Iterative formula to find the reciprocal of a given number N is
Xn+1 =xn [2+Nxn-1]
xn+1 =xn [2+Nxn]
xn+1 =xn [2-Nxn]
xn+1 =xn [2+Nxn+1]
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2xn. = 1. 2. ( xn + a xn. ) . 3. Newton's equation y3 − 2y − 5 = 0 has a root near y = 2. Starting with y0 = 2, compute ...
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