Scientific Computing

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CS205 Homework #5 Solutions Problem 1 [Heath 5.5, p.248] 1. Show that the iterative method x k +1 = x k - 1 f ( x k ) - x k f ( x k - 1 ) f ( x k ) - f ( x k - 1 ) is mathematically equivalent to the secant method for solving a scalar nonlinear equa- tion f ( x ) = 0. 2. When implemented in finite-precision floating-point arithmetic, what advantages or disadvantages does the formula given in part (1) have compared with the formula for the secant method (given in the notes and in Heath, section 5.5.4)? Solutions 1. Starting with the secant method update formula is given as x k +1 = x k - f ( x k )( x k - x k - 1 ) f ( x k ) - f ( x k - 1 ) = x k ( f ( x k ) - f ( x k - 1 )) - f ( x k )( x k - x k - 1 ) f ( x k ) - f ( x k - 1 ) = x k f ( x k ) - x k f ( x k - 1 ) - x k f ( x k ) + x k - 1 f ( x k ) f ( x k ) - f ( x k - 1 ) = x k - 1 f ( x k ) - x k f ( x k - 1 ) f ( x k ) - f ( x k - 1 ) 2. When we’re near the solution x k - 1 and x k are close to each other and thus their difference is near zero. Similarly f ( x k - 1 ) and f ( x k ) are close to each other so their difference is near zero. Thus we have an indefinite form in secant computing ( x k - x k - 1 ) / ( f ( x k ) - f ( x k - 1 )). However, since we’re near the root we have f ( x k ) near zero so
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