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Newton's Method : If xn is an approximation a solution of f ( x ) = 0 and if given by, f ′ ( xn ) ≠ 0 the next approximation is given by
xn+1 = xn - f(xn)/f'(xn)
It has to lead to the question of while do we stop? How several times do we go through this procedure? One of the more common stopping points in the procedure is to continue till two successive approximations agree upon a given number of decimal places.
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Ravi is a teacher of Class 4 in a municipal school in Delhi. When the new school year started, he opened the textbook and started teaching the children how to write 4-digit numbers
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