5 3! 5! y7 y9 7! 9! + 11! We decompose the eqations into 5 simplified expressions as follows: 1. y=r(x) = x x π/180 2. p(y, k) = yk 3. f(k)= k! 4. q(y,k) = P(y.k) f(k) 5. S(y,n)=y-q(y,3)+q(y, 5)-q(y, 7) + q(y, 9) -q(y, 11)+...+q(y,n) S(y, n) = y- + y3 where k is an integer smaller than or equal to n Write C++ functions to calculate each of the above expressions. In your main, you prompt the user to enter x and n, control the user input, then call function S(x, n) to compute expression 5. Function S(x, n) will call function r(x) to get y, then it calls other functions as needed to complete the computations and display your result at the end. e.g. Value of x: 30 Number of interations n: 12 S(30,12): 0.5 Method 1 int fact, sign=-1; float y = x*M_PI/180, p, sum = 0; for(int i=1;i

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Concept explainers

Declaring and Defining the function

A set of statements, which performs a particular task is called a function. The program gains modularity by using function. The application is quite easy to understand, read, and edit when they are created with functions. The developer/programmer creates a function and uses main() to call the function. Once the function gets executed, the control is transferred to the main program.

Question

Exercise #3, Functions
Given a positive integer n and a floating point number x between 0.0 and 360.0. The objective of
this exercise is to write a C++ program to compute the below math equations:
y = xx π/180
уз y5 y7
39
ylı
3! 5! 7! 9! 11!
We decompose the eqations into 5 simplified expressions as follows:
1. y=r(x) = x x π/180
2. p(y, k) = yk
where k is an integer smaller than or equal to n
3. f(k)= k!
4. q(y, k) =
5.
S(y,n)=y-q(y, 3) + q (y, 5)-q(y, 7) + q(y, 9)-q(y, 11) +±q(y,n)
p(y,k)
f(k)
S(y, n) y- +
Method 1
int fact, sign=-1;
float y = x*M_PI/180, p, sum =0;
for(int i=1;i<=n; i+=2)
{
Write C++ functions to calculate each of the above expressions. In your main, you prompt the user
to enter x and n, control the user input, then call function S(x, n) to compute expression 5.
Function S(x, n) will call function r(x) to get y, then it calls other functions as needed to complete
the computations and display your result at the end. e.g.
Value of x: 30
Number of interations n: 12
S(30,12): 0.5
p= 1;
fact= 1;
for(int j=1; j<=i; j++)
{
Two other programmers calculated the above expression using different methods, in Method 1 (see
below), the programmer used two loops, in Method 2 the programmer used one single loop.
Method 2
int fact, sign=-1;
float y = x*M_PI/180, p, sum = 0;
for(int i=1; i<n; i+=2)
{
p= p*y;
fact=fact*j;
}
+
sign=-1*sign;
sum += sign*p/fact;
+...+
}
yn
n!
if(i==1){
p= y;
fact= 1;
}
else {
p= p*y*y;
fact fact* (i-1)*(i);
}
sign=-1*sign;
sum += sign*p/fact;
Try to understand the above methods, test them, then compare the three implementations (yours, method1, and
method2) to find out: which one is the best, what are the advantages and disadvantages of each method.
By the way, when you test the three methods: They all should GIVE THE SAME RESULT. This is one way to verify that
your implementation is correct.
e.g. S(30,12)= 0.5
S(45,12)= 0.707107
S(60,12)=0.866025

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