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You can change your ad preferences anytime. Solucionario metodos numericos para ingenieros chapra. Upcoming SlideShare. Like this document? Why not share! Embed Size px. Start on. Show related SlideShares at end. WordPress Shortcode. Published in: Education , Technology. Full Name Comment goes here. Are you sure you want to Yes No. Browse by Genre Available eBooks Show More. Erick Poveda. Jose Adrian Silva Baldelomar. No Downloads. Views Total views.

Actions Shares. Embeds 0 No embeds. No notes for slide. Solucionario metodos numericos para ingenieros chapra 1. Chapra Tufts University 2. For the period from ending June 1: 3.

The numerical results are: step v 12 absolute relative error 2 The problem stems primarily from the fact that we are subtracting two nearly equal numbers in the denominator. Such subtractive cancellation is worsened by the fact that the denominator is squared.

The complete Taylor series to the third-order term is! This is due to the fact that cubics have zero fourth and higher derivatives. A smaller step size is required to obtain convergence. This occurs because the underlying function is a cubic equation that has zero fourth and higher derivatives. Note the 2nd and 3rd order Taylor Series functions are the same.

From the plots below, we see that the answer is the 4th Order Taylor Series expansion. The second iteration is Using bisection, the first iteration is The second iteration is 2 2 5. Thus, after ten iterations, the false position method is converging at a very slow pace and is still far from the root in the vicinity of 1.

Discussion: This is a classic example of a case where false position performs poorly and is inferior to bisection. Insight into these results can be gained by examining the plot that was developed in part a. Because of the shape of the present function, the opposite is true. First, it can be solved for the linear x, 7. An alternative is to solve for the second-order x, 9. The result can be checked by substituting it back into the original function, The guess of 0.

Therefore, the first iteration results in a prediction of 2. At these points the function is very flat and hence, the Newton-Raphson results in a very high value Thereafter, the methods slowly converge on the nearest roots. Explanation of results: The results are explained by looking at a plot of the function. Both guesses are in a region where the function is relatively flat.

Because the two guesses are on opposite sides of a minimum, both are sent to different regions that are far from the initial guesses. Due to the concavity of the slope, the next iteration will always diverge.

The following graph illustrates how the divergence evolves. Hence, the solution is cast far from the roots in the vicinity of the original guess. Roots seem to occur at about 40o and 50o. The solution can be checked by substituting it back into the equations to give 60 4. The major difference is that the elimination is only implemented for the left-hand side coefficients.

The result is This is left for an exercise. Results may be inaccurate. Thus, for this case, the condition number tends to exaggerate the impact of ill-conditioning. First iteration: You just clipped your first slide! Clipping is a handy way to collect important slides you want to go back to later. Now customize the name of a clipboard to store your clips. Visibility Others can see my Clipboard. Cancel Save.

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