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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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Differentiate following functions. (a) f ( x ) = 15x 100 - 3x 12 + 5x - 46 (b) h ( x ) = x π - x √2 Solution (a) f ( x ) = 15x 100 - 3x 12 + 5x - 46 I
The points A,B,C and D represent the numbers Z1,Z2,Z3 and Z4.ABCD is rhombus;AC=2BD.if Z2=2+i ,Z4=1-2i,find Z1 and Z3 Ans) POI of diagonals: (3-i)/2. Using concept of rotation:
Parametric Equations and Curves Till to this point we have looked almost completely at functions in the form y = f (x) or x = h (y) and approximately all of the formulas that w
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