3 3 votes closed with the note: recategorize Newton-Raphson method is used to compute a root of the equation $x^2 - 13 = 0$ with 3.5 as the initial value. The approximation after one iteration is 3.575 3.676 3.667 3.607 Numerical Methods gatecse-2010 numerical-methods newton-raphson easy non-gatecse + – gatecse 8.3k views comment Share Follow Print See all 3 Comments 3 3 Comments reply akash.dinkar12 commented Aug 6, 2018 reply Follow flag follow this one... 1 1 replyShare mili_dhara commented Jul 9, 2025 reply Follow flag How f'(xn) = 2xn^2, it should be 2xn. 0 0 replyShare juDson_Abhi commented Aug 5, 2025 reply Follow flag Out of syllabus 0 0 replyShare Please log in or register to add a comment.
3 3 votes x=3.5 - ((3.5^2)-13)/(2*3.5) so x=3.607 it is from the next term formula of newton raphson method Bhagirathi answered Sep 21, 2014 Bhagirathi comment Share Follow See all 4 Comments 4 4 Comments reply anchitjindal07 commented Aug 30, 2017 reply Follow flag What is the formula, Sir? 0 0 replyShare Bikram commented Aug 31, 2017 reply Follow flag This one http://nptel.ac.in/courses/122104019/numerical-analysis/Rathish-kumar/ratish-1/f3node6.html 0 0 replyShare anchitjindal07 commented Aug 31, 2017 reply Follow flag I have seen this formula, but it looks like Bhagirathi is using some other formula 0 0 replyShare Rohan7980 commented Feb 2, 2018 reply Follow flag same formula is used, x1= x0-f(x0)/f'(x0) 0 0 replyShare Please log in or register to add a comment.