The software, mathematica 9.
This equation indicates that a good error estimate would be Comment out the disp statement displaying and in newton. Conveniently, r1 either goes to zero or remains bounded.
If the sequence converges, r1 should remain below 1, or at least its average should remain below 1. Convergence will never be indicated when because errorEstimate is non-negative. Use newton to fill in the following table, where you can compute the absolute value of the true error because you can easily guess the exact solution.
You have already done these cases using the original convergence criterion in Exercise 3, so you can get those values from that exercise.
A reliable error estimate is almost as important as the correct solution-if you don't know how accurate a solution is, what can you do with it? This modification of the stopping criterion is very nice when settles down to a constant value quickly.
In real problems, a great deal of care must be taken because can cycle among several values, some larger than 1, or it can take a long time to settle down. In the rest of this lab, you should continue using newton.
More to the point, if you know that the solution lies in some interval and on that interval, then the Newton iteration will converge to the solution, starting from any point in the interval. When there are zeros of the derivative nearby, Newton's method can display highly erratic behavior and may or may not converge.
In the last part of the previous exercise, you saw a case where there are several roots, with zeros of the derivative between them, and moving the initial guess to the right moved the chosen root to the left. In this and the following exercise, you will be interested in the sequence of iterates, not just the final result.
Re-enable the disp statement displaying the values of the iterates in newton. Write a function m-file for the cosmx function used in the previous lab.
Be sure to calculate both the function and its derivative, as we did for f1, f2, f3 and f4. What is the solution and how many iterations did it take? If it took more than ten iterations, go back and be sure your formula for the derivative is correct.
Note thatso there are several zeros of the derivative between the initial guess and the root. You should observe that it takes the maximum number of iterations and seems not to converge. To do this, include in the call: Does it locate a solution in fewer than iterations? How many iterations does it take?
Does it get the same root as before? There is no real pattern to the numbers, and it is pure chance that finally put the iteration near the root.
Is the final estimated error smaller than the square of the immediately preceeding estimated error? You have just observed a common behavior: Another possible behavior is simple divergence to infinity. The following exercise presents a case of divergence to infinity.
You should find it diverges in a monotone manner, so it is clear that the iterates are unbounded. What are values of iterates 95 through ?
This behavior can be proved, but the proof is not required for this lab. No root Sometimes people become so confident of their computational tools that they attempt the impossible. What would happen if you attempted to find a root of a function that had no roots?
Basically, the same kind of behavior occurs as when there are zeros of the derivative between the initial guess and the root.
Intermediate prints will no longer be needed in this lab. Comment out the disp statements in newton. Leave the error statement intact. Complex functions But the function does have roots!
The roots are complex, but Matlab knows how to do complex arithmetic. Actually the roots are imaginary, but it is all the same to Matlab.
All Matlab needs is to be reminded to use complex arithmetic. If you have used the letter i as a variable in your Matlab session, its special value as the square root of -1 has been obscured.The convergence of the bisection method is very slow. Although the error, in general, does not decrease monotonically, the average rate of convergence is 1/2 and so, slightly changing the definition of order of convergence, it is possible to say that the method converges linearly with rate 1/2.
The Secant Method One drawback of Newton’s method is that it is necessary to evaluate f0(x) at various points, Because is the rate of convergence, the left side must converge to a positive constant Cas k!1.
(Bisection) Let fbe a continuous function on the interval [a;b] that changes sign on. CS Numerical Methods Nonlinear Equations Eric Shaffer Some slides adapted from: Convergence Rate For Bisection.
Bisection Method. Newton’s Method. Newton’s Method. Newton’s Method: Example. o Convergence rate of Newton's method for simple root is therefore. Lecture 7: Root finding I 3/12 3 Bisection If q=1, as it is for bisection, we say the convergence is linear, and we call α the rate of convergence.
For q>1 the convergence is said to be superlinear, and specifically, if q=2 the convergence is quadratic. For superlinear. Nonlinear Equations Rate of convergence MATH (Numerical Methods) WK 03 – Nonlinear equations 24 / 32 This preview has intentionally blurred sections.
Sign up to view the full version. Common Numerical Root Finding Methods: To improve the slow convergence of the bisection method, has a lower and uncertain convergence rate compared to the secant method. The emphasis on bracketing the root may sometimes restrict the false position method in difficult situations while solving highly nonlinear equations.