Chemical Engineering Hand Written_Notes_Part_11

Chemical Engineering Hand Written_Notes_Part_11 - 24 2...

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24 2. FUNDAMENTALS OF FUNCTIONAL ANALYSIS is also a solution of equation (6.2). Here we consider two well known examples of such formulation. Example 9 . Iterative schemes for solving single variable nonlinear algebraic equation: Consider one variable nonlinear equation f ( x ) = x 3 + 7 x 2 + x sin( x ) e x = 0 This equation can be rearranged to formulate an iteration sequence as (1.3) x ( k +1) = exp( x ( k ) ) £ x ( k ) ¤ 3 7 £ x ( k ) ¤ 2 sin( x ( k ) ) Alternatively, using Newton-Raphson method for single variable nonlinear equa- tions, iteration sequence can be formulated as x ( k +1) = x ( k ) f ( x ( k ) ) [ df ( x ( k ) ) /dx ] = x ( k ) £ x ( k ) ¤ 3 + 7 £ x ( k ) ¤ 2 + x ( k ) sin( x ( k ) ) exp( x ( k ) ) 3 [ x ( k ) ] 2 + 14 x ( k ) + sin( x ( k ) ) + x ( k ) cos( x ( k ) ) exp( x ( k ) ) (1.4) Both the equations (1.3) and (1.4) are of the form given by equation (1.2). Example 10 . Solving coupled nonlinear algebraic equations by suc- cessive substitution Consider three coupled nonlinear algebraic equations (1.5) F ( x, y, z, w ) = f 1 ( x, y, z, w ) f 2 ( x, y, z, w ) f 3 ( x, y, z, w ) f 4 ( x, y, z, w ) =
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