Exercises C-Programming

Transcription

1 Exercises C-Programming Claude Fuhrer 0 November 016 Contents 1 Serie 1 1 Min function Triangle surface Triangle surface Multiple assignements Serie 4 5 Numerical integration using the rectangles method Root of a function

2 1 SERIE 1 1 Serie 1 Exercise 1: Min function Write a program which read two integer from the keyboard and displays the smallest (i.e. minumum) value entered. Exercise : Triangle surface 1 Write a program which ask the user to enter the base and height of a triangle and display its area. Exercise 3: Triangle surface One can compute the area of a triangle given by its 3 sides using the Heron formula. This formula states that if s 1, s and s 3 are the length of the three sides, then the area is given by: area = p (p s 1 ) (p s ) (p s 3 ) where the variable p is the half perimeter, that is: p = s 1 + s + s 3 Tu be able to compute the square root, you should first include the following into your program: #include <math.h> and then call the function sqrt(). Exercise 4: Multiple assignements Consider the following program. 1 #include <stdio.h> #include <stdlib.h> 3 4 int main()

3 1 SERIE 1 5 { 6 int a, b, c; 7 8 a = 4; 9 b = ; 10 c = 1; 11 printf ("a = %i, b = %i, c = %i\n", a, b, c); 1 13 a = b + ; 14 b = a - b + c; 15 c = c + a; 16 printf ("a = %i, b = %i, c = %i\n", a, b, c); a = * b; 19 c = a - c; 0 b = b + / a; 1 printf ("a = %i, b = %i, c = %i\n", a, b, c); 3 c = + a; 4 a = a - b; 5 b = b + c / 3; 6 printf ("a = %i, b = %i, c = %i\n", a, b, c); 7 } 3

4 SERIE Serie Exercise 5: Numerical integration using the rectangles method Let be f(x) a function bounded over the interval [a, b]. For our example, one would take f(x) = sin(x) + sin( x ). One would compute the area delimited by the function f on the interval [0, π]. Mathematically this operation is called an integration and is written: area = π 0 sin(x) + sin( x )dx = 4 But, for this exercise, one would not use the standard analysis tools, but try to compute the integral using a numerical method. For that goal, one would decompose the function in small rectangles, compute the area of each rectangle and then sum all these rectangles to obtain an approximation of the real integral value. The graph of the function is given by: and the rectangle we woud compute are: 4

5 SERIE The algorithm may be written as: Data: The function f(x) and the bounds of the interval [a, ] Result: A numerical approximation of the interval begin numberofsteps 10; // Width of rectangles; step (upperbound- lowerbound) / numberofsteps; totalarea 0; x i lowerbound; while x i < upperbound do y i = f(x i ); rectanglearea y i step; totalarea = totalarea + rectanglearea; // Next x i value; x i = x i + step; The value of the integral is into the variable totalarea; Algorithm 1: Integration with the rectangles method To enhance the precision of the computation, it is possible to augment the number of steps, that is to reduce the width of the rectangles. 5

6 SERIE Exercise 6: Root of a function For that exercise, one would compute one root (i.e. zero) of a function over a given interval. For that goal one would use a very important theorem of analysis, called Bolzano s theorem. This theorem states that: Theorem 1 If a function f(x) is continuous over the interval [a, b] and has the value f(a) and f(b) at the bounds of the interval, then it takes any value between f(a) and f(b) within this interval. One important corollary of this theorem is that is the sign of f(a) is different of the sign of f(b) that the function f takes the value 0 over the interval [a, b]. Consider the function f(x) = sin(x) cos(3x). The graph of this function is given by: sin(x) - cos(3x) One can see that this function takes the value 0 between 0 and 0.5. But what is the exact value of x such that f(x) = 0? To solve numerically this problem, one would use a method called the bissection method. This method takes the interval [a, b] and divide it into two part. One can then check if the zero of the function is between the bounds [a, a+b a+b ] or between [, b]. The new interval is than halved iteratively until one has found an interval sufficiently small for the precision we wanted to reach. One can describe the algorithm of this method as in algorithm on page 7. Now that the algorithm is clear, it is simple to implement this approximation method. For an example, of solution, see the code of the program bissection.c. 6

7 SERIE Data: The function f(x) and the bounds of the initial interval [a, b] Result: An interval containing the zero of the function [a, b ] such b a < ε begin a a; b b; middle (a + b)/; while b a > ε do if sign(f(a )) sign(f(middle)) then b middle; else a middle; middle (a + b )/ Algorithm : The bissection method 7