Sine series summation

This is a discussion on Sine series summation within the C Programming forums, part of the General Programming Boards category; Okay so, Im not going to ask that you give me specific code or anything, Im just looking for a ...

  1. #1
    Registered User
    Join Date
    Dec 2008
    Posts
    1

    Sine series summation

    Okay so, Im not going to ask that you give me specific code or anything, Im just looking for a little guidance. Im new to C and frankly im pretty lost.

    I have to do the following quesion:

    Write your own function, double sine1(double x); which returns the value of the sine of x by evaluating the series expansion sin(x)= x- x^3/3! + x^5/5! - X^7/7"!.... etc. where x is in radians. You can calculate each new term from the previous term.[Hint: the termx^n in is –x^2/n(n-1) times the term in x^n-2]. You will have to stop the loop either after a fixed number of terms, or by breaking out when the current term becomes sufficiently small. Write a program to test your function by comparing your values with those returned by the standard math library function sin(). Your program should print out a comparison table listing x, sine1(x), sin(x) and the difference sine1(x) – sin(x) for a range of angles. Record a copy of your sample output.

    My lecturer no doubt is looking something similar to how the following question was done:

    Write your own function, double expo(double x);
    which returns the value of the exponential function e^x
    based on the series expansion e^x=1+x+x^2/2!+x^3/3!+ ... etc.
    You can calculate each new term from the previous term
    [hint: the nth term is x/(n-1) times the (n-1)th term].
    Test your function by comparing your values with those
    returned by the standard math library function exp()
    for a range of parameter values.
    */

    Code:
    #include <stdio.h>
    #include <stdlib.h>
    #include <math.h>
    double expo(double x);
    
    int main()
    {
        double y, myexp,ratio;
    
        for (y=0.0;y<2.0;y=y+0.1)
        {
            myexp = expo(y);
            ratio = myexp/exp(y);
            printf("%8.2f  %20.10f  %20.10f\n",y,myexp,ratio);
        }
        return 0;
    }
    double expo(double x)
    {
        double term,result;
        int n;
        term=1.0;
        result=1.0;
        for (n=2;n<20;n=n+1)
        {
            term =term * x/(n-1);
            if(term/result<1e-1)break;
            result=result + term;
        }
        printf("%4d",n);
        return result; }
    Ive been playing around for a while and im just getting nowhere. My knowledge is C is obviously very shakey, but I really need help! Would anyone even perhaps explain (in bullet point form, something brief) the exponential code was constructed? Particularly that within bold. Hopefully from that I can tackle my own question!

    Thanks.

  2. #2
    Fountain of knowledge.
    Join Date
    May 2006
    Posts
    794
    You are basically be asked to find the value of.

    x- x^3/3! + x^5/5! - X^7/7

    So....


    I would do something like this, it may have some problems but it might get you on the
    right lines?

    Code:
    
    for( n=3; n<somebigvalueforn; n=n+2){//you are stepping through the terms here
      z=n;//z is the 'significant number in the term, you need a 'copy' of it.
      term=x; //x is the angle
    
      while(z>1){ you are calculating them here.
          term=x*term/z;
          z--;
      }
    
     running_total+=term
    
    }
    
    result=running_total+x;
    I think basically you needed another loop?
    Might be a prob with division returning a double?
    Try it.

    You could start from 1 not 3 I guess, looks nicer perhaps.
    Last edited by esbo; 12-17-2008 at 07:03 PM.

  3. #3
    Registered User carrotcake1029's Avatar
    Join Date
    Apr 2008
    Posts
    398
    If you use the Maclaurin series for sine, you could modify my cosine function that you could probably find if you searched. Here is what I discovered to make it easy.

    You actually dont need to calculate factorials. Look at this expression
    Code:
    x^3/3!
    Is that not the same as
    Code:
    x/3 * x/2 * x/1
    <bighint>There you go</bighint>

  4. #4
    Fountain of knowledge.
    Join Date
    May 2006
    Posts
    794
    Quote Originally Posted by carrotcake1029 View Post
    If you use the Maclaurin series for sine, you could modify my cosine function that you could probably find if you searched. Here is what I discovered to make it easy.

    You actually dont need to calculate factorials. Look at this expression
    Code:
    x^3/3!
    Is that not the same as
    Code:
    x/3 * x/2 * x/1
    <bighint>There you go</bighint>
    Yes it is, I was not sure if you could use ! as I rarely need to use it.
    Same goes for ^
    Makes it lot easier I guess

Popular pages Recent additions subscribe to a feed

Similar Threads

  1. Sine (Sin) Algorithm Help
    By StaticKyle in forum C Programming
    Replies: 52
    Last Post: 05-13-2008, 09:37 AM
  2. Implement of a Fast Time Series Evaluation Algorithm
    By BiGreat in forum C Programming
    Replies: 7
    Last Post: 12-04-2007, 02:30 AM
  3. Sine approximation Help
    By What111 in forum C Programming
    Replies: 8
    Last Post: 11-04-2007, 06:22 PM
  4. Computing the sine of an angle
    By Bearcat in forum C Programming
    Replies: 1
    Last Post: 02-13-2002, 12:46 PM
  5. sine C program with Power Series
    By Unregistered in forum C Programming
    Replies: 1
    Last Post: 10-08-2001, 11:46 AM

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