You must have JavaScript enabled in your browser to utilize the functionality of this website. Join over 1. Page 1. Save View my saved documents Submit similar document. Share this Facebook. Decimal Search. Extracts from this document This method will fail if x 0 is on a turning point, or if the graph has a discontinuity: E. Found what you're looking for?
Not the one? Search for your essay title The Gradient Fraction I will now incorporate more graphs into my investigation. Functions Coursework - A2 Maths Therefore two roots here are undetected.
Arctic Research Maths Coursework These three calculations are Pythagoras theorem, trigonometry of a right angled triangle and non-right angled triangle and the sine and cosine rules.
Using Decimal search 0,1. See more essays. Over , pieces of student written work Annotated by experienced teachers Ideas and feedback to improve your own work. Save Sign up now Want to read the rest? Read more The above preview is unformatted text. Looking for expert help with your Maths work? Take me to free Study Guides. Or get inspiration from these FREE essays:. This method for finding a solution is Newton's method.
As we'll see, Newton's method can be a very efficient method to approximate a solution to an equation — when it works. We can now describe Newton's method algebraically. Newton's method in the above example is much faster than the bisection algorithm! In only 4 iterations we have 11 decimal places of accuracy!
The number of decimal places of accuracy approximately doubles with each iteration! To how many decimal places is the approximate solution accurate?
The number of decimal places accuracy roughly triples with each iteration! Compare the convergence to what you obtained with the bisection method in exercise 5. While Newton's method can give fantastically good approximations to a solution, several things can go wrong. Figure 1. The approximations approach the actual root The approximations are derived by looking at tangent lines to the graph of.
To find the next approximation, we use Figure. Since the derivative is Using Figure with and a calculator that displays 10 digits , we obtain. To find the next approximation, we use Figure with and the value of stored on the calculator. We find that. The function has one root over the interval.
Hint Use Figure. Solution For From Figure , we know that. Figure 3. Hint For Figure reduces to. Therefore, we cannot continue the iterative process. The approximations may approach a different root. If the function has more than one root, it is possible that our approximations do not approach the one for which we are looking, but approach a different root see Figure. This event most often occurs when we do not choose the approximation close enough to the desired root. The approximations may fail to approach a root entirely.
In Figure , we provide an example of a function and an initial guess such that the successive approximations never approach a root because the successive approximations continue to alternate back and forth between two values. Figure 4. If the initial guess is too far from the root sought, it may lead to approximations that approach a different root.
Solution For the derivative is Therefore,. Figure 5. The approximations continue to alternate between and never approach the root of. Finding a Limit for an Iterative Process Let and let For all let Find the values Make a conjecture about what happens to this list of numbers as If the list of numbers approaches a finite number then satisfies and is called a fixed point of. Solution If then. Figure 6. This iterative process approaches the value. Hint Consider the point where the lines and intersect.
Iterative Processes and Chaos Iterative processes can yield some very interesting behavior. Figure 7. The Mandelbrot set is a well-known example of a set of points generated by the iterative chaotic behavior of a relatively simple function.
Figure 8. A cobweb diagram for is presented here. The sequence of values results in an 8-cycle. Any process in which a list of numbers is generated by defining an initial number and defining the subsequent numbers by the equation for some function is an iterative process.
Solution fails, works. Solution a. Solution We need to be twice continuously differentiable. Solution There is no solution to the equation. Solution Newton: 11 iterations, secant: 16 iterations. Solution Newton: three iterations, secant: six iterations. Solution Newton: five iterations, secant: eight iterations. Previous: 4.
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