Tuesday, April 9, 2013

Answer In Deg And Rad - C And C++ | Dream.In.Code


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    4 Replies - 65 Views - Last Post: Yesterday, 09:46 PM Rate Topic: -----

    #1 seraz ?Icon User is offline

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    Posted Yesterday, 08:32 PM

    this is my coding...
    for this coding my result should be 6 iteration only...
    im not very sure whether my answer in rad or deg...
    ples someone help me...
    
 #include<stdio.h> #include<math.h> #include<stdlib.h> #define true 1; #define false 0;  main() { 	int i,m,OK,it; 	double c,d,TOL,fx[100],dfx[100],a[100],x[100]; 	x[0]=2.3,i=0; 	TOL=(pow(10,-32)); 	m=2; 	 	fx[0]=pow((pow(sin(x[0]),2)-pow(x[0],2)+1),2); 	d=fabs(fx[0]); 	printf("%d	%.12e	%.12e\n",i,x[0],d); 	 	OK=false; 	it=0; 	i=0; 	while(!OK) 	{ 		 		it=it+1; 		fx[i]=pow((pow(sin(x[i]),2)-pow(x[i],2)+1),2); 		dfx[i]=2*(pow(sin(x[i]),2)-pow(x[i],2)+1)*(2*sin(x[i])*cos(x[i])-2*x[i]); 	 		x[i+1]=x[i]-(m*(fx[i]/dfx[i])); 		fx[i+1]=pow((pow(sin(x[i+1]),2)-pow(x[i+1],2)+1),2); 	 		c=fabs(fx[i+1]); 		printf("%d	%.12e	%.12e\n",i+1,x[i+1],c); 		 	if(c<TOL) OK=true;  i=i+1;  	} } 

    This post has been edited by Skydiver: Yesterday, 08:42 PM


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    Replies To: answer in deg and rad

    #2 Skydiver ?Icon User is online

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    Re: answer in deg and rad

    Posted Yesterday, 08:54 PM

    Try increasing your tolerance to 10-30 instead of 10-32.

    This post has been edited by Skydiver: Yesterday, 08:55 PM


    #3 seraz ?Icon User is offline

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    Re: answer in deg and rad

    Posted Yesterday, 09:25 PM

    View PostSkydiver, on 07 April 2013 - 08:54 PM, said:

    Try increasing your tolerance to 10-30 instead of 10-32.

    i cannot do that...
    because it already fit in my project...
    i want to observe the number of iteration with same TOL but different test function


    #4 Skydiver ?Icon User is online

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    Re: answer in deg and rad

    Posted Yesterday, 09:38 PM

    Well, base on the values of C, I don't see how you will ever get a value of c less than 10-32 in 6 iterations.

    Here's the output I'm seeing:

    
 0       2.300000000000e+000     1.394218647024e+001 1       1.632475867168e+000     4.472626765749e-001 2       1.435079952458e+000     6.046710267708e-003 3       1.405195569536e+000     3.057005611515e-006 4       1.404492036112e+000     9.272594171137e-013 5       1.404491648215e+000     8.590695257857e-026 6       1.404491648215e+000     1.109335647967e-031 7       1.404491648215e+000     1.972152263053e-031 8       1.404491648215e+000     1.109335647967e-031 9       1.404491648215e+000     1.972152263053e-031 10      1.404491648215e+000     1.109335647967e-031 11      1.404491648215e+000     1.972152263053e-031 12      1.404491648215e+000     1.109335647967e-031 13      1.404491648215e+000     1.972152263053e-031 : until it overruns the buffers you declared. 

    How did you determine that 6 iterations was going to be sufficient?

    This post has been edited by Skydiver: Yesterday, 09:43 PM


    #5 Skydiver ?Icon User is online

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    Re: answer in deg and rad

    Posted Yesterday, 09:46 PM

    Perhaps use GMP or MPIR as your number and math library instead of the standard C/C++ library to get more precision and perhaps get past the repetition that is happening above.


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