help: nonlinear secant method program

This is a discussion on help: nonlinear secant method program within the C++ Programming forums, part of the General Programming Boards category; hi, im a beginner in using C++... i have a project due in a few days and i've been trying ...

  1. #1
    Registered User
    Join Date
    Oct 2009
    Posts
    2

    Exclamation help: nonlinear secant method program

    hi, im a beginner in using C++...

    i have a project due in a few days and i've been trying to do the program for a week now... we're supposed to make a program that would allow the user to enter any polynomial equation (always = 0), asks the user the interval and number of iterations and solve it using secant method...

    for example, the user entered the polynomial equation of 0.5x^4 + 2x^2 - x + 4 = 0, interval [-0.25, 0], no of iterations is 5, the result should be the following:

    http://i37.tinypic.com/339qefn.jpg

    the screen should look like this when program is run:

    http://i33.tinypic.com/3325hs4.jpg

    i already have a code for the secant method formula part for computing the new x sub n, however i don't have a clue as to how i would make a code enabling the user to enter the polynomial equation to be solved and substituting the new x (which is the new computed x sub n obtained from the secant formula) to the polynomial equation…

    the question is what I want possible?... is there any kind of library that I need for wat I want?..

    hope u guys can help coz honestly im no good in programming… I'm getting crazy with this project coz due is in 2 days time (sobs)… please help…

    for additional information, for the secant part of the program… the code is the following:
    Code:
    for(int i=1;i<=n;i++)
    {
          c[i+1]=c[i]-(f(c[i]) * (c[i]-c[i-1])) / (f(c[i])-f(c[i-1]));
          cout <<"c(" << i+1 <<") = "<< c[i+1] <<"\n";
    }

  2. #2
    Registered User
    Join Date
    Oct 2009
    Location
    While(1)
    Posts
    377
    Look at this

    Code:
    #include <iostream>
    #include <math.h>
    
    class Polynomial {
    private:
      float m_variable;
      float m_interval;
    
    public:
      Polynomial() : m_variable(0), m_interval(0) {}
      Polynomial(const float variable,
                 const float interval) : m_variable(variable),
                                         m_interval(interval) {}
    
      void SetVariable(const float variable) { m_variable = variable; }
      void SetInterval(const float interval) { m_interval = interval; }
    
      void Calculate();
    };
    
    void Polynomial::Calculate() {
      for (int counter = 0; counter < 5; ++counter) {
        const float result = (pow((0.5 * m_variable), 4)) +
                             (pow(2 * m_variable, 2))     -
                             m_variable                   +
                             4;
        printf("Xn == %f ==  || F(Xn) == %f\n", m_variable, result);
        m_variable += m_interval;
      }
    }
    
    int main() {
      Polynomial obj_poly;
      obj_poly.SetVariable(2.5);
      obj_poly.SetInterval(-0.25);
      obj_poly.Calculate();
    
      return 0;
    }

  3. #3
    Registered User
    Join Date
    Oct 2009
    Posts
    2
    thanks... i'll look into the code u gave me...

    oh yeah... i better explain how the values in the example i gave came about so that u guys cud understand wat i want to do more...

    1) the polynomial equation that the user will enter will be the f(x) which is always equal to 0 (hence the example equation i used before was equal to 0)
    2) the interval that the user will enter is the 1st x and 2nd x (respectively) which the program will evaluate giving a new value that will be shown under the column of f(x sub n)
    [in the example, the interval was [-0.25 , 0] and these were substituted to the polynomial equation thus giving the values of 3.876953 (for x=-0.25) and 4 (for x=0)]
    3) the new x that will be used for evaluation is obtained through the secant formula which is:
    http://i33.tinypic.com/1zfsbhc.jpg
    [in the example, the 3rd x in the column under x sub n is obtained through the secant formula:
    http://i35.tinypic.com/o7j6s7.jpg
    this new x will be evaluated (substituting it to the polynomial equation and getting a value of 2325.374361]
    4) the 3rd step is repeated until the number of iterations is satisfied (in the example, the intervals, which are -0.25 and 0 are iteration 0 and iteration 1 respectively)

    is what i want possible?..

    by the way, i was able to make a program that does wat i want with the exception of entering the polynomial equation... this program solves the equation x - 2 + ln x = 0... however i dont know how to construct the part wherein the program could do the number of iterations that the user wants... i hope this gives you an idea...

    Code:
    #include <stdio.h>
    #include <math.h>
    double f(double);
    int main()
    {
        double k, x1, x2;
        int n;
        double y;
        
        printf("Enter the interval [x1,x2]: ");
        scanf("%lf, %lf", &x2, &x1);
        
      
        printf("Checking the boundaries:\n");
        printf("Xn\t\tf(Xn)\n");
        printf("%.6lf\t%.6lf\n", x1, f(x1));
        printf("%.6lf\t%.6lf\n", x2, f(x2));
        
        for (n = 1; n<=4; n++)
        {
            y = (x1 - x2) / (f(x1) - f(x2)) * f(x1);
            x2 = x1;
            x1 = x1 - y;
            printf("%.6lf\t%.6lf\n", x1, f(x1));
        }
        
    
        scanf("%lf", &k);
        return 0;
    }
    
    double f(double x)
    {   
        return((x-2)+log(x));
    }
    please if u cud, help me... i wud really really appreciate it...

  4. #4
    Registered User
    Join Date
    Oct 2009
    Location
    While(1)
    Posts
    377
    Quote Originally Posted by lady_lyssa View Post
    thanks... i'll look into the code u gave me...

    oh yeah... i better explain how the values in the example i gave came about so that u guys cud understand wat i want to do more...

    1) the polynomial equation that the user will enter will be the f(x) which is always equal to 0 (hence the example equation i used before was equal to 0)
    2) the interval that the user will enter is the 1st x and 2nd x (respectively) which the program will evaluate giving a new value that will be shown under the column of f(x sub n)
    [in the example, the interval was [-0.25 , 0] and these were substituted to the polynomial equation thus giving the values of 3.876953 (for x=-0.25) and 4 (for x=0)]
    3) the new x that will be used for evaluation is obtained through the secant formula which is:
    http://i33.tinypic.com/1zfsbhc.jpg
    [in the example, the 3rd x in the column under x sub n is obtained through the secant formula:
    http://i35.tinypic.com/o7j6s7.jpg
    this new x will be evaluated (substituting it to the polynomial equation and getting a value of 2325.374361]
    4) the 3rd step is repeated until the number of iterations is satisfied (in the example, the intervals, which are -0.25 and 0 are iteration 0 and iteration 1 respectively)

    is what i want possible?..

    by the way, i was able to make a program that does wat i want with the exception of entering the polynomial equation... this program solves the equation x - 2 + ln x = 0... however i dont know how to construct the part wherein the program could do the number of iterations that the user wants... i hope this gives you an idea...

    Code:
    #include <stdio.h>
    #include <math.h>
    double f(double);
    int main()
    {
        double k, x1, x2;
        int n;
        double y;
        
        printf("Enter the interval [x1,x2]: ");
        scanf("%lf, %lf", &x2, &x1);
        
      
        printf("Checking the boundaries:\n");
        printf("Xn\t\tf(Xn)\n");
        printf("%.6lf\t%.6lf\n", x1, f(x1));
        printf("%.6lf\t%.6lf\n", x2, f(x2));
        
        for (n = 1; n<=4; n++)
        {
            y = (x1 - x2) / (f(x1) - f(x2)) * f(x1);
            x2 = x1;
            x1 = x1 - y;
            printf("%.6lf\t%.6lf\n", x1, f(x1));
        }
        
    
        scanf("%lf", &k);
        return 0;
    }
    
    double f(double x)
    {   
        return((x-2)+log(x));
    }
    please if u cud, help me... i wud really really appreciate it...
    hey i gave u the above code if u can modify that to suite ur needs that will be better or what u can do
    try to modify the above code then i can help u to reach ur goal ...
    or somethin better u can suggest me that will be better

  5. #5
    3735928559
    Join Date
    Mar 2008
    Location
    RTP
    Posts
    839
    you need to write a parser.

    if you're only dealing with regular polynomials, your job is fairly easy. consider each independent variable and constant to be a token. scan over the input string for the independent variables. their exponents should be adjacent to them (separated by some exponentiated operator, ^, perhaps) to the right. each term should be separated by additions. use atof or some other ascii-to-numeric conversion routine to convert the text into numeric types.

    pretty straightforward.

  6. #6
    Registered User MacNilly's Avatar
    Join Date
    Oct 2005
    Posts
    306
    There is an easier way to input the terms of the polynomial. Well, easier for the programmer and that's whats really important, isn't it?

    Enter number of terms: 3
    Enter a1, n1: 2, 2
    Enter a2, n2: 4, 1
    Enter a3, n3: 4, 0

    That would be 2x^2 + 4x + 4.

    Something like that would be a heck of a lot easier to write than a real equation parser. If you do want to write a parser you'll have to make a polynomial class first. Also, the strtod() C library function will keep track of your pointer position of your string for you.
    Last edited by MacNilly; 10-09-2009 at 12:43 PM.
    GCC on Fedora Core 10

Popular pages Recent additions subscribe to a feed

Similar Threads

  1. Replies: 4
    Last Post: 02-21-2008, 09:39 AM
  2. Robust method for storing data outside of a program
    By goatslayer in forum C++ Programming
    Replies: 17
    Last Post: 09-19-2007, 03:08 PM
  3. Basic:switch Program
    By time789 in forum C Programming
    Replies: 9
    Last Post: 07-25-2007, 06:21 AM
  4. Replies: 3
    Last Post: 03-04-2005, 01:46 PM
  5. My program, anyhelp
    By @licomb in forum C Programming
    Replies: 14
    Last Post: 08-14-2001, 10:04 PM

Tags for this Thread


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21