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# c program to implement newton raphson method for finding roots of a polynomial

 P: 21 c program for newton raphson algorithm for finding roots of a polynomial Oct 9 '06 #1
6 Replies

 10K+ P: 13,264 c program for newton raphson algorithm for finding roots of a polynomial Why won't you try to write it yourself and then post at a more specific problem? Oct 9 '06 #2

 P: 21 how can i implement a c progran for newton raphson algorithm using a for loop? Oct 9 '06 #3

 Expert Mod 5K+ P: 8,916 how can i implement a c progran for newton raphson algorithm using a for loop? Please dont double post. Please do have an attempt yourself first. Do you know how the newton raphson method works? Oct 9 '06 #4

 P: 21 #include /Calculate j(10) from j(0) int main() { int n; double j=0.6; printf("forward iterations\n"); printf("initial estimate:j(0)=%g\n",j); for(n=1,n<=10,n++) j=1-(j*n); printf("j(%d)=%g\n",n,j); } return 0; } this code is not helping me find roots of( x*x-2).What do I need to change? Oct 10 '06 #5

 P: 32 Ok, let me see if I understand your code. I'm a numerical methods and modeling engineer so I should understand the Newton-Raphson Method to be: j( i+1 ) = j( i ) - [ j( i )^2 - 2 ] / [ 2*j( i ) ] for the function f( j ) = (j*j - 2). So, you should step i forward until f( j ) is within your tolerance of 0. Then your answer should be your last j. Here's some pseudocode: i = 0 while( absoluteValue( f( j ) ) > toleranceValue) { j = j - [ f( j ) ] / [ df( j ) ]; i++; } printOut("Root of Function is at %g After %d Iterations" j, i ); When you write this out in whatever language you want, should work. - Miles Oct 10 '06 #6

 Expert Mod 5K+ P: 8,916 Please DON'T double post (hmmm I seem to be repeating myself here) Read This Oct 10 '06 #7